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# ANEQ 328 Study Guide For Exam 3 ANEQ328-001

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This 21 page Study Guide was uploaded by Destinee on Friday April 22, 2016. The Study Guide belongs to ANEQ328-001 at Colorado State University taught by Milton Thomas in Spring 2016. Since its upload, it has received 224 views. For similar materials see Foundation in Animal Genetics in Animal Science at Colorado State University.

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Date Created: 04/22/16

Exam 3 Study Guide This study guide contains material covered in class from 4/5/16-4/26/16. This study guide contains important vocab terms, fill in the blanks, and questions that focus on the main points learned in class. All definitions and answers to these questions can be found within the notes weeks 12 through 14. Although an answer key is provided, I strongly suggest filling out the study guide by yourself first, and then use the answer key to check your answers. Happy Studying! Week 12 Review (4/5/16-4/7/16) 1.) Define the following terms: Population Artificial Selection Natural Selection Allele Frequency Genotypic Frequency Allele Fixation Law of Hardy Weinberg Equilibrium 2.) What is the foundational information needed to understand inheritance? 3.) Compare and contrast artificial and natural selection. 4.) Compare and contrast allele frequency and genotypic frequency. 5.) Use the following data to solve for the allelic frequency of those that are Normal within the population. Using the example of HERDA (Hereditary Equine Regional Dermal Asthenia) which is the loss of strength in the skin we can solve for the allele frequency of the disease within a population of horses. o Where the genotype NN signifies the horse is normal and doesn’t have the condition. o Where the genotype N/HRD signifies the horse is normal but is a carrier of the condition. o Where the genotype HRD/HRD signifies the horse has HERDA. In this scenario we will treat NN (Normal) as homozygous dominant (p), N/HRD (Normal but a carrier) as heterozygous, and HRD/HRD (Has HERDA) as homozygous recessive (q). Number of Phenotype Genotype Number of Number of Total Horses “N” “HRD” Number of (p) (q) Alleles 7 Normal NN 14 0 14 2 Normal N/HRD 2 2 4 (Carrier) 1 Has HC or HRD/HRD 0 2 2 HRD Total: 16 4 20 6.) Use the following data to solve for the allelic frequency of those that have HERDA within the population. Using the example of HERDA (Hereditary Equine Regional Dermal Asthenia) which is the loss of strength in the skin we can solve for the allele frequency of the disease within a population of horses. o Where the genotype NN signifies the horse is normal and doesn’t have the condition. o Where the genotype N/HRD signifies the horse is normal but is a carrier of the condition. o Where the genotype HRD/HRD signifies the horse has HERDA. In this scenario we will treat NN (Normal) as homozygous dominant (p), N/HRD (Normal but a carrier) as heterozygous, and HRD/HRD (Has HERDA) as homozygous recessive (q). Number of Phenotype Genotype Number of Number of Total Horses “N” “HRD” Number of (p) (q) Alleles 7 Normal NN 14 0 14 2 Normal N/HRD 2 2 4 (Carrier) 1 Has HC or HRD/HRD 0 2 2 HRD Total: 16 4 20 2 2 7.) Using the data below and the equation p +2pq+q =1, find the frequency of those that are Normal (homozygous dominant individuals), normal but carriers (heterozygous individuals), and those that have HERDA (homozygous recessive individuals) within the population. Using the example of HERDA (Hereditary Equine Regional Dermal Asthenia) which is the loss of strength in the skin we can solve for the allele frequency of the disease within a population of horses. o Where the genotype NN signifies the horse is normal and doesn’t have the condition. o Where the genotype N/HRD signifies the horse is normal but is a carrier of the condition. o Where the genotype HRD/HRD signifies the horse has HERDA. In this scenario we will treat NN (Normal) as homozygous dominant individuals (p ), N/HRD (Normal but a carrier) as heterozygous individuals (2pq), and HRD/HRD (Has HERDA) as homozygous recessive individuals (q ). Number of Horses Phenotype Genotype Number of Genotypes 7 Normal N/N 7 (p ) 2 Normal (Carrier) N/HRD 2 (2pq) 1 HC or HRD (has HRD/HRD 1 Disease) 2 (q ) Total: 10 8.) Why are allele and genotypic frequencies important? 9.) What does the law of Hardy Weinberg Equilibrium (HWE) state? 10.) What are the five conditions a population must meet in order to remain at Hardy Weinberg Equilibrium? 11.) What is the equation used to solve for Hardy Weinberg Equilibrium? 12.) Conduct a Punnett using the equation p +2pq+q =1. 