Notification settings The new notifications

    246
    Views

    Do somebody know what's wrong with my pepper ,they got white dot stuff .

    Reply

    Questionskatrisn Post a question • 1 people followed • 0 replies • 246 views • 2017-12-30 07:39 • came from similar tags

    242
    Views

    My garlic leaves became yellow

    Reply

    QuestionsJayden Post a question • 1 people followed • 0 replies • 242 views • 2017-12-30 07:35 • came from similar tags

    226
    Views

    Yuichi Mori (CEO, Mebiol) soil free farming on ted talk

    ExperienceJohn replied • 2 people followed • 1 replies • 226 views • 2017-12-30 07:28 • came from similar tags

    247
    Views

    How can my tomato became like these?

    QuestionsMohdtariq replied • 2 people followed • 1 replies • 247 views • 2017-12-16 20:59 • came from similar tags

    268
    Views

    My peppers in my backyard have disease, the fruit became dark, and so many.

    Reply

    QuestionsJayden Post a question • 0 people followed • 0 replies • 268 views • 2017-12-13 07:09 • came from similar tags

    256
    Views

    My eggplants have disease,Does someone know how to fix it?

    Reply

    QuestionsWilliam Post a question • 0 people followed • 0 replies • 256 views • 2017-12-06 22:52 • came from similar tags

    263
    Views

    What's wrong with my tomato, how to cure this kind of disease?

    Reply

    QuestionsWilliam Post a question • 0 people followed • 0 replies • 263 views • 2017-12-06 22:52 • came from similar tags

    223
    Views

    How do you raise chicken in your backyard for eggs?

    QuestionsMichael replied • 3 people followed • 2 replies • 223 views • 2017-12-01 18:30 • came from similar tags

    114
    Views

    Dawn to Dusk - Finishing Soybean Harvest

    Daily LifeHowfarmswork Published the article • 0 comments • 114 views • 2017-12-01 18:11 • came from similar tags

     

     
     
     
     
     
     


     
     
     
     
     
    126
    Views

    We bring home an 8370R from Sloan's to try out!

    Daily LifeHowfarmswork Published the article • 0 comments • 126 views • 2017-12-01 18:11 • came from similar tags

     
     
     
     
     
     
     
     


    109
    Views

    I ride down with Bob to the grain elevator on the Mississippi River to where all of our crops go

    Daily LifeHowfarmswork Published the article • 0 comments • 109 views • 2017-12-01 18:11 • came from similar tags

    After The Farm - Where The Crops Get Sold
     
     
    After The Farm - Where The Crops Get Sold
     
     


    113
    Views

    We wean the calves off their mothers

    Daily LifeHowfarmswork Published the article • 0 comments • 113 views • 2017-12-01 18:11 • came from similar tags

     

     
     
     
     


     
     
     
    139
    Views

    2nd Year Visa Farm Work Irrigation Hand wanted(Weekly Wage + Food + accomodation)

    Jobskatrisn Published the article • 0 comments • 139 views • 2017-11-28 19:48 • came from similar tags

    Looking for two workers to work and live on a farm situated near Nevertire, NSW as an irrigation farm hand.




    Work includes starting/stopping irrigation pipes and other general farm duties. Work often requires working early in the morning and occasionally in the evenings. No previous experience required, driver’s license preferred.




    Work will commence on the 1st of December and will be required to work for 3 months. Payment will be a weekly wage, food and accommodation included.




    This job is perfect for backpackers looking for farm work to obtain there 2nd year visa.




    Please send resume and/or 300 words as to why you would be a suitable candidate to  view all
    Looking for two workers to work and live on a farm situated near Nevertire, NSW as an irrigation farm hand.




    Work includes starting/stopping irrigation pipes and other general farm duties. Work often requires working early in the morning and occasionally in the evenings. No previous experience required, driver’s license preferred.




    Work will commence on the 1st of December and will be required to work for 3 months. Payment will be a weekly wage, food and accommodation included.




    This job is perfect for backpackers looking for farm work to obtain there 2nd year visa.




    Please send resume and/or 300 words as to why you would be a suitable candidate to 
    180
    Views

    Experienced Calf Rearer wanted in Wyuna Victoria,Australia

    Jobsflowerjob Published the article • 0 comments • 180 views • 2017-11-28 19:27 • came from similar tags

    New calf rearing facility in the Wyuna area is looking to employ experienced calf rearers.

    Primary duties will include feeding and overseeing the care of calves aged 0-6 months. However, day to day duties will also include assistance with farm maintenance.

    Previous experience in farming services and rearing livestock is essential.

    We require new employees to start work as soon as possible.

    We look forward to receiving expressions of interest from anyone who is enthusiastic and motivated to work hard as the business expands.

    Please contact Tony Gahan on view all
    New calf rearing facility in the Wyuna area is looking to employ experienced calf rearers.

    Primary duties will include feeding and overseeing the care of calves aged 0-6 months. However, day to day duties will also include assistance with farm maintenance.

    Previous experience in farming services and rearing livestock is essential.

    We require new employees to start work as soon as possible.

    We look forward to receiving expressions of interest from anyone who is enthusiastic and motivated to work hard as the business expands.

    Please contact Tony Gahan on
    180
    Views

    Flower Farm worker position is waiting for you

    Jobsflowerjob Published the article • 0 comments • 180 views • 2017-11-28 19:27 • came from similar tags

    Farm laborer required for an immediate start on flower farm on Central Coast 2 hours north of Sydney. Wage is $22.86 per hour plus superannuation. Basic accommodation on farm is available at no charge or there are caravan parks in the area. 


    LABOURING DUTIES: digging trenches, laying pipes out, replacing greenhouse materials, getting ground ready for steaming and general greenhouse maintenance. Farm work can be very cold and wet,or very hot and dry. It's physical work and can be long hours. Attention to detail and high concentration levels required in some jobs.

    This job is only for someone who can start immediately, sorry no phone calls taken.


    please send resume stating previous farm experience to:Flower Farm worker view all
    Farm laborer required for an immediate start on flower farm on Central Coast 2 hours north of Sydney. Wage is $22.86 per hour plus superannuation. Basic accommodation on farm is available at no charge or there are caravan parks in the area. 


    LABOURING DUTIES: digging trenches, laying pipes out, replacing greenhouse materials, getting ground ready for steaming and general greenhouse maintenance. Farm work can be very cold and wet,or very hot and dry. It's physical work and can be long hours. Attention to detail and high concentration levels required in some jobs.

    This job is only for someone who can start immediately, sorry no phone calls taken.


    please send resume stating previous farm experience to:Flower Farm worker
    156
    Views

    Looking for 100workers to pick ONIONS in Gatton,QLD

    Jobsdominic kwon Published the article • 0 comments • 156 views • 2017-11-28 19:27 • came from similar tags

    Looking for ONION 100pickersDay by day pays.




    • Location : Gatton, QLD

    • Price : $45 per a Bin

    • Meterials : a scissors, 10to15 baskets and gloves.

    (Pls see the photes on this ad)




    • Contacts

    Mobile : ******6006 + click to reveal 

    Email : Elqu******@******con + click to reveal 

    Line : Elque12 view all
    Looking for ONION 100pickersDay by day pays.




    • Location : Gatton, QLD

    • Price : $45 per a Bin

    • Meterials : a scissors, 10to15 baskets and gloves.

    (Pls see the photes on this ad)




    • Contacts

    Mobile : ******6006 + click to reveal 

    Email : Elqu******@******con + click to reveal 

    Line : Elque12
    142
    Views

    Vegan Fertilizer for Container Citrus Tree

    ExperienceJohn Published the article • 0 comments • 142 views • 2017-11-24 03:39 • came from similar tags

     
    Save Time and Jump to the specific Questions:
     02:43 What are the invertebrates used with aquaponics? 
    03:24 How do I get tree collards? 
    06:02 Have you ever used hugelkultur or Takakura Composting?
     10:35 Are coco fiber sheets the same as coco fiber lining?
     12:05 How should I treat wooden wine boxes to grow in outside?
     13:37 What kind of watering system should install to grow tree collards? 
    14:57 Veganic Fertilizer for Container Citrus Tree? 
    21:12 What advice do you have growing in the summer in Austin, Texas?
     25:00 Have you ever considered beneficial predatory mites?
     27:38 How about a video on LED lights or selling LED lights? 
    29:11 Why can't a plant be trained to grow from zone to zone? 
    32:22 Can I add limestone to my tomato plants?
     
     
     
      view all
     
    Save Time and Jump to the specific Questions:
     02:43 What are the invertebrates used with aquaponics? 
    03:24 How do I get tree collards? 
    06:02 Have you ever used hugelkultur or Takakura Composting?
     10:35 Are coco fiber sheets the same as coco fiber lining?
     12:05 How should I treat wooden wine boxes to grow in outside?
     13:37 What kind of watering system should install to grow tree collards? 
    14:57 Veganic Fertilizer for Container Citrus Tree? 
    21:12 What advice do you have growing in the summer in Austin, Texas?
     25:00 Have you ever considered beneficial predatory mites?
     27:38 How about a video on LED lights or selling LED lights? 
    29:11 Why can't a plant be trained to grow from zone to zone? 
    32:22 Can I add limestone to my tomato plants?
     
     
     
     


    247
    Views

    How can a Urban Farmer Makes $1000 a Week Growing Vegetables in Rental Home

    Reply

    ExperienceJohn Post a question • 1 people followed • 0 replies • 247 views • 2017-11-24 03:39 • came from similar tags

    179
    Views

    Module 5: Ranking the Animals -5.7 Interpretation of Breeding Values and Accuracies

    ExperienceEmily Published the article • 0 comments • 179 views • 2017-11-21 07:24 • came from similar tags

    Ranking the Animals - In Summary
     Before you move on to the assignments covering the whole module there is an animated video to watch, covering the most important topics that you learnt in this module on Ranking the animals. 

    In a clip of about 10 minutes, you will get a summary of the principle of estimating breeding values (0:00 to 4:37), of the principle of accuracies of EBVs (4:37 to 6:51), and of the effect of using information on relatives on the accuracy of the estimated breeding value (6:51 to 10:45). 

    Animation Estimated Breeding Value
     

     
     
    video subtitle:
     
    The success of animal breeding depends on the success of identifying the genetically best animals so that they can be used as parents to produce the next generation.
    How can that be achieved? How can the genetically best animals be identified?
    Let us consider a sheep farmer who wants to select a ram to mate with his ewes.
    The most important breeding goal of the farmer is to increase the body weight of his sheep.
    So which ram should he choose?
    The challenge the farmer faces is how to identify the genetically best ram for breeding.
    In other words, he needs to get a good impression of the genetic value of the ram as father of the next generation.
    This is also called the breeding value of an animal.
    If we would know the breeding values of the rams, we could rank them accordingly, and then choose the best ram for breeding.
    But how can we obtain such a ranking of the animals?
    To rank the animals, we would need to get insight in the breeding values of the rams.
    The problem is that we cannot measure the true breeding value, but we can make an estimate of it.
    The most basic method of ranking the rams would be by recording their own body weight, and rank them accordingly.
    As the aim is to rank the rams such that the superior animals can be identified,
    the weight of the individual rams is expressed relative to the average weight of all rams.
    This way, animals that are heavier than average will obtain a positive value, and animals that are lighter will obtain a negative value.
    Obviously, given the fact that the farmer wants to increase the weight of his sheep,
    the interest is in the rams with positive values, and the larger the value the better.
    Now we have ranked the rams according to their body weight, and we have determined which ones are better than average and how much.
    But we still do not know whether this represents the ranking according to the genetic potential, or the breeding value of the rams.
    What should we do to translate these relative performances of the rams into their estimated breeding values?
    We need to find out how well the recorded body weights represent the genetic potential of the rams.
    To achieve that we can make use of the heritability.
    The heritability of a trait indicates how much of the variation among phenotypes in a population, so in this case body weight,
    is due to additive genetic differences between the animals in the population, and how much is due to the environment.
    The heritability is in fact a regression coefficient of the true breeding value of animals on their phenotype, where both are again expressed as deviations from the population average.
    So the estimated breeding value equals a regression coefficient times the phenotype, expressed as deviation from the population average.
    So although we do not know the true genetic potential, we can estimate the breeding value, in short EBV,
    because we know the phenotype and the regression coefficient, which is in this case the heritability.
    This procedure of using the phenotype and a regression coefficient to estimate the breeding value is the core principle of breeding value estimation.
    Now let us go back to our sheep farmer.
    He is considering using a ram that weighs 90 kilograms, but first he wants to know its estimated breeding value.
    We know that the average ram in the population weighs 80 kilograms, and the heritability for body weight is 0.45.
    Because in this case of own performance the regression coefficient of the true genetic potential of the rams on their phenotype is equal to the heritability,
    we can estimate the breeding value of the ram as 0.45 times (90 minus 80) is equal to plus 4.5 kilograms.
    This means that the genetic potential of the ram is estimated to be 4.5 kilograms higher than the population average.
    Note that the EBV is expressed in units of the trait, in this case in kilograms.
    Using the same approach, we can estimate the breeding values of each of the rams in the population, and rank them accordingly.
    Some rams will have negative EBV, because their estimated genetic potential is lower than the population average.
    These rams obviously will not be rams of choice for breeding, given that the breeding goal is to increase the body weight in the next generation.
    This was quite a simple way to estimate a breeding value. But how accurate is an estimated breeding value?
    The accuracy largely depends on the heritability of the trait that you want to estimate breeding values for.
    The higher the heritability of a trait, the better the genetic potential of an animal can be estimated from its phenotype,
    and the better ranking of animals on their EBVs represents the true genetic ranking of the animals.
    Unfortunately, for traits of low heritability it is not so easy to accurately estimate the breeding value of an animal.
    For such traits, the ranking according to EBV based on own performance will not necessarily represent the ranking according to the true breeding value of the animal.
    To get some insight in this matter, imagine that you could create a plot with the true breeding value on the y-axis and the estimated breeding value on the x-axis.
    Each EBV is linked to its accompanying true breeding value in a data-point.
    If the EBVs are estimated accurately, the data points will be nicely on a straight line connecting each EBV to the true breeding value of equal size.
    Unfortunately, it is very rarely the case that breeding values are estimated with such a high accuracy.
    So why not? Why are we not always able to accurately estimate the breeding value of an animal?
    We already briefly discussed the role of the heritability.
    If the heritability is low, then the variation in phenotypes is only for a small proportion due to genetic differences between the animals.
    So if we only use the information on the animals themselves to estimate the breeding value,
    then the variation in EBV also will not only be due to genetic differences among the animals, but also due to environmental circumstances.
    If you would now create such a plot of the estimated against the true breeding value, the points in your plot will not be on one line, but instead will form a cloud.
    So what can we do to improve the accuracy of EBV?
    There are two factors that can be improved: the quality of the phenotypes and the use of information of relatives.
    We will not go into detail on how to improve phenotype recording, but we will show how we can use information of relatives.
    For that, we can make use of the fact that related animals share part of their genetic potential.
    The closer the relationship, the larger the shared part of their genetic potential.
    So if we would also know the body weight of the father of the ram,
    it would help to create a more accurately estimated breeding value for the ram, because the father and the ram share 50% of their genes.
    If, in addition, we would also know the body weights of 5 brothers of the ram, it would add even more,
    because the ram and his brothers also share 50% of their genetic information.
    And body weights of 25 half-brothers again would add, even though the genetic relationship between the ram and his half-brothers is not as strong as with his full brothers, or his father.
    In other words, not all information is equally valuable.
    The closer the genetic relationship, the more valuable the information for estimating the breeding value of an animal.
    However, there is only one father and there may be multiple brothers.
    And multiple brothers (or sisters) are more informative for estimating the genetic potential of an animal than a single parent.
    But offspring are even more valuable, because there can be multiple of them.
    The more offspring are available, the more accurately we can estimate the breeding value of the animal.
    If we have information on many offspring, we can even approach the 100% accuracy!
    Then the EBV becomes equal to the tbv. That would be the ideal situation.
    As we have seen, not all information is equally valuable for estimating breeding values.
    How can we account for that when estimating breeding values?
    The answer is in calculating the regression coefficient.
    Let us consider the situation where we have information from 10 half-sibs on body weight.
    Suppose they are 5 kilograms heavier than average.
    The formula for calculating the regression coefficient is in this case...
    Therefore, the EBV of the ram based on these half-sibs is...
    Now we have two EBVs: one based on the own body weight and one based on half-sibs.
    The question is now whether we can do better than that and combine the information.
    We should not just sum these two EBVs. The solution is multiple regression.
    The b-values cannot be calculated with the equations in the appendix anymore, because there is an overlap in information.
    The rams itself and his half-sibs share a quarter of the genes and we need to account for that in calculating the b-values.
    Therefore, we use an Excel sheet specifically designed to calculate b-values in case of multiple regression.
    With this sheet the calculated b-values are 0.41 for own body weight and 0.33 for the half-sibs.
    So when we fill in those values, we can obtain the EBV of the ram.
    Generally speaking, the more phenotypic information there is on related animals, the more accurately we can estimate the breeding values.
    In summary, the success of identifying the genetically best animals depends on the heritability of the trait,
    and on the amount and quality of information we have on the phenotypic performance of the animal and its close relatives.
    Those factors determine the accuracy of the estimated breeding value.
    So the success of animal breeding depends on the collection of good quality data on many animals,
    obviously in addition to keeping records of genetic relationships between the animals.
     
     
     
    Working with Breeding Values and Accuracies
     
    In breeding, we use the estimated breeding values to rank animals. Which aspects are taken into account in calculating the EBVs fully depends on the breeding goal that we determined earlier. Which traits have we defined as important traits for this population in the future? Which traits do we select the animals on, to improve the future generations? Which are the best animals to breed with?

    You have now seen that you can calculate estimated breeding values, as well as accuracies, in different situations. 

    Now take a few moments to think about the following question, and write down a short answer for yourself:

    Should we select the animal with highest EBV or with the highest accuracy?

    Now look back at the answer you chose at the beginning of this module, whether you would use information on the animals themselves, their parents, their siblings, or their offspring to select the best animals for breeding. In this module you learned how to make a decision like this based on the accuracy you can expect from different information sources.  

    Breeding values and accuracies

    A poultry breeding company is interested in selecting roosters for egg weight. The information that is available comes from 40 hens which are full sibs of the rooster. The heritability of egg weight is 0.40. We assume that the common environment is zero. 
     

     
      view all
    Ranking the Animals - In Summary
     Before you move on to the assignments covering the whole module there is an animated video to watch, covering the most important topics that you learnt in this module on Ranking the animals. 

    In a clip of about 10 minutes, you will get a summary of the principle of estimating breeding values (0:00 to 4:37), of the principle of accuracies of EBVs (4:37 to 6:51), and of the effect of using information on relatives on the accuracy of the estimated breeding value (6:51 to 10:45). 

    Animation Estimated Breeding Value
     


     
     
    video subtitle:
     
    The success of animal breeding depends on the success of identifying the genetically best animals so that they can be used as parents to produce the next generation.
    How can that be achieved? How can the genetically best animals be identified?
    Let us consider a sheep farmer who wants to select a ram to mate with his ewes.
    The most important breeding goal of the farmer is to increase the body weight of his sheep.
    So which ram should he choose?
    The challenge the farmer faces is how to identify the genetically best ram for breeding.
    In other words, he needs to get a good impression of the genetic value of the ram as father of the next generation.
    This is also called the breeding value of an animal.
    If we would know the breeding values of the rams, we could rank them accordingly, and then choose the best ram for breeding.
    But how can we obtain such a ranking of the animals?
    To rank the animals, we would need to get insight in the breeding values of the rams.
    The problem is that we cannot measure the true breeding value, but we can make an estimate of it.
    The most basic method of ranking the rams would be by recording their own body weight, and rank them accordingly.
    As the aim is to rank the rams such that the superior animals can be identified,
    the weight of the individual rams is expressed relative to the average weight of all rams.
    This way, animals that are heavier than average will obtain a positive value, and animals that are lighter will obtain a negative value.
    Obviously, given the fact that the farmer wants to increase the weight of his sheep,
    the interest is in the rams with positive values, and the larger the value the better.
    Now we have ranked the rams according to their body weight, and we have determined which ones are better than average and how much.
    But we still do not know whether this represents the ranking according to the genetic potential, or the breeding value of the rams.
    What should we do to translate these relative performances of the rams into their estimated breeding values?
    We need to find out how well the recorded body weights represent the genetic potential of the rams.
    To achieve that we can make use of the heritability.
    The heritability of a trait indicates how much of the variation among phenotypes in a population, so in this case body weight,
    is due to additive genetic differences between the animals in the population, and how much is due to the environment.
    The heritability is in fact a regression coefficient of the true breeding value of animals on their phenotype, where both are again expressed as deviations from the population average.
    So the estimated breeding value equals a regression coefficient times the phenotype, expressed as deviation from the population average.
    So although we do not know the true genetic potential, we can estimate the breeding value, in short EBV,
    because we know the phenotype and the regression coefficient, which is in this case the heritability.
    This procedure of using the phenotype and a regression coefficient to estimate the breeding value is the core principle of breeding value estimation.
    Now let us go back to our sheep farmer.
    He is considering using a ram that weighs 90 kilograms, but first he wants to know its estimated breeding value.
    We know that the average ram in the population weighs 80 kilograms, and the heritability for body weight is 0.45.
    Because in this case of own performance the regression coefficient of the true genetic potential of the rams on their phenotype is equal to the heritability,
    we can estimate the breeding value of the ram as 0.45 times (90 minus 80) is equal to plus 4.5 kilograms.
    This means that the genetic potential of the ram is estimated to be 4.5 kilograms higher than the population average.
    Note that the EBV is expressed in units of the trait, in this case in kilograms.
    Using the same approach, we can estimate the breeding values of each of the rams in the population, and rank them accordingly.
    Some rams will have negative EBV, because their estimated genetic potential is lower than the population average.
    These rams obviously will not be rams of choice for breeding, given that the breeding goal is to increase the body weight in the next generation.
    This was quite a simple way to estimate a breeding value. But how accurate is an estimated breeding value?
    The accuracy largely depends on the heritability of the trait that you want to estimate breeding values for.
    The higher the heritability of a trait, the better the genetic potential of an animal can be estimated from its phenotype,
    and the better ranking of animals on their EBVs represents the true genetic ranking of the animals.
    Unfortunately, for traits of low heritability it is not so easy to accurately estimate the breeding value of an animal.
    For such traits, the ranking according to EBV based on own performance will not necessarily represent the ranking according to the true breeding value of the animal.
    To get some insight in this matter, imagine that you could create a plot with the true breeding value on the y-axis and the estimated breeding value on the x-axis.
    Each EBV is linked to its accompanying true breeding value in a data-point.
    If the EBVs are estimated accurately, the data points will be nicely on a straight line connecting each EBV to the true breeding value of equal size.
    Unfortunately, it is very rarely the case that breeding values are estimated with such a high accuracy.
    So why not? Why are we not always able to accurately estimate the breeding value of an animal?
    We already briefly discussed the role of the heritability.
    If the heritability is low, then the variation in phenotypes is only for a small proportion due to genetic differences between the animals.
    So if we only use the information on the animals themselves to estimate the breeding value,
    then the variation in EBV also will not only be due to genetic differences among the animals, but also due to environmental circumstances.
    If you would now create such a plot of the estimated against the true breeding value, the points in your plot will not be on one line, but instead will form a cloud.
    So what can we do to improve the accuracy of EBV?
    There are two factors that can be improved: the quality of the phenotypes and the use of information of relatives.
    We will not go into detail on how to improve phenotype recording, but we will show how we can use information of relatives.
    For that, we can make use of the fact that related animals share part of their genetic potential.
    The closer the relationship, the larger the shared part of their genetic potential.
    So if we would also know the body weight of the father of the ram,
    it would help to create a more accurately estimated breeding value for the ram, because the father and the ram share 50% of their genes.
    If, in addition, we would also know the body weights of 5 brothers of the ram, it would add even more,
    because the ram and his brothers also share 50% of their genetic information.
    And body weights of 25 half-brothers again would add, even though the genetic relationship between the ram and his half-brothers is not as strong as with his full brothers, or his father.
    In other words, not all information is equally valuable.
    The closer the genetic relationship, the more valuable the information for estimating the breeding value of an animal.
    However, there is only one father and there may be multiple brothers.
    And multiple brothers (or sisters) are more informative for estimating the genetic potential of an animal than a single parent.
    But offspring are even more valuable, because there can be multiple of them.
    The more offspring are available, the more accurately we can estimate the breeding value of the animal.
    If we have information on many offspring, we can even approach the 100% accuracy!
    Then the EBV becomes equal to the tbv. That would be the ideal situation.
    As we have seen, not all information is equally valuable for estimating breeding values.
    How can we account for that when estimating breeding values?
    The answer is in calculating the regression coefficient.
    Let us consider the situation where we have information from 10 half-sibs on body weight.
    Suppose they are 5 kilograms heavier than average.
    The formula for calculating the regression coefficient is in this case...
    Therefore, the EBV of the ram based on these half-sibs is...
    Now we have two EBVs: one based on the own body weight and one based on half-sibs.
    The question is now whether we can do better than that and combine the information.
    We should not just sum these two EBVs. The solution is multiple regression.
    The b-values cannot be calculated with the equations in the appendix anymore, because there is an overlap in information.
    The rams itself and his half-sibs share a quarter of the genes and we need to account for that in calculating the b-values.
    Therefore, we use an Excel sheet specifically designed to calculate b-values in case of multiple regression.
    With this sheet the calculated b-values are 0.41 for own body weight and 0.33 for the half-sibs.
    So when we fill in those values, we can obtain the EBV of the ram.
    Generally speaking, the more phenotypic information there is on related animals, the more accurately we can estimate the breeding values.
    In summary, the success of identifying the genetically best animals depends on the heritability of the trait,
    and on the amount and quality of information we have on the phenotypic performance of the animal and its close relatives.
    Those factors determine the accuracy of the estimated breeding value.
    So the success of animal breeding depends on the collection of good quality data on many animals,
    obviously in addition to keeping records of genetic relationships between the animals.
     