2 2 2 13.) Using Hardy-Weinberg Equilibrium (p +2pq+q =1) find the allele and genotypic frequency of the Hampshire pigs given that 910 pigs are belted and 90 black. Keep in mind that belted is dominant to black. Week 13 Review (4/12/1616-4/14/16) 1.) Define the following terms: Population Value Trait Simple Trait Complex Trait Mean Normal Distribution (Bell Curve) Variance (???? ) Standard Deviation (????) Standard Error Covariance Correlation (r) Regression (b) Phenotypic Value Population Mean Genotypic Value Environmental Effect Breeding Value Gene Combination Value Progeny Difference Producing Ability Permanent Environment Effect Temporary Environmental Effect 2.) Graph a general bell curve graph and label the x and y axis appropriately. 3.) A bell curve graph demonstrates the different standard deviations of the mean, what are they? 4.) What is the equation used to calculate mean? 5.) Using the following data, calculate the mean of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 6.) What unit is used to represent the mean? 7.) What is the equation used to calculate variance? 8.) Using the following data, calculate the variance of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 9.) What is the equation used to calculate standard deviation? 10.) Using the following data, calculate the standard variation of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 11.) What is the equation used to calculate standard error? 12.) Using the following data, calculate the standard error of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 13.) What is the equation used to calculate covariance? 14.) Using the following data, calculate the covariance of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 15.) What are the characteristics of covariation? 16.) What is the equation used to calculate correlation? 17.) Using the following data, calculate the correlation of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 18.) What is the equation used to calculate regression? 19.) Using the following data, calculate the regression of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 20.) What is the equation used to calculate the genetic model of quantitative traits? 21.) What is the equation used to calculate the genetic model of quantitative traits including breeding value and gene combination value? 22.) What is the equation used to calculate progeny difference? 23.) When looking at an EPD the sire and dam can be considered _________ traits while birth date are considered ___________ traits. 24.) What is the equation used to calculate the genetic model of repeated quantitative traits? Week 14 Review (4/19/16-4/21/16) 1.) Define the following terms: Narrow Sense (h ) Broad Sense (H ) Heritability Inheritance Repeatability Contemporary Group 2.) What is the equation used to calculate narrow sense? 3.) What is the equation used to calculate broad sense? 4.) What level determines if a trait will be inherited? 5.) Compare and contrast heritability and inheritance. 6.) What is the equation used to calculate repeatability? Exam 3 Study Guide Answer Key Week 12 Review (4/5/16-4/7/16) 1.) Define the following terms: Population o A breeding group of individuals. Artificial Selection o When humans make the selection in the breeding process of animals based on desirable traits. Natural Selection o When breeding occurs naturally without humans interfering or “hand-selecting” for desirable traits. Ex. Inter-se mating and random mating. Allele Frequency o The frequency of an allele or alleles in a population. Can be calculated using the equation p+q=1. p equals the frequency of the dominant alleles. q equals the frequency of the recessive alleles. Genotypic Frequency o The frequency of a specific genotype within a population. 2 2 Can be calculated using the equation p +2pq+q =1 p equals the frequency of the homozygous dominant individuals. 2pq equals the frequency of the heterozygous individuals. 2 q equals the frequency of the homozygous recessive individuals. Allele Fixation o The point at which a particular allele becomes the only allele at its locus in a population—the frequency of the allele becomes 1. Law of Hardy Weinberg Equilibrium (HWE) o A law that states the frequency of alleles and genotypes within a population will remain constant from generations to generations. 2.) What is the foundational information needed to understand inheritance? The genotypes of the parents. How genes interact. Mendel’s Principles Genotype and Allele Frequency 3.) Compare and contrast artificial and natural selection. Artificial Selection o When humans make the selection in the breeding process of animals based on desirable traits. Natural Selection o When breeding occurs naturally without humans interfering or “hand-selecting” for desirable traits. Ex. Inter-se mating and random mating. 