     
     
    Working with Breeding Values and Accuracies
     
    In breeding, we use the estimated breeding values to rank animals. Which aspects are taken into account in calculating the EBVs fully depends on the breeding goal that we determined earlier. Which traits have we defined as important traits for this population in the future? Which traits do we select the animals on, to improve the future generations? Which are the best animals to breed with?

    You have now seen that you can calculate estimated breeding values, as well as accuracies, in different situations. 

    Now take a few moments to think about the following question, and write down a short answer for yourself:

    Should we select the animal with highest EBV or with the highest accuracy?

    Now look back at the answer you chose at the beginning of this module, whether you would use information on the animals themselves, their parents, their siblings, or their offspring to select the best animals for breeding. In this module you learned how to make a decision like this based on the accuracy you can expect from different information sources.  

    Breeding values and accuracies

    A poultry breeding company is interested in selecting roosters for egg weight. The information that is available comes from 40 hens which are full sibs of the rooster. The heritability of egg weight is 0.40. We assume that the common environment is zero. 
     

     
     
    118
    Views

    Module 5: Ranking the Animals -5.6 Breeding Value Estimation Using Genomic Information

    ExperienceEmily Published the article • 0 comments • 118 views • 2017-11-21 07:15 • came from similar tags

    Using genomic information

    Breeding values are estimated based on phenotypic information. After an animal is born it may take some time before we can obtain this phenotypic information. Think of reproduction traits where the information cannot be measured until the animals are adults. At a young age, the information is limited to the performance of the parents which limits the accuracy of the EBV. It would be very interesting to have a methodology that increases the accuracy of the EBV already at younger age, without having to wait. 

    Genomic selection

    With genomic selection it is possible to estimate the breeding value of an animal quite accurately without the need to wait for own performance or performance of a large number of offspring. Genomic selection is based on estimating the associations between genetic marker genotypes (SNP) and phenotypes on a group of animals that have information for both. These associations can subsequently be used to predict the so-called genomic breeding values (gEBV) for animals that have been genotyped for the SNP, but that don’t (yet) have own phenotypes. 

    Using Genomic Information
     

     
     
      view all
    Using genomic information

    Breeding values are estimated based on phenotypic information. After an animal is born it may take some time before we can obtain this phenotypic information. Think of reproduction traits where the information cannot be measured until the animals are adults. At a young age, the information is limited to the performance of the parents which limits the accuracy of the EBV. It would be very interesting to have a methodology that increases the accuracy of the EBV already at younger age, without having to wait. 

    Genomic selection

    With genomic selection it is possible to estimate the breeding value of an animal quite accurately without the need to wait for own performance or performance of a large number of offspring. Genomic selection is based on estimating the associations between genetic marker genotypes (SNP) and phenotypes on a group of animals that have information for both. These associations can subsequently be used to predict the so-called genomic breeding values (gEBV) for animals that have been genotyped for the SNP, but that don’t (yet) have own phenotypes. 

    Using Genomic Information
     


     
     
     
    124
    Views

    Module 5: Ranking the Animals -5.5 Breeding Value Estimation Using BLUP

    ExperienceEmily Published the article • 0 comments • 124 views • 2017-11-21 07:15 • came from similar tags

    What is BLUP?

    In real life, we do not want to estimate breeding values for each animal by calculating regression coefficients. Furthermore, animals may perform in different environmental units such as different farms. Therefore, in real life, BLUP is used for breeding value estimation. BLUP stands for Best Linear Unbiased Prediction. It is a method to estimate breeding values while making use of the additive genetic relationships between animals, thereby simultaneously correcting the phenotypes for systematic effects. In this course, we will only explain the very basics of the method to show you what the possibilities are in animal breeding. You will also get an assignment to familiarize yourself with the output of BLUP.

    What does BLUP do?

    This all may sound quite complicated, but in a way BLUP does follow the simple genetic model P = A + E, and, in addition, provides estimates for systematic environmental effects. For example, if animals on one farm are fed much better than on another farm then ranking animals based on their weight would benefit the animals from the farm with the better nutrition. However, genetically the animals on both farms may be similar. Without taking this systematic influence of farm of origin into account, it is likely that the top ranking animals would mainly originate from the farm with the better feed. To be able to compare the performance of the animals more on their genetic potential it is important to take this farm effect into account and this is what BLUP does (if you provide the information about on which farm each animal was housed). 

    The principle of BLUP is to determine the average weight of the animals on each farm and subtract the difference from the animals on the farm with the highest weight. So if animals on farm 1 weigh 100 kg on average and on farm 2 they weigh 120 kg on average, then you disadvantage the animals of farm 2 by subtracting 20 kg from their weight. 

    Information is needed

    Systematic effects can be estimated, provided that the information is present. For example, registration of the farm should be available for each animal. Other systematic effects that are often estimated are:

    Sex (males are heavier than females, for example)
    Country
    Birth year 
    Birth season
    Year
    Month of phenotype recording
    If applicable, barn or pen number (housing location)
    Treatment (in case of an experiment)

    Critical issues in BLUP

    Critical issue in correcting for systematic effects is that it only works well if genotypes are sufficiently spread across systematic environmental influences. So the animals on both farms need to be related, for example because the same fathers were used or because the fathers used on each farm were brothers. If the animals on both farms are unrelated, then part of the reason of the difference in weight may be a difference in genetic potential. And that is what you want to estimate so you do not want to lose that by correcting the weight. Artificial insemination allows for genetic links between farms because the same sires are used in many farms. In farm animal species where natural mating is common practice, such as in beef cattle and sheep, it often is impossible to estimate systematic farm effects accurately because lack of exchange of animals between farms results in poor genetic links between farms. In species where the sires are brought to their mates on various locations, as can be the case in horse or dog breeding, genetic links will not be a limiting factor, provided that the sires are used frequently.
     
      view all
    What is BLUP?

    In real life, we do not want to estimate breeding values for each animal by calculating regression coefficients. Furthermore, animals may perform in different environmental units such as different farms. Therefore, in real life, BLUP is used for breeding value estimation. BLUP stands for Best Linear Unbiased Prediction. It is a method to estimate breeding values while making use of the additive genetic relationships between animals, thereby simultaneously correcting the phenotypes for systematic effects. In this course, we will only explain the very basics of the method to show you what the possibilities are in animal breeding. You will also get an assignment to familiarize yourself with the output of BLUP.

    What does BLUP do?

    This all may sound quite complicated, but in a way BLUP does follow the simple genetic model P = A + E, and, in addition, provides estimates for systematic environmental effects. For example, if animals on one farm are fed much better than on another farm then ranking animals based on their weight would benefit the animals from the farm with the better nutrition. However, genetically the animals on both farms may be similar. Without taking this systematic influence of farm of origin into account, it is likely that the top ranking animals would mainly originate from the farm with the better feed. To be able to compare the performance of the animals more on their genetic potential it is important to take this farm effect into account and this is what BLUP does (if you provide the information about on which farm each animal was housed). 

    The principle of BLUP is to determine the average weight of the animals on each farm and subtract the difference from the animals on the farm with the highest weight. So if animals on farm 1 weigh 100 kg on average and on farm 2 they weigh 120 kg on average, then you disadvantage the animals of farm 2 by subtracting 20 kg from their weight. 