4.) Compare and contrast allele frequency and genotypic ratio. Allele Frequency o The frequency of an allele or alleles in a population. Can be calculated using the equation p+q=1. p equals the frequency of the dominant alleles. q equals the frequency of the recessive alleles. Genotypic Frequency o The frequency of a specific genotype within a population. Can be calculated using the equation p +2pq+q =12 p equals the frequency of the homozygous dominant individuals. 2pq equals the frequency of the heterozygous individuals. q equals the frequency of the homozygous recessive individuals. 5.) Use the following data to solve for the allelic frequency of those that are Normal within the population. Number of Phenotype Genotype Number of Number of Total Horses “N” “HRD” Number of (p) (q) Alleles 7 Normal NN 14 0 14 2 Normal N/HRD 2 2 4 (Carrier) 1 Has HC or HRD/HRD 0 2 2 HRD Total: 16 4 20 Use the equation p+q=1, where p equals the frequency of the dominant alleles and q equals the frequency of the recessive alleles, to find the frequency of those that are Normal within the population. o Since the number of normal in this case represents the homozygous dominants and the heterozygous (since heterozygous are also dominant) we just need to solve for p which can be found by using the Number that are N. The total “Number of N” is 16, meaning there are 16 “normal” alleles within this population of horses. Now we take that number 16 and divide it by the total number of alleles which from the table is 20. So 16/20 gives us .80 or 80%, so p=.80. So 80% of this horse population carries the normal dominant alleles. 6.) Use the following data to solve for the allelic frequency of those that have HERDA within the population. Number of Phenotype Genotype Number of Number of Total Horses “N” “HRD” Number of (p) (q) Alleles 7 Normal NN 14 0 14 2 Normal N/HRD 2 2 4 (Carrier) 1 Has HC or HRD/HRD 0 2 2 HRD Total: 16 4 20 Use the equation p+q=1, where p equals the frequency of the dominant alleles and q equals the frequency of the recessive alleles, to find the frequency of those that have HERDA within the population. o Since the number of HRD in this case represents the homozygous recessives we need to solve for q which can be found by using the Number of HRD. The total “Number of HRD” is 4, meaning there are 4 “diseased” alleles within this population of horses. Since we know that p is .80 from solving it for it earlier we can rearrange the equation p+q=1 to 1-p=q to help us find q. So 1-.80=.20, so q=.20. So 20% of this horse population carries the homozygous recessive alleles. q can also be calculated like how we found p above. We would take the total “Number of HRD” which is 4 and divide it by the total number of alleles which is 20. So 4/20 equal 20%. 2 2 7.) Using the data below and the equation p +2pq+q =1, find the frequency of those that are Normal (homozygous dominant individuals), normal but carriers (heterozygous individuals), and those that have HERDA (homozygous recessive individuals) within the population. Remember the following: o p equals the frequency of the homozygous dominant individuals. o 2pq equals the frequency of the heterozygous individuals. o q equals the frequency of the homozygous recessive individuals. 2 First, lets start by finding p which is the frequency of the homozygous dominant individuals which are the Normal phenotype with the genotype N/N. There are 7 individuals that have the N/N genotype. Thus, p can be calculated by taking the number of individuals that are homozygous dominant which is the Normal phenotype, 7 divided by the total population which is 10. So 7/10 equals .7 or 70% which equals p . Second, lets find 2pq, which is the frequency of the heterozygous individuals which are the normal (carrier) phenotype with the genotype N/HRD. There are 2 individuals that have the N/HRD genotype. Thus, 2pq can be calculated by taking the number of individuals that are heterozygous which is the normal (carrier) phenotype, 2 divided by the total population which is 10. So 2/10 equals .2 or 20%, which equals 2pq. Next, lets find q which the frequency of the homozygous recessive individuals which is the HC or HRD phenotype with the genotype HRD/HRD. There is 1 individual that has the 2 HRD/HRD genotype. Thus, q can be calculated by taking the number of individuals that are homozygous recessive which is the HC or HRD phenotype, 1 divided by the total population which is 10. So 1/10 equals .1 or 10% which 2 equals q . We can double check out work by plugging in our 2 2 values that we got for p , 2pq, and q into the equation p +2pq+q =1. 