    Information is needed

    Systematic effects can be estimated, provided that the information is present. For example, registration of the farm should be available for each animal. Other systematic effects that are often estimated are:

    Sex (males are heavier than females, for example)
    Country
    Birth year 
    Birth season
    Year
    Month of phenotype recording
    If applicable, barn or pen number (housing location)
    Treatment (in case of an experiment)

    Critical issues in BLUP

    Critical issue in correcting for systematic effects is that it only works well if genotypes are sufficiently spread across systematic environmental influences. So the animals on both farms need to be related, for example because the same fathers were used or because the fathers used on each farm were brothers. If the animals on both farms are unrelated, then part of the reason of the difference in weight may be a difference in genetic potential. And that is what you want to estimate so you do not want to lose that by correcting the weight. Artificial insemination allows for genetic links between farms because the same sires are used in many farms. In farm animal species where natural mating is common practice, such as in beef cattle and sheep, it often is impossible to estimate systematic farm effects accurately because lack of exchange of animals between farms results in poor genetic links between farms. In species where the sires are brought to their mates on various locations, as can be the case in horse or dog breeding, genetic links will not be a limiting factor, provided that the sires are used frequently.
     
     
    294
    Views

    Module 5: Ranking the Animals -5.4 Breeding Value Estimation Using Multiple Information Sources

    ExperienceEmily Published the article • 0 comments • 294 views • 2017-11-21 07:15 • came from similar tags

    Tutorial Clip 
     
    To calculate the accuracy of an estimated breeding value based on a single type of information we can use the formulas presented in 5.3. With multiple information sources the calculations are more complicated as explained on the previous page. These calculations have been implemented in an excel tool called stselind.xls developed by Julius van der Werf (https://jvanderw.une.edu.au/). The next tutorial will introduce, step by step, how to use stselind.xls.
     

     
    Video subtitle:
     
    The aim of this video is to explain how the Excel sheet Selection_index can be used
    to calculate the regression coefficients, otherwise known as b-values,
    and to calculate the breeding value and to calculate its accuracy
    when there are multiple information sources.
    With this Excel-sheet you can play around a bit
    to investigate the impact of multiple information sources
    on the accuracy of estimated breeding values.
    First, let’s take a look at the Excel sheet.
    Here at the top, in the cells highlighted in orange,
    we can fill in the values for the heritability of the trait,
    the repeatability and the c squared, which is the proportion of variance
    explained by the common environmental effect.
    If the repeatability and c squared are not given,
    we can set them to a value of zero.
    Further down, in the cells highlighted in blue,
    we can fill in how many records we have for each type of information:
    own records, records of the dam, the sire,
    records of full-sibs, records of half-sibs and records of progeny.
    If we want to calculate the b-values and the accuracy for a new set of parameters,
    we can click on Run
    and the calculated b-values will appear in the column called ‘Index weight’
    and the value for accuracy will appear in the purple cell.
    The column ‘value of variate’ is not used in this course.
    The yellow cells labelled P-matrix and G-matrix are outside the scope of this course.
    The yellow cells with b-values and accuracy
    are the same values as you see in the purple
    and in the pink cells.
    Let’s consider a situation where we want to calculate the b-values
    and accuracy of the estimated breeding value
    for litter size for a sow with one own observation,
    an observation of the dam, 4 full-sibs and 20 half-sibs.
    Let’s first fill in the heritability of 0.11;
    we set the repeatability and c squared to zero.
    Let’s now fill in the values where the sow has only an own observation.
    If we click on Run, we obtain a b-value of 0.11
    and the accuracy is 0.33.
    We expect to see an accuracy of 0.33 since 0.33 is the square root of the heritability,
    as we have seen before.
    Let’s now say that we have a record of the dam of the sow for litter size
    so we add that value to our sheet.
    We click on Run, and we see that the b-value for the own records decreases a little bit.
    This happens because there is some overlap in the information of the sow and her dam
    since they share 50% of their DNA.
    We can see that the accuracy increases from 0.33 to 0.36
    because we have more information
    and therefore, the estimated breeding value is more accurate.
    When we add 4 records of full-sibs,
    we see that the accuracy increases to 0.45.
    The four full-sisters share 50% of their DNA with the sow of interest
    and therefore, the information on their litter size
    increases the accuracy of the estimated breeding value.
    Then, when we add 20 half-sibs,
    we see that the accuracy increases to 0.5
    since the half-sisters share 25% of their DNA with the sow of interest.
    So the accuracy of the estimated breeding value increases every time
    when we add additional information of relatives .
    In summary, the Excel sheet ‘Selection_index.xlsx’
    can be used to calculate the accuracy of the estimated breeding values
    for any combination of information available.
    Furthermore, the b-values can be used to weigh
    the different information sources when estimating the breeding value of an animal.
      view all
    Tutorial Clip 
     
    To calculate the accuracy of an estimated breeding value based on a single type of information we can use the formulas presented in 5.3. With multiple information sources the calculations are more complicated as explained on the previous page. These calculations have been implemented in an excel tool called stselind.xls developed by Julius van der Werf (https://jvanderw.une.edu.au/). The next tutorial will introduce, step by step, how to use stselind.xls.
     


     
    Video subtitle:
     
    The aim of this video is to explain how the Excel sheet Selection_index can be used
    to calculate the regression coefficients, otherwise known as b-values,
    and to calculate the breeding value and to calculate its accuracy
    when there are multiple information sources.
    With this Excel-sheet you can play around a bit
    to investigate the impact of multiple information sources
    on the accuracy of estimated breeding values.
    First, let’s take a look at the Excel sheet.
    Here at the top, in the cells highlighted in orange,
    we can fill in the values for the heritability of the trait,
    the repeatability and the c squared, which is the proportion of variance
    explained by the common environmental effect.
    If the repeatability and c squared are not given,
    we can set them to a value of zero.
    Further down, in the cells highlighted in blue,
    we can fill in how many records we have for each type of information:
    own records, records of the dam, the sire,
    records of full-sibs, records of half-sibs and records of progeny.
    If we want to calculate the b-values and the accuracy for a new set of parameters,
    we can click on Run
    and the calculated b-values will appear in the column called ‘Index weight’
    and the value for accuracy will appear in the purple cell.
    The column ‘value of variate’ is not used in this course.
    The yellow cells labelled P-matrix and G-matrix are outside the scope of this course.
    The yellow cells with b-values and accuracy
    are the same values as you see in the purple
    and in the pink cells.
    Let’s consider a situation where we want to calculate the b-values
    and accuracy of the estimated breeding value
    for litter size for a sow with one own observation,
    an observation of the dam, 4 full-sibs and 20 half-sibs.
    Let’s first fill in the heritability of 0.11;
    we set the repeatability and c squared to zero.
    Let’s now fill in the values where the sow has only an own observation.
    If we click on Run, we obtain a b-value of 0.11
    and the accuracy is 0.33.
    We expect to see an accuracy of 0.33 since 0.33 is the square root of the heritability,
    as we have seen before.
    Let’s now say that we have a record of the dam of the sow for litter size
    so we add that value to our sheet.
    We click on Run, and we see that the b-value for the own records decreases a little bit.
    This happens because there is some overlap in the information of the sow and her dam
    since they share 50% of their DNA.
    We can see that the accuracy increases from 0.33 to 0.36
    because we have more information
    and therefore, the estimated breeding value is more accurate.
    When we add 4 records of full-sibs,
    we see that the accuracy increases to 0.45.
    The four full-sisters share 50% of their DNA with the sow of interest
    and therefore, the information on their litter size
    increases the accuracy of the estimated breeding value.
    Then, when we add 20 half-sibs,
    we see that the accuracy increases to 0.5
    since the half-sisters share 25% of their DNA with the sow of interest.
    So the accuracy of the estimated breeding value increases every time
    when we add additional information of relatives .
    In summary, the Excel sheet ‘Selection_index.xlsx’
    can be used to calculate the accuracy of the estimated breeding values
    for any combination of information available.
    Furthermore, the b-values can be used to weigh
    the different information sources when estimating the breeding value of an animal.
     
    102
    Views

    Module 5: Ranking the Animals -5.3 Calculation of Accuracies of Breeding Values

    ExperienceEmily Published the article • 0 comments • 102 views • 2017-11-21 07:09 • came from similar tags

    Calculating Accuracies

    Maximum accuracy

    The formulas for calculating accuracies of selection for the same information source are given in the table below. From the table you can see that with only information on the parents, or only on the grand-parents, the accuracy of the EBV can never be as large as what can be achieved with own performance. The maximum accuracy that can be achieved can be determined by assuming a very large n. If n becomes very large, then the maximum rIH that can be achieved with full sib information is equal to 1412, which equals 0.707.

    Common environment

    If there is information on a group of related animals, for example a group of half sibs, then the animal and its sibs may share the same environment: a common environment. The variance explained by the common environmental effect is indicated by c2. Common environmental effects make it more difficult to disentangle the effect of genetics and environment. The common environmental effect has a negative influence on the accuracy.
     
      view all
    Calculating Accuracies

    Maximum accuracy

    The formulas for calculating accuracies of selection for the same information source are given in the table below. From the table you can see that with only information on the parents, or only on the grand-parents, the accuracy of the EBV can never be as large as what can be achieved with own performance. The maximum accuracy that can be achieved can be determined by assuming a very large n. If n becomes very large, then the maximum rIH that can be achieved with full sib information is equal to 1412, which equals 0.707.

    Common environment

    If there is information on a group of related animals, for example a group of half sibs, then the animal and its sibs may share the same environment: a common environment. The variance explained by the common environmental effect is indicated by c2. Common environmental effects make it more difficult to disentangle the effect of genetics and environment. The common environmental effect has a negative influence on the accuracy.
     
     
    184
    Views

    Module 5: Ranking the Animals -5.4 Breeding Value Estimation Using Multiple Information Sources

    ExperienceEmily Published the article • 0 comments • 184 views • 2017-11-21 07:09 • came from similar tags

    Using Multiple Information Sources

    Selection index

    In real life, we often have information on different types of relatives for the animal who’s breeding value we want to estimate. In these cases, we would like to combine phenotypic information in an optimal way to estimate the breeding value. Again, we can use regression but now we extend to multiple regression. Within animal breeding this is called a selection index.

    A cow may have information on her own milk production and the milk production of her dam. To estimate the breeding value by regression we would write the following equation:

    Avoid double counting of phenotypic information

    When we combine the phenotype of the animal and the phenotype of the dam in estimating the breeding value we see that the regression coefficients 0.284 and 0.107 were different from the regression with only one source of information. With only the own performance the regression is the heritability (0.3) or with only the phenotype of the dam the regression is half the heritability (0.15).

    Why this difference? The reason is that the information of the animal itself and of the dam contain overlapping information about the same genes, those genes from the dam that were passed to the daughter. To avoid double counting of this information, the regression coefficients in a selection index with multiple information sources are lower than when only a single source of information is available.

    To calculate the EBV from multiple sources, still the same 3 steps as for single source need to be taken:

    Determine the phenotypic deviations of your information sources
    Determine the regression coefficients
    Combine the previous two to estimate the breeding value

    With phenotypic information of multiple relatives, the EBV is the sum of phenotypic deviations times the multiple regression coefficients.

    EBV=b1*(P1-mean)+b2*(P2-mean)+.....+bn*(Pn-mean)

     

     
     

     
     
    Video subtitle:
     
    In a previous clip, we showed how we can estimate breeding values.
    We can, for example, use a record on own performance or on a group of offspring.
    However, as you may have noticed, the estimated breeding values for those two cases may differ.
    With own performance the estimate was +1.36 eggs,
    whereas with offspring the estimate was +3.6 eggs.
    Now we have a problem: which estimate should we use?
    Ideally, we would like to have only a single estimated breeding value, with the maximum possible accuracy.
    In this clip, I will show how we can estimate a single breeding value
    by combining multiple information sources.
    In the previous clip, you saw that breeding values can be estimated using simple regression.
    The estimated breeding value is the product of a regression coefficient, denoted by b,
    and the difference between the value of the information source and the population mean.
    The regression coefficient is the slope of the regression line.
    With own performance, for example, the regression coefficient is equal to heritability.
    Suppose heritability of egg production is 0.34,
    and that we have a chicken producing 80 eggs.
    Mean egg production of the population is 76 eggs.
    Then the estimated breeding value of the chicken is 0.34 x (80 – 76) = +1.36 eggs.
    With multiple information sources, we can extend this approach
    by using multiple regression rather than simple regression.
    Let’s do an example.
    Suppose we have two information sources: an own performance record
    and the average egg production of six offspring.
    Because we have two information sources, our regression equation now has two components;
    one for own performance with regression coefficient b1,
    and another for offspring performance with regression coefficient b2.
    This is an example of multiple regression, because we regress on more than one information source.
    The regression coefficients b1 and b2 can be calculated using specific software.
    Those b-values depend on heritability, on the number of records,
    and on the relationships between the animals.
    The details of calculating the b-values are outside the scope of this clip.
    Let’s continue with the example of the chicken that produced 80 eggs.
    Suppose that her six offspring produced on average 81 eggs.
    Using the software and a heritability of 0.34, we find that b1 = 0.25 and b2 = 0.54.
    Thus the estimated breeding value of the chicken equals 0.25 x (80-76) + 0.54 x (81-76) = which together is +3.7 eggs.
    In this way, we can get a single estimated breeding value
    that optimally combines both sources of information.
    If you want to combine multiple information sources, you cannot use simple regression.
    Let’s look at what happens if we would use simple regression.
    From the previous clip, you may remember that the b-value for own performance was 0.34,
    and that the b-value for six offspring was 0.72.
    If we were to use those b-values together,
    then the estimated breeding value would be 4.96 eggs.
    This value is much larger than the 3.7 eggs we get from multiple regression.
    This illustrates that simple regression leads to double counting of information.
    Double counting occurs because our information sources are not independent.
    In this case, the offspring carry half of the alleles of the individual,
    and both information sources therefore overlap.
    Multiple regression takes this overlap into account.
    Now you have seen how we can combine both information sources.
    But what happens with the accuracy?
    Let’s first look at the accuracies we had with a single information source.
    With an own performance record, the accuracy equals the square-root of heritability,
    which is 0.58.
    With records on six offspring, the accuracy can be calculated with an equation, and equals 0.60.
    When we combine both information sources, we can find the accuracy using specific software.
    The result shows that accuracy equals 0.72.
    This value is higher than the accuracies for a single information source.
    In fact, when you combine information sources with multiple regression
    you get the maximum possible accuracy.
    The previous has shown how we can combine two information sources using multiple regression.
    This multiple regression approach can be generalized to any number of information sources,
    as illustrated by this equation.
    In conclusion, we can estimate breeding values using regression.
    With a single information source, we can use simple regression.
    With multiple information sources, we have to use multiple regression.
    By using multiple regression, we combine all information in the optimal way,
    and maximize the accuracy of the estimated breeding value. view all
    Using Multiple Information Sources

    Selection index

    In real life, we often have information on different types of relatives for the animal who’s breeding value we want to estimate. In these cases, we would like to combine phenotypic information in an optimal way to estimate the breeding value. Again, we can use regression but now we extend to multiple regression. Within animal breeding this is called a selection index.

    A cow may have information on her own milk production and the milk production of her dam. To estimate the breeding value by regression we would write the following equation:

    Avoid double counting of phenotypic information

    When we combine the phenotype of the animal and the phenotype of the dam in estimating the breeding value we see that the regression coefficients 0.284 and 0.107 were different from the regression with only one source of information. With only the own performance the regression is the heritability (0.3) or with only the phenotype of the dam the regression is half the heritability (0.15).

    Why this difference? The reason is that the information of the animal itself and of the dam contain overlapping information about the same genes, those genes from the dam that were passed to the daughter. To avoid double counting of this information, the regression coefficients in a selection index with multiple information sources are lower than when only a single source of information is available.

    To calculate the EBV from multiple sources, still the same 3 steps as for single source need to be taken:

    Determine the phenotypic deviations of your information sources
    Determine the regression coefficients
    Combine the previous two to estimate the breeding value

    With phenotypic information of multiple relatives, the EBV is the sum of phenotypic deviations times the multiple regression coefficients.

    EBV=b1*(P1-mean)+b2*(P2-mean)+.....+bn*(Pn-mean)

     

     
     


     
     
    Video subtitle:
     
    In a previous clip, we showed how we can estimate breeding values.
    We can, for example, use a record on own performance or on a group of offspring.
    However, as you may have noticed, the estimated breeding values for those two cases may differ.
    With own performance the estimate was +1.36 eggs,
    whereas with offspring the estimate was +3.6 eggs.
    Now we have a problem: which estimate should we use?
    Ideally, we would like to have only a single estimated breeding value, with the maximum possible accuracy.
    In this clip, I will show how we can estimate a single breeding value
    by combining multiple information sources.
    In the previous clip, you saw that breeding values can be estimated using simple regression.
    The estimated breeding value is the product of a regression coefficient, denoted by b,
    and the difference between the value of the information source and the population mean.
    The regression coefficient is the slope of the regression line.
    With own performance, for example, the regression coefficient is equal to heritability.
    Suppose heritability of egg production is 0.34,
    and that we have a chicken producing 80 eggs.
    Mean egg production of the population is 76 eggs.
    Then the estimated breeding value of the chicken is 0.34 x (80 – 76) = +1.36 eggs.
    With multiple information sources, we can extend this approach
    by using multiple regression rather than simple regression.
    Let’s do an example.
    Suppose we have two information sources: an own performance record
    and the average egg production of six offspring.
    Because we have two information sources, our regression equation now has two components;
    one for own performance with regression coefficient b1,
    and another for offspring performance with regression coefficient b2.
    This is an example of multiple regression, because we regress on more than one information source.
    The regression coefficients b1 and b2 can be calculated using specific software.
    Those b-values depend on heritability, on the number of records,
    and on the relationships between the animals.
    The details of calculating the b-values are outside the scope of this clip.
    Let’s continue with the example of the chicken that produced 80 eggs.
    Suppose that her six offspring produced on average 81 eggs.
    Using the software and a heritability of 0.34, we find that b1 = 0.25 and b2 = 0.54.
    Thus the estimated breeding value of the chicken equals 0.25 x (80-76) + 0.54 x (81-76) = which together is +3.7 eggs.
    In this way, we can get a single estimated breeding value
    that optimally combines both sources of information.
    If you want to combine multiple information sources, you cannot use simple regression.
    Let’s look at what happens if we would use simple regression.
    From the previous clip, you may remember that the b-value for own performance was 0.34,
    and that the b-value for six offspring was 0.72.
    If we were to use those b-values together,
    then the estimated breeding value would be 4.96 eggs.
    This value is much larger than the 3.7 eggs we get from multiple regression.
    This illustrates that simple regression leads to double counting of information.
    Double counting occurs because our information sources are not independent.
    In this case, the offspring carry half of the alleles of the individual,
    and both information sources therefore overlap.
    Multiple regression takes this overlap into account.
    Now you have seen how we can combine both information sources.
    But what happens with the accuracy?
    Let’s first look at the accuracies we had with a single information source.
    With an own performance record, the accuracy equals the square-root of heritability,
    which is 0.58.
    With records on six offspring, the accuracy can be calculated with an equation, and equals 0.60.
    When we combine both information sources, we can find the accuracy using specific software.
    The result shows that accuracy equals 0.72.
    This value is higher than the accuracies for a single information source.
    In fact, when you combine information sources with multiple regression
    you get the maximum possible accuracy.
    The previous has shown how we can combine two information sources using multiple regression.
    This multiple regression approach can be generalized to any number of information sources,
    as illustrated by this equation.
    In conclusion, we can estimate breeding values using regression.
    With a single information source, we can use simple regression.
    With multiple information sources, we have to use multiple regression.
    By using multiple regression, we combine all information in the optimal way,
    and maximize the accuracy of the estimated breeding value.
    107
    Views

    Module 5: Ranking the Animals -5.3 Calculation of Accuracies of Breeding Values

    ExperienceEmily Published the article • 0 comments • 107 views • 2017-11-21 07:00 • came from similar tags

    Accuracy of Estimated Breeding Values

    The accuracy of an EBV gives an indication of how well the EBV resembles the true breeding value. It is therefore an indication of the value of the EBV as a selection criterion. The accuracy of the breeding value estimation represents the correlation between the EBV and the true breeding value and has a value between 0 (inaccurate) and 1 (100% accurate).

    Accuracy of Estimated Breeding Values
     

     
    Video subtitle:
     
    In the previous videos we discussed how breeding values can be estimated.
    But how accurate are estimated breeding values?
    Or in other words, how close is the estimated breeding value to the true breeding value?
    If the estimated breeding value is inaccurate,
    we may select the wrong animals as parents for the next generation.
    In this video I will explain the accuracy of estimated breeding values
    and how it can be calculated.
    So the first question is how can we measure the accuracy of estimated breeding values?
    This figure shows that the estimated breeding values deviate from the true breeding values:
    the so-called prediction errors.
    To measure how close the estimated breeding values are to the true breeding values,
    we can use the correlation between estimated and true breeding values.
    This correlation is called the accuracy and denoted as rIH.
    The accuracy can vary between 0 and 1.
    In these two examples, the accuracy is 0.51 in the left figure and 0.95 in the right figure.
    In other words, the breeding values are more accurate in the right figure than in the left figure.
    However, these figures assume that we know the true breeding value of an animal,
    while in real life we cannot measure the true breeding value of each animal.
    Fortunately, we can mathematically derive the accuracy of estimated breeding values.
    For instance in the case of own performance, like milk production of this cow,
    we can derive that the accuracy is equal to the square root of the heritability.
    So the accuracy is higher when the heritability is higher.
    In this table, you find equations to calculate the accuracy for a number of situations:
    own performance, information of parents or grandparents
    and information of full-sibs, half-sibs, or progeny.
    Let’s have a closer look at two examples.
    The first example is the case that we want to calculate the accuracy
    of a rooster based on his paternal half-sisters or half-sibs.
    The equation contains three variables: the number of half-sibs, the heritability
    and the ratio of the common environmental variance to the phenotypic variance.
    This figure shows that the accuracy increases with the number of half-sibs and with a higher heritability.
    In this case, the maximum accuracy is 0.5,
    because we only get information about the breeding value of the sire of this rooster
    and we do not have any information about his dam and his own mendelian sampling term.
    The second example is the case when we want to calculate the accuracy of the
    breeding value of this Hereford bull for weaning weight based on his progeny.
    This equation contains only two variables, the number of progeny and the heritability.
    This figure shows that the accuracy increases when the number of progeny increases
    and when the heritability is higher.
    Furthermore, we see that the maximum accuracy is approaching one if we have many progeny,
    because we get full information about the breeding value of this bull.
    So in summary, the accuracy is the correlation between the estimated and the true breeding values.
    The accuracy increases with heritability and with the amount of information available.
    And we now have a set of equations to calculate the accuracy of estimated breeding values in a number of situations.
    These equations can be used to design and optimize your breeding program.
     