2 2 p =.70 2pq=.20 2 q =.10 .70+.20+.10 does indeed equal 1. 8.) Why are allele and genotypic frequencies important? Gene frequencies can be used to predict the gene frequency of offspring. Genetic change can allow us to measure change in gene frequency. 9.) What does the law of Hardy Weinberg Equilibrium (HWE) state? A law that states the frequency of alleles and genotypes within a population will remain constant from generations to generations. 10.) What are the five conditions a population must meet in order to remain at Hardy Weinberg Equilibrium? No random mating No natural selection No genetic mutations No migration/emigration No genetic drift 11.) What is the equation used to solve for Hardy Weinberg Equilibrium? p +2pq+q =12 o p equals the frequency of the homozygous dominant individuals. o 2pq equals the frequency of the heterozygous individuals. o q equals the frequency of the homozygous recessive individuals. 12.) Conduct a Punnett using the equation p +2pq+q =1. 2 13.) Using Hardy-Weinberg Equilibrium (p +2pq+q =1) find the allele and genotypic frequency of the Hampshire pigs given that 910 pigs are belted and 90 black. Keep in mind that belted is dominant to black. o First, lets find the total since 910 are belted and 90 are black that makes 1,000 Hampshire pigs total. We know that 910 are belted which is homozygous dominant and 90 are black which is homozygous recessive. Out of those 910 pigs we don’t know how many for certain are homozygous dominant or heterozygous, since both genotypes are displaying a belted phenotype. But since we know that 90 are black and thus homozygous recessive we can start by first solving for q . To find q we need to take the number of homozygous recessive, 90, and divide it by the total population of 1000. 90/1000 which equals .09 which is q . 2 Now lets solve for q. To do this we just need to take the square root of what we got for q which is .09. So √.09 equals .3 which is q. Now that we have q we can solve for p. To solve for p we can adjust the equation p+q=1 to 1- q=p. So 1-.3=.7 which equals p. Now lets solve for p . To do so we take our p vaule that we found and square it. So .7 equals .49, which equals p . Now that we know p and q we can solve for 2pq. To do so just multiply the values we found for p and q to two. So (2*.7*.3) gives us .42 which equals 2pq. Now we can calculate the phenotypic frequencies using the values we just found. Since p equals the frequency of the homozygous dominant individuals which is this case are the belted pigs. Our p value that we found was .49. So we take .49 and multiply it by the total population of pigs which is 1,000. So .49*1,000=490. So 490 pigs are belted (homozygous dominant). Since 2pq equals the frequency of the heterozygous individuals which in this case are the belted pigs that have a heterozygous genotype. Our 2pq value we found was .42. So we take .42 and multiply it by the total population of pigs which is 1,000. So .42*1,000=420. So 420 pigs are belted but have a heterozygous genotype. 2 Since q equals the frequency of the homozygous recessive individuals in this case the number of black pigs. Our q value we found was .09. So we take .09 and multiply it by the total population of pigs which is 1,000. So .09*1,000=90. So 90 pigs are black. Week 13 Review (4/12/1616-4/14/16) 1.) Define the following terms: Population o A breeding population of animals. Value o A numeric value applied to an individual as opposed to a population. Phenotypic Trait Trait o A genetically determined characteristics. Trait doesn’t always equal phenotype. Simple Trait o When a gene is controlled by 1 or a few loci. o Can be qualitative or categorical. Ex. The coat color of a horse is only controlled by a few loci. Complex Trait o When a gene is controlled by many loci. o Can be quantitative or polygenic. Ex. Milk production in cows in controlled by many loci. Mean o The average of a set of numbers. To calculate: Just add up all the numbers, then divide by how many numbers there are. Normal Distribution (Bell Curve) o The statistical distribution that appears graphically as a symmetric, bell-shaped curve. 