     
     
      view all
    Accuracy of Estimated Breeding Values

    The accuracy of an EBV gives an indication of how well the EBV resembles the true breeding value. It is therefore an indication of the value of the EBV as a selection criterion. The accuracy of the breeding value estimation represents the correlation between the EBV and the true breeding value and has a value between 0 (inaccurate) and 1 (100% accurate).

    Accuracy of Estimated Breeding Values
     


     
    Video subtitle:
     
    In the previous videos we discussed how breeding values can be estimated.
    But how accurate are estimated breeding values?
    Or in other words, how close is the estimated breeding value to the true breeding value?
    If the estimated breeding value is inaccurate,
    we may select the wrong animals as parents for the next generation.
    In this video I will explain the accuracy of estimated breeding values
    and how it can be calculated.
    So the first question is how can we measure the accuracy of estimated breeding values?
    This figure shows that the estimated breeding values deviate from the true breeding values:
    the so-called prediction errors.
    To measure how close the estimated breeding values are to the true breeding values,
    we can use the correlation between estimated and true breeding values.
    This correlation is called the accuracy and denoted as rIH.
    The accuracy can vary between 0 and 1.
    In these two examples, the accuracy is 0.51 in the left figure and 0.95 in the right figure.
    In other words, the breeding values are more accurate in the right figure than in the left figure.
    However, these figures assume that we know the true breeding value of an animal,
    while in real life we cannot measure the true breeding value of each animal.
    Fortunately, we can mathematically derive the accuracy of estimated breeding values.
    For instance in the case of own performance, like milk production of this cow,
    we can derive that the accuracy is equal to the square root of the heritability.
    So the accuracy is higher when the heritability is higher.
    In this table, you find equations to calculate the accuracy for a number of situations:
    own performance, information of parents or grandparents
    and information of full-sibs, half-sibs, or progeny.
    Let’s have a closer look at two examples.
    The first example is the case that we want to calculate the accuracy
    of a rooster based on his paternal half-sisters or half-sibs.
    The equation contains three variables: the number of half-sibs, the heritability
    and the ratio of the common environmental variance to the phenotypic variance.
    This figure shows that the accuracy increases with the number of half-sibs and with a higher heritability.
    In this case, the maximum accuracy is 0.5,
    because we only get information about the breeding value of the sire of this rooster
    and we do not have any information about his dam and his own mendelian sampling term.
    The second example is the case when we want to calculate the accuracy of the
    breeding value of this Hereford bull for weaning weight based on his progeny.
    This equation contains only two variables, the number of progeny and the heritability.
    This figure shows that the accuracy increases when the number of progeny increases
    and when the heritability is higher.
    Furthermore, we see that the maximum accuracy is approaching one if we have many progeny,
    because we get full information about the breeding value of this bull.
    So in summary, the accuracy is the correlation between the estimated and the true breeding values.
    The accuracy increases with heritability and with the amount of information available.
    And we now have a set of equations to calculate the accuracy of estimated breeding values in a number of situations.
    These equations can be used to design and optimize your breeding program.
     
     
     
     
    100
    Views

    Module 5: Ranking the Animals -5.3 Calculation of Accuracies of Breeding Values

    ExperienceEmily Published the article • 0 comments • 100 views • 2017-11-21 07:00 • came from similar tags

    Accuracy

    If we would be able to estimate the breeding value with 100% accuracy, the estimated breeding value and the true breeding value (TBV) would be the same. If we were to plot the TBV against the EBV, then all data points would fit perfectly in line. The more the data points deviate from the regression line, the less certain you can be that the EBV indeed is representing the true breeding value: we say that the EBV are less accurate. 

    A measure of the EBV accuracy is the correlation between the EBV and the true genetic deviation and has a value between 0 (inaccurate) and 1 (100% accurate). If the correlation between the estimated and true breeding values is 1, then you have managed to create the perfect EBV and the plot of EBV and TBV form a straight line. The more the correlation deviates from 1, the less accurate the EBV are. In that case, EBV and TBV form a cloud around the regression line.
     

     
     
     
    Box : plots of true (TBV) and estimated breeding values (EBV) with the perfect regression line if EBV = TBV. On the left is an example of less accurate EBV, indicated by the cloud of data points with correlation between EBV and TBV of 0.76. On the right the EBV are estimated much more accurately and are almost the same as the TBV, resulting in a correlation between EBV and TBV of 0.98. 

    In real life, we cannot produce a graph like in this figure because we do not know the true breeding value. But what we can do is estimate the accuracy of the estimated breeding value by equations that have been derived mathematically. view all
    Accuracy

    If we would be able to estimate the breeding value with 100% accuracy, the estimated breeding value and the true breeding value (TBV) would be the same. If we were to plot the TBV against the EBV, then all data points would fit perfectly in line. The more the data points deviate from the regression line, the less certain you can be that the EBV indeed is representing the true breeding value: we say that the EBV are less accurate. 

    A measure of the EBV accuracy is the correlation between the EBV and the true genetic deviation and has a value between 0 (inaccurate) and 1 (100% accurate). If the correlation between the estimated and true breeding values is 1, then you have managed to create the perfect EBV and the plot of EBV and TBV form a straight line. The more the correlation deviates from 1, the less accurate the EBV are. In that case, EBV and TBV form a cloud around the regression line.
     

     
     
     
    Box : plots of true (TBV) and estimated breeding values (EBV) with the perfect regression line if EBV = TBV. On the left is an example of less accurate EBV, indicated by the cloud of data points with correlation between EBV and TBV of 0.76. On the right the EBV are estimated much more accurately and are almost the same as the TBV, resulting in a correlation between EBV and TBV of 0.98. 

    In real life, we cannot produce a graph like in this figure because we do not know the true breeding value. But what we can do is estimate the accuracy of the estimated breeding value by equations that have been derived mathematically.
    103
    Views

    6.5 The Effect of Mating and Mating and Genetic Gain(Module 6: Response to Selection )

    ExperienceEmily Published the article • 0 comments • 103 views • 2017-11-21 06:53 • came from similar tags

    In the previous sections, you learnt how to predict the response to selection. By performing selection, we determine which animals will become parents of the next generation, and which animals will not reproduce at all. However, until now we have not looked into the combinations of parents. This is what we call mating. The central question here is: Which male fits best to a certain selected female animal? Which criteria do we use?

    Watch the clip by dr. Piter Bijma to learn more about the process of mating. 
     

     
    Video subtitle:
     
     
    I guess everyone has some idea of the meaning of “mating”.
    But what precisely does “mating” mean in animal breeding?
    That is what I will explain in this clip.
    Mating refers to the pairing of the selected sires and dams.
    So mating occurs after selection.
    With selection, we decide which individuals become parents of the next generations.
    With mating, in contrast, we decide how the individuals that have already been selected as parents are combined to produce offspring.
    Here you see four dogs, two males and two females.
    Suppose we have selected those four dogs as parents of the next generation,
    and now we want to mate them.
    We could, for example, mate Jim with Jill, and Boris with Casey.
    Alternatively, we could mate Jim with Casey and Boris with Jill.
    Now what would be a good strategy to mate the selected parents?
    A common strategy is “random mating”,
    where sires and dams are just combined at random.
    Random mating is used a lot, and is often a good choice.
    However, we can do a bit better.
    With random mating, for example, we may accidentally mate closely related individuals,
    which would result in highly inbred offspring.
    It is better to avoid such matings.
    If Jim and Jill descend from the same father, for example, it is better not to mate them.
    In this way we can avoid highly inbred offspring.
    We can go a bit further and use mating to minimize the inbreeding of the offspring.
    This is called minimum-coancestry mating.
    With minimum-coancestry mating, the least related sires and dams are mated.
    In this table you see the relationships between the sires and dams from the dog example.
    The relationship is lowest between Jim and Casey, and between Boris and Jill.
    Therefore, to minimize inbreeding in the offspring, we should mate Jim with Casey and Boris with Jill.
    So far, we have only considered inbreeding.
    However, sometimes we want to choose matings based on traits.
    This is the case for so-called optimum traits, where intermediate values are preferred.
    For such traits we can use compensatory mating.
    Here you see teat length in dairy cattle as an example.
    Cows with very short teats are difficult to milk,
    whereas cows with very long teats have higher chance of mastitis.
    So farmers prefer teats of intermediate length.
    In this table, you see compensatory mating for teat length.
    Cows with very short teats are mated to bulls that give daughters with very long teats, and vice versa.
    In other words, we choose a bull that compensates the trait of the cow.
    In this way, we can reduce the chance of getting offspring with extreme teat length.
    In conclusion, mating refers to the paring of the sires and dams
    that have been selected as parents for the next generation.
    While random mating is common, mating can be used to avoid high inbreeding or to compensate for weaknesses of the parents.
      view all
    In the previous sections, you learnt how to predict the response to selection. By performing selection, we determine which animals will become parents of the next generation, and which animals will not reproduce at all. However, until now we have not looked into the combinations of parents. This is what we call mating. The central question here is: Which male fits best to a certain selected female animal? Which criteria do we use?

    Watch the clip by dr. Piter Bijma to learn more about the process of mating. 
     


     
    Video subtitle:
     
     
    I guess everyone has some idea of the meaning of “mating”.
    But what precisely does “mating” mean in animal breeding?
    That is what I will explain in this clip.
    Mating refers to the pairing of the selected sires and dams.
    So mating occurs after selection.
    With selection, we decide which individuals become parents of the next generations.
    With mating, in contrast, we decide how the individuals that have already been selected as parents are combined to produce offspring.
    Here you see four dogs, two males and two females.
    Suppose we have selected those four dogs as parents of the next generation,
    and now we want to mate them.
    We could, for example, mate Jim with Jill, and Boris with Casey.
    Alternatively, we could mate Jim with Casey and Boris with Jill.
    Now what would be a good strategy to mate the selected parents?
    A common strategy is “random mating”,
    where sires and dams are just combined at random.
    Random mating is used a lot, and is often a good choice.
    However, we can do a bit better.
    With random mating, for example, we may accidentally mate closely related individuals,
    which would result in highly inbred offspring.
    It is better to avoid such matings.
    If Jim and Jill descend from the same father, for example, it is better not to mate them.
    In this way we can avoid highly inbred offspring.
    We can go a bit further and use mating to minimize the inbreeding of the offspring.
    This is called minimum-coancestry mating.
    With minimum-coancestry mating, the least related sires and dams are mated.
    In this table you see the relationships between the sires and dams from the dog example.
    The relationship is lowest between Jim and Casey, and between Boris and Jill.
    Therefore, to minimize inbreeding in the offspring, we should mate Jim with Casey and Boris with Jill.
    So far, we have only considered inbreeding.
    However, sometimes we want to choose matings based on traits.
    This is the case for so-called optimum traits, where intermediate values are preferred.
    For such traits we can use compensatory mating.
    Here you see teat length in dairy cattle as an example.
    Cows with very short teats are difficult to milk,
    whereas cows with very long teats have higher chance of mastitis.
    So farmers prefer teats of intermediate length.
    In this table, you see compensatory mating for teat length.
    Cows with very short teats are mated to bulls that give daughters with very long teats, and vice versa.
    In other words, we choose a bull that compensates the trait of the cow.
    In this way, we can reduce the chance of getting offspring with extreme teat length.
    In conclusion, mating refers to the paring of the sires and dams
    that have been selected as parents for the next generation.
    While random mating is common, mating can be used to avoid high inbreeding or to compensate for weaknesses of the parents.

     
    103
    Views

    How Can We Calculate the Response to Selection?