2 Variance (???? ) o Differences among individuals within a population. Standard Deviation (????) o The average deviation from the mean. Standard Error o A measure of the statistical accuracy of an estimate, equal to the standard deviation divided by the square root of the number in the population. Covariance o How two traits or values vary together in a population. Correlation (r) o A measure of strength of the relationship between two variables. Regression (b) o The expected or average change in one variable (y) per unit change in another (x). Phenotypic Value o The performance of an individual animal for a specific trait. Population Mean o The average phenotypic value for the specific trait for all animals in the population. Genotypic Value o The genotypic values of the individual for the specific trait. Environmental Effect o The environmental effects on the individual’s performance for the trait. Breeding Value o The value of the individual as a parent, (the sum of the independent genotypes). Gene Combination Value o An individual’s genotypic value of gene interaction. Progeny Difference (PD) o The expected difference between the mean performance of the individual’s progeny and the mean performance of all progeny. Permanent Environmental Effect o An environmental effect that permanently influences an individual’s performance for a repeated trait. Temporary Environmental Trait o An environmental effect that influences a single performance record of an individual but does not permanently affect an individual’s performance potential for a repeated trait. 2.) Graph a general bell curve graph and label the x and y axis appropriately. Frequency Category 3.) A bell curve graph demonstrates the different standard deviations of the mean, what are they? About 66% of the animal population will fall within the Standard Deviation of the mean. About 95% of the animal population will fall with 2 Standard Deviations of the mean. 99% of the animal population will fall within 3 Standard Deviations of the mean. 4.) What is the equation used to calculate mean? To calculate the mean just add up all the numbers, then divide by how many numbers there are. 5.) Using the following data, calculate the mean of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 Weaning Weight Mean Calculation (500 + 475 + 425 ) 1400 3 = 3 = 466.67???? Yearling Weight Mean Calculation (1100 + 1050 + 1150) 3300 = = 1100???? 3 3 6.) What unit is used to represent the mean? ???? 7.) What is the equation used to calculate variance? 2 ∑(????−????????)2 2 (????1−????????) +(2 −????????) +(3 −???? ) …… Equation: ???? = ????−1 or ???? = ????−1 8.) Using the following data, calculate the variance of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 We know that the number in our population is 3 calves, so n=3. o Weaning Weight Variance Calculation 2 2 2 2 2 2 (500 − 500) + (475 − 500) + (525 − 500) (????) + −25 ) + 25 ) 0 + 625 + 625 1250 2 2 3 − 1 = 2 = 2 = 2 = 625???????????? = ???????? o Yearling Weight Variance Calculation (1100 − 1100) + (1050 − 1100) + (1150 − 11????0)+ −50)2+ 50)2 0 + 2500 + 25005000 = = = = 2500????????????= ????????2 3 − 1 2 2 2 9.) What is the equation used to calculate standard deviation? Equation: √???? 2 10.) Using the following data, calculate the standard variation of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 2 2 2 o Based on previous calculations we know that ???? ???? = 625???????????? and ???? ???? = 2500???????????? . Weaning Weight Standard Deviation Calculation ????????2 = 625???????????? = 62√ = 25????????????=???? ???? Yearling Weight Standard Deviation Calculation ????????2 = 2500???????????? = √2500 = 50???????????? = ???? ???? 11.) What is the equation used to calculate standard error? ???? Equation: or ???????????????????????????????? ???????????????????????????????????? √???? √???????????????????????? ???????? ???????????????????????????????????????? 12.) Using the following data, calculate the standard error of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 o Based on previous calculations we know that ???? = 25???????????? and ???? = 50????????????????and the number in our population is 3 calves, so n=3. Weaning Weight Standard Error Calculation 25 ????????= 25???????????? = = 15???????????? = ????????.500 ± 15???????????? √ 3 Yearling Weight Standard Error Calculation 50 ????????= 50???????????? = = 30???????????? = ????????.1100 ± 30???????????? √ 3 13.) What is the equation used to calculate covariance? (???? −???? )( −???? )+ ???? −???? )( −???? )+ ???? −???? )( −???? ). Equation: 1 ???? 1 ???? 2 ???? 2 ???? 3 ???? 3 ???? ????−1 14.) Using the following data, calculate the covariance of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 We know that the number in our population is 3 calves, so n=3. Weaning Weight and Yearling Weight Covariance Calculation (500 − 500 1100 − 1100 + 475 − 500 1050 − 1100 + 525 − 500 1150 − 1100 ) = 2500 = 1250???????????? 3 − 1 2 15.) What are the characteristics of covariation? 1.) Positive or Negative Number 2.) Correlation shows the strength of a relationship. 3.) Regression show how much (amount). 16.) What is the equation used to calculate correlation? Equation: ???????????? = (Cov(x,y)) ???????? ???? 