    ExperienceEmily Published the article • 0 comments • 103 views • 2017-11-21 06:53 • came from similar tags

    Now that you’ve seen and learnt about the different aspects of the formula to predict response to selection or genetic gain, let’s take an example and do the calculation together. Try to really understand and repeat what dr. Mario Calus does, so that you are able to do the calculations by yourself in the exercises that will follow after this video!
     

    video subtitle:
    Maximizing genetic gain is one of the focal points of a breeding program.
    But how is genetic gain calculated?
    Today we will demonstrate this by means of an example.
    Let us first repeat the formula used to compute the genetic gain,
    which is often also referred to as the response to selection per year.
    The response to selection, denoted as capital R, is computed as the product of the selection intensity i,
    the accuracy of selection r<sub>IH</sub>, and the genetic standard deviation (sigma A).
    Finally, the genetic gain per year is obtained by dividing the response to selection by the generation interval L in years.
    Let us consider the following example about running speed in horses, on a 2,000 metre long track.
    In this case, the breeding goal is the racing time across 2,000 metres.
    Let us assume that we select the 10% best offspring for breeding,
    the genetic standard deviation is 3 seconds, the accuracy of selection is 0.24, and the generation interval is 10 years.
    Now, the question is: what is the value of ∆G?
    As a first step, we compute the response per generation, assuming that the accuracy is 1.
    Now we need the value of the selection intensity, given that the proportion of selected animals is 10%.
    In a table that gives the selection intensity as function of the proportion of selected animals,
    we can find that this is 1.76.
    Considering the value of the genetic standard deviation of 3 seconds, we can compute that the response per generation is 5.28 seconds.
    As a second step, we replace the accuracy of 1 by its actual value of 0.24.
    Keeping the other input parameters as before, we now get a value of 1.27 seconds per generation.
    So we can see that having a relatively low accuracy of 0.24 yields a relatively low response,
    compared to the maximum possible response with an accuracy of 1.
    As a third and final step, we want to compute the genetic gain per year.
    This involves dividing the response by the generation interval.
    In this case, this is 10 years.
    So, finally, we obtain that the genetic gain is 0.127 seconds per year.
    So, in this example,
    the assumed selection approach is expected to decrease the racing time in 2,000 metre races by 0.127 seconds per year.
     
      view all
    Now that you’ve seen and learnt about the different aspects of the formula to predict response to selection or genetic gain, let’s take an example and do the calculation together. Try to really understand and repeat what dr. Mario Calus does, so that you are able to do the calculations by yourself in the exercises that will follow after this video!
     


    video subtitle:
    Maximizing genetic gain is one of the focal points of a breeding program.
    But how is genetic gain calculated?
    Today we will demonstrate this by means of an example.
    Let us first repeat the formula used to compute the genetic gain,
    which is often also referred to as the response to selection per year.
    The response to selection, denoted as capital R, is computed as the product of the selection intensity i,
    the accuracy of selection r<sub>IH</sub>, and the genetic standard deviation (sigma A).
    Finally, the genetic gain per year is obtained by dividing the response to selection by the generation interval L in years.
    Let us consider the following example about running speed in horses, on a 2,000 metre long track.
    In this case, the breeding goal is the racing time across 2,000 metres.
    Let us assume that we select the 10% best offspring for breeding,
    the genetic standard deviation is 3 seconds, the accuracy of selection is 0.24, and the generation interval is 10 years.
    Now, the question is: what is the value of ∆G?
    As a first step, we compute the response per generation, assuming that the accuracy is 1.
    Now we need the value of the selection intensity, given that the proportion of selected animals is 10%.
    In a table that gives the selection intensity as function of the proportion of selected animals,
    we can find that this is 1.76.
    Considering the value of the genetic standard deviation of 3 seconds, we can compute that the response per generation is 5.28 seconds.
    As a second step, we replace the accuracy of 1 by its actual value of 0.24.
    Keeping the other input parameters as before, we now get a value of 1.27 seconds per generation.
    So we can see that having a relatively low accuracy of 0.24 yields a relatively low response,
    compared to the maximum possible response with an accuracy of 1.
    As a third and final step, we want to compute the genetic gain per year.
    This involves dividing the response by the generation interval.
    In this case, this is 10 years.
    So, finally, we obtain that the genetic gain is 0.127 seconds per year.
    So, in this example,
    the assumed selection approach is expected to decrease the racing time in 2,000 metre races by 0.127 seconds per year.
     
     
    88
    Views

    Causes and Consequences of Trade-Offs(6.4 Trade-Offs in Predicting Genetic Gain )

    ExperienceEmily Published the article • 0 comments • 88 views • 2017-11-21 06:53 • came from similar tags

    Now that you are able to perform calculations with the formula to predict genetic gain, let’s take it one step further and dive into the elements of the formula and their relationships. The formula is of course used to predict genetic gain in a given situation, but can also be used to optimise genetic gain by varying the different elements of the formula. However, the different elements of the formula are not independent of each other. In the following clip, dr. Mario Calus will tell you more about this. 
     

     
    video subtitle:
     
     
    The success of a breeding program is largely determined by the genetic gain per year, denoted by ∆G.
    In previous clips, different components involved in ∆G and its computation have been discussed.
    The challenge is that several trade-offs exist between the elements of ∆G.
    In this lecture we will focus on those trade-offs
    and on how to balance them to optimize genetic gain per year.
    Let’s start with a brief recollection of the formula for ∆G.
    Three of its components, indicated here in blue, are under the breeders’ control.
    These are the selection intensity i, the accuracy rIH, and the generation interval L.
    The fourth component, the genetic standard deviation, is assumed to have a fixed value.
    Optimizing ∆G is a matter of balancing the selection intensity, the accuracy, and the generation interval.
    Now we will show an example of a trade-off between the accuracy and the selection intensity.
    Consider selection for a carcass traits in pigs,
    with a heritability of 30% and a generation interval of 2 years.
    The breeding program aims to select 10 animals.
    Measuring this carcass trait for an animal, requires that the animal is slaughtered.
    But of course, we don’t want to slaughter selection candidates.
    Therefore, the trait is measured on a number of half-sibs.
    For an individual, we can compute the expected accuracy of its breeding value,
    based on the heritability and the number of half-sibs tested.
    The formula as shown here, was previously mentioned in module 5.
    The figure in this slide shows the relationship between the number of half-sibs tested
    and the accuracy, which plateaus towards a value of one half.
    All the half-sibs of the selection candidate are kept in a testing facility,
    which has a maximum capacity of 5000 individuals.
    So this means that the number of selection candidates
    multiplied by the number of half-sibs tested per candidate, should not exceed 5000.
    In the figure we show the trade-off between the number of selection candidates
    and the number of half-sibs tested per candidate.
    So for instance we can test 50 half-sibs of 100 candidates,
    or we can test 100 half-sibs of 50 candidates.
    So, the question is: which numbers will give us the highest genetic gain?
    To answer this question, we evaluate the genetic gain for different numbers of selection candidates,
    making use of all the other parameters that we know.
    In this table we show the expected genetic gain when the number of selection candidates is:
    50, 100, 200, or 500.
    We can compute and plot the genetic gain for every possible number of selection candidates.
    From this plot, we can see that the highest genetic gain is achieved when testing 21 half-sibs for each of a total of 236 selection candidates.
    In addition to trade-offs between the selection intensity and the accuracy,
    trade-offs can also exist between the accuracy and the generation interval.
    Consider for instance a situation in meat sheep,
    where a breeding value of a female sheep for a carcass trait may depend on measurements in its progeny.
    The accuracy of the breeding value will increase
    with every new litter reaching the age of slaughter.
    At the same time, however, the generation interval will increase as well, which reduces the genetic gain per year.
    Let’s briefly summarize the main messages of this lecture.
    The genetic gain depends on three parameters that are under the breeders’ control:
    the selection intensity, the accuracy of selection and the generation interval.
    Several trade-offs may appear between these parameters.
    Optimizing genetic gain per year in a breeding program
    therefore requires careful balancing of these parameters.
     
      view all
    Now that you are able to perform calculations with the formula to predict genetic gain, let’s take it one step further and dive into the elements of the formula and their relationships. The formula is of course used to predict genetic gain in a given situation, but can also be used to optimise genetic gain by varying the different elements of the formula. However, the different elements of the formula are not independent of each other. In the following clip, dr. Mario Calus will tell you more about this. 
     


     
    video subtitle:
     
     
    The success of a breeding program is largely determined by the genetic gain per year, denoted by ∆G.
    In previous clips, different components involved in ∆G and its computation have been discussed.
    The challenge is that several trade-offs exist between the elements of ∆G.
    In this lecture we will focus on those trade-offs
    and on how to balance them to optimize genetic gain per year.
    Let’s start with a brief recollection of the formula for ∆G.
    Three of its components, indicated here in blue, are under the breeders’ control.
    These are the selection intensity i, the accuracy rIH, and the generation interval L.
    The fourth component, the genetic standard deviation, is assumed to have a fixed value.
    Optimizing ∆G is a matter of balancing the selection intensity, the accuracy, and the generation interval.
    Now we will show an example of a trade-off between the accuracy and the selection intensity.
    Consider selection for a carcass traits in pigs,
    with a heritability of 30% and a generation interval of 2 years.
    The breeding program aims to select 10 animals.
    Measuring this carcass trait for an animal, requires that the animal is slaughtered.
    But of course, we don’t want to slaughter selection candidates.
    Therefore, the trait is measured on a number of half-sibs.
    For an individual, we can compute the expected accuracy of its breeding value,
    based on the heritability and the number of half-sibs tested.
    The formula as shown here, was previously mentioned in module 5.
    The figure in this slide shows the relationship between the number of half-sibs tested
    and the accuracy, which plateaus towards a value of one half.
    All the half-sibs of the selection candidate are kept in a testing facility,
    which has a maximum capacity of 5000 individuals.
    So this means that the number of selection candidates
    multiplied by the number of half-sibs tested per candidate, should not exceed 5000.
    In the figure we show the trade-off between the number of selection candidates
    and the number of half-sibs tested per candidate.
    So for instance we can test 50 half-sibs of 100 candidates,
    or we can test 100 half-sibs of 50 candidates.
    So, the question is: which numbers will give us the highest genetic gain?
    To answer this question, we evaluate the genetic gain for different numbers of selection candidates,
    making use of all the other parameters that we know.
    In this table we show the expected genetic gain when the number of selection candidates is:
    50, 100, 200, or 500.
    We can compute and plot the genetic gain for every possible number of selection candidates.
    From this plot, we can see that the highest genetic gain is achieved when testing 21 half-sibs for each of a total of 236 selection candidates.
    In addition to trade-offs between the selection intensity and the accuracy,
    trade-offs can also exist between the accuracy and the generation interval.
    Consider for instance a situation in meat sheep,
    where a breeding value of a female sheep for a carcass trait may depend on measurements in its progeny.
    The accuracy of the breeding value will increase
    with every new litter reaching the age of slaughter.
    At the same time, however, the generation interval will increase as well, which reduces the genetic gain per year.
    Let’s briefly summarize the main messages of this lecture.
    The genetic gain depends on three parameters that are under the breeders’ control:
    the selection intensity, the accuracy of selection and the generation interval.
    Several trade-offs may appear between these parameters.
    Optimizing genetic gain per year in a breeding program
    therefore requires careful balancing of these parameters.
     
     
    113
    Views

    6.6 Long Term Genetic Contributions and Inbreeding > Breeding and Inbreeding

    ExperienceEmily Published the article • 0 comments • 113 views • 2017-11-21 06:53 • came from similar tags

    In the previous sections, we looked at selection, at predicting genetic gain and at the process of mating. The process of selection determines which alleles are transmitted to the next generation. Therefore, the choices made in a breeding program will always have consequences for the allele frequencies at the population level. The following clip by dr. Piter Bijma deals with inbreeding at the population level. 

    In order to be able to understand this, it is important to recall what inbreeding is about. If you would like to refresh your knowledge on the concept and the consequences of inbreeding, we recommend you go back to module 3 and watch the clips on “Concept of inbreeding” and “Consequences of inbreeding”.
     
     
    Video: Inbreeding at the Population Level
     

     
    Video subtitle:
     
     
    This clip deals with inbreeding at the population level.
    In a previous clip, you have seen that mating of relatives leads to inbreeding.
    In this pedigree, for example, the dogs Jim and Jill are related, so their offspring Donald is inbred.
    Because inbreeding reduces health and fertility, we want to avoid it as much as possible.
    However, can we always fully avoid inbreeding?
    In this clip, I will explain the mechanisms that determine inbreeding at the population level.
    This will show that inbreeding cannot be avoided.
    Let us investigate inbreeding at the population level using an example.
    Here you see two students, a male and a female.
    To see whether they are related, we have to investigate their pedigree.
    Together, both students have 4 parents, 8 grandparents, 16 great-grand parents, and so on.
    So if we go up the pedigree, the number of ancestors quickly becomes very large,
    and soon it will become larger than the population size.
    This means that some of the ancestors must be the same individual.
    In other words, because the number of ancestors doubles each generation, the two students must have ancestors in common.
    Therefore, they must be related.
    Now suppose both students like each other, and produce an offspring.
    Because the students are related, the offspring must be inbred.
    This example, therefore, demonstrates that inbreeding cannot be avoided.
    It also implies that the average inbreeding coefficient of individuals in the population must increase over time.
    In other words, the average inbreeding level in a population increases over time.
    So now we know that inbreeding increases over time.
    But how can we measure this increase, and what determines the speed of the increase?
    On the population level, the increase of inbreeding is measured per generation, and expressed as the so-called “rate of inbreeding”.
    The rate of inbreeding is denoted by the symbol ∆F.
    Because inbreeding has negative consequences, breeders try to keep the rate below 1% per generation.
    To keep the rate of inbreeding below 1%, we have to understand the factors that determine it.
    So what determines the rate of inbreeding?
    The most important factor is the population size.
    That is, the number of breeding males and females per generation.
    We can illustrate this with a simple example: suppose we have a population of N parents.
    Together, these N parents carry 2N alleles.
    Now suppose we make an offspring by randomly sampling two alleles out of the 2N alleles available.
    Then what is the probability that this offspring is inbred?
    The offspring is inbred only if it carries two identical alleles.
    With 2N alleles, this probability is equal to 1/(2N).
    You can see this as follows: whatever allele you sample as the first allele,
    the probability that the second allele is the same simply equals 1/(2N).
    Therefore, in a simple population, the rate of inbreeding is one divided by two times the population size.
    Hence, smaller populations tend to show more inbreeding.
    Beware that this equation holds only for simple populations,
    where each parent has an equal chance of contributing offspring to the next generation.
    However, also in more complex cases, the rate of inbreeding tends to be larger when the population is smaller.
    So how many parents do we need to restrict the rate of inbreeding below 1%?
    If we have a simple population, we need at least 50 parents per generation.
    However, for more complex populations this value may be much larger.
    A second important factor that determines inbreeding is the variation in the contribution of parents.
    Some parents may contribute many offspring, while others contribute few.
    Such variation increases the rate of inbreeding.
    Most livestock species, for example, have two sexes, and the contribution of a parent differs between the sexes.
    In cattle, for example, a selected bull usually gets more offspring than a selected cow.
    A bull, therefore, makes a greater contribution to the next generation than a cow,
    and this increases the rate of inbreeding compared to a species of a single sex.
    A second reason for variation in contributions is the presence of elite animals.
    The winner of the dog show, for example, may get many more offspring than an ordinary dog.
    Again, this increases the rate of inbreeding.
    Now let us look a bit further into the relationship between inbreeding and contributions.
    Suppose we have an ancestor who contributes 10% of the genes in the current population.
    How much does this ancestor contribute to the inbreeding?
    To answer this question, let us take a sire and dam from the current population, and make an offspring.
    Now we can calculate the probability that both genes in the offspring descend from the ancestor.
    On average, 10% of the genes in the sire and dam come from the ancestor.
    Therefore, the probability that both genes of the offspring descend from the ancestor is equal to the square of 10%, which is 1%.
    This is illustrated by the area of the red square on the slide.
    This result shows that the inbreeding due to an ancestor depends on the square of the contribution of this ancestor.
    With a little mathematics, you can show that the total rate of inbreeding
    is equal to the sum of the squared contributions of all ancestors, multiplied by a quarter.
    Let us use an example to illustrate the consequences of this equation.
    Suppose we have 10 dogs, 5 males and 5 females.
    First consider the case where each dog has the same contribution.
    In this case, each dog contributes 10%.
    If we substitute this value into the equation, we get a rate of inbreeding of 2.5% per generation.
    Now consider the case where one elite sire makes a large contribution.
    Suppose this sire contributes 46% of the genes, the four other sires each contribute 1%, and the dams still contribute 10% each.
    Substituting those values into the equation, gives a rate of inbreeding of 6.65%;
    almost three times greater than with equal contributions.
    Hence, this example illustrates that unequal contributions increase the rate of inbreeding.
    In conclusion: inbreeding cannot be avoided,
    small populations tend to show more inbreeding,
    and unequal contributions of parents increase the rate of inbreeding.
      view all
    In the previous sections, we looked at selection, at predicting genetic gain and at the process of mating. The process of selection determines which alleles are transmitted to the next generation. Therefore, the choices made in a breeding program will always have consequences for the allele frequencies at the population level. The following clip by dr. Piter Bijma deals with inbreeding at the population level. 

    In order to be able to understand this, it is important to recall what inbreeding is about. If you would like to refresh your knowledge on the concept and the consequences of inbreeding, we recommend you go back to module 3 and watch the clips on “Concept of inbreeding” and “Consequences of inbreeding”.
     
     
    Video: Inbreeding at the Population Level
     


     
    Video subtitle:
     
     
    This clip deals with inbreeding at the population level.
    In a previous clip, you have seen that mating of relatives leads to inbreeding.
    In this pedigree, for example, the dogs Jim and Jill are related, so their offspring Donald is inbred.
    Because inbreeding reduces health and fertility, we want to avoid it as much as possible.
    However, can we always fully avoid inbreeding?
    In this clip, I will explain the mechanisms that determine inbreeding at the population level.
    This will show that inbreeding cannot be avoided.
    Let us investigate inbreeding at the population level using an example.
    Here you see two students, a male and a female.
    To see whether they are related, we have to investigate their pedigree.
    Together, both students have 4 parents, 8 grandparents, 16 great-grand parents, and so on.
    So if we go up the pedigree, the number of ancestors quickly becomes very large,
    and soon it will become larger than the population size.
    This means that some of the ancestors must be the same individual.
    In other words, because the number of ancestors doubles each generation, the two students must have ancestors in common.
    Therefore, they must be related.
    Now suppose both students like each other, and produce an offspring.
    Because the students are related, the offspring must be inbred.
    This example, therefore, demonstrates that inbreeding cannot be avoided.
    It also implies that the average inbreeding coefficient of individuals in the population must increase over time.
    In other words, the average inbreeding level in a population increases over time.
    So now we know that inbreeding increases over time.
    But how can we measure this increase, and what determines the speed of the increase?
    On the population level, the increase of inbreeding is measured per generation, and expressed as the so-called “rate of inbreeding”.
    The rate of inbreeding is denoted by the symbol ∆F.
    Because inbreeding has negative consequences, breeders try to keep the rate below 1% per generation.
    To keep the rate of inbreeding below 1%, we have to understand the factors that determine it.
    So what determines the rate of inbreeding?
    The most important factor is the population size.
    That is, the number of breeding males and females per generation.
    We can illustrate this with a simple example: suppose we have a population of N parents.
    Together, these N parents carry 2N alleles.
    Now suppose we make an offspring by randomly sampling two alleles out of the 2N alleles available.
    Then what is the probability that this offspring is inbred?
    The offspring is inbred only if it carries two identical alleles.
    With 2N alleles, this probability is equal to 1/(2N).
    You can see this as follows: whatever allele you sample as the first allele,
    the probability that the second allele is the same simply equals 1/(2N).
    Therefore, in a simple population, the rate of inbreeding is one divided by two times the population size.
    Hence, smaller populations tend to show more inbreeding.
    Beware that this equation holds only for simple populations,
    where each parent has an equal chance of contributing offspring to the next generation.
    However, also in more complex cases, the rate of inbreeding tends to be larger when the population is smaller.
    So how many parents do we need to restrict the rate of inbreeding below 1%?
    If we have a simple population, we need at least 50 parents per generation.
    However, for more complex populations this value may be much larger.
    A second important factor that determines inbreeding is the variation in the contribution of parents.
    Some parents may contribute many offspring, while others contribute few.
    Such variation increases the rate of inbreeding.
    Most livestock species, for example, have two sexes, and the contribution of a parent differs between the sexes.
    In cattle, for example, a selected bull usually gets more offspring than a selected cow.
    A bull, therefore, makes a greater contribution to the next generation than a cow,
    and this increases the rate of inbreeding compared to a species of a single sex.
    A second reason for variation in contributions is the presence of elite animals.
    The winner of the dog show, for example, may get many more offspring than an ordinary dog.
    Again, this increases the rate of inbreeding.
    Now let us look a bit further into the relationship between inbreeding and contributions.
    Suppose we have an ancestor who contributes 10% of the genes in the current population.
    How much does this ancestor contribute to the inbreeding?
    To answer this question, let us take a sire and dam from the current population, and make an offspring.
    Now we can calculate the probability that both genes in the offspring descend from the ancestor.
    On average, 10% of the genes in the sire and dam come from the ancestor.
    Therefore, the probability that both genes of the offspring descend from the ancestor is equal to the square of 10%, which is 1%.
    This is illustrated by the area of the red square on the slide.
    This result shows that the inbreeding due to an ancestor depends on the square of the contribution of this ancestor.
    With a little mathematics, you can show that the total rate of inbreeding
    is equal to the sum of the squared contributions of all ancestors, multiplied by a quarter.
    Let us use an example to illustrate the consequences of this equation.
    Suppose we have 10 dogs, 5 males and 5 females.
    First consider the case where each dog has the same contribution.
    In this case, each dog contributes 10%.
    If we substitute this value into the equation, we get a rate of inbreeding of 2.5% per generation.
    Now consider the case where one elite sire makes a large contribution.
    Suppose this sire contributes 46% of the genes, the four other sires each contribute 1%, and the dams still contribute 10% each.
    Substituting those values into the equation, gives a rate of inbreeding of 6.65%;
    almost three times greater than with equal contributions.
    Hence, this example illustrates that unequal contributions increase the rate of inbreeding.
    In conclusion: inbreeding cannot be avoided,
    small populations tend to show more inbreeding,
    and unequal contributions of parents increase the rate of inbreeding.