17.) Using the following data, calculate the correlation of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 o Based on previous calculations we know that Cov(x,y ( ))= 1250lbs and ???? ???? 25???????????? and ???? = 50????????????. ???? Weaning Weight and Yearling Weight Standard Correlation Calculation (Cov x,y))= 1250lbs 1250 (???? = 25????????????)(???? = 50????????????) = (25)(50) = 1 ???? ???? 18.) What is the equation used to calculate regression? Equation: ????????∗???? = (Cov x,y ) ????????2 19.) Using the following data, calculate the regression of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 Based on previous calculations we know that Co( x,y( )) = 1250lbs and ????????2 = 625????????????. Weaning Weight and Yearling Weight Standard Regression Calculation (Cov x,y)) = 1250lbs 1250 2 = = 2 ???????? = 625???????????? 625 For every 1lb increase in weaning weight, yearling weight increases an average of 2 lbs. 20.) What is the equation used to calculate the genetic model of quantitative traits? Equation: P=????+G+E P= Phenotypic Value The performance of an individual animal for a specific trait. ????= Population mean The average phenotypic value for the specific trait for all animals in the population. G= Genotypic value The genotypic values of the individual for the specific trait. E= Environmental Effect The environmental effects on the individual’s performance for the trait. 21.) What is the equation used to calculate the genetic model of quantitative traits including breeding value and gene combination value? Equation: P=????+BV+GCV+E P= Phenotypic Value The performance of an individual animal for a specific trait. ????= Population mean The average phenotypic value for the specific trait for all animals in the population. BV= Breeding Value The value of the individual as a parent, (the sum of the independent genotypes). Ex. A= +5, a= -5 AA= Value of 10 aa= Value of -10 Aa= Value of 0 GCV= Gene Combination Value An individual’s genotypic value of gene interaction. E= Environmental Effect The environmental effects on the individual’s performance for the trait. 22.) What is the equation used to calculate progeny difference? Progeny Difference (PD) The expected difference between the mean performance of the individual’s progeny and the mean performance of all progeny. Equation: PD= ½BV BV= Breeding Value The value of the individual as a parent, (the sum of the independent genotypes). 23.) When looking at an EPD the sire and dam can be considered genetic traits while birth date are considered environmental traits. 24.) What is the equation used to calculate the genetic model of repeated quantitative traits? Producing Ability (PA) The performance potential of an individual for a repeated trait. Equation: PA= ????+BV+GCV+E +E p t ????= Population mean The average phenotypic value for the specific trait for all animals in the population. BV= Breeding Value The value of the individual as a parent, (the sum of the independent genotypes). Ex. A= +5, a= -5 AA= Value of 10 Week 12 Review (4/19/16-4/21/16) 1.) Define the following terms: Narrow Sense (h 2) o The proportion of phenotype that is due to genetic variation. Broad Sense (H )2 o A measure of the strength of relationship between performance (phenotypic values) and genotypic values for a trait in a population. Heritability o A statistical term that describes the phenotypic variance due to genetic variance. Inheritance o What segregated from parents to progeny. Repeatability o A trait we can measure multiple times. Contemporary Group o Group of animals that where managed together (i.e. have the same age, same breed, etc.) 2.) What is the equation used to calculate narrow sense? ???? 2 ???? ???????? Equation: h (hereditability)=???????? ???? o P= Phenotypic Value The performance of an individual animal for a specific trait. o BV= Breeding Value The value of the individual as a parent, (the sum of the independent genotypes). 3.) What is the equation used to calculate broad sense? ???? ???? ????2????????+???? 2???????????? Equation: H = 2 = 2 ???? ???? ???? ???? o BV= Breeding Value The value of the individual as a parent, (the sum of the independent genotypes). o P= Phenotypic Value The performance of an individual animal for a specific trait. o GCV= Gene Combination Value An individual’s genotypic value of gene interaction 4.) What level determines if a trait will be inherited? Low<Moderate<High o If the value of h is between 0 and .19 then the trait has a low chance of being inherited. Ex. Calving Interval in Beef Cattle o If the value of h is between .20 and .39 then the trait has a moderate chance of being inherited. 2 o If the value of h is greater than .40 then the trait has a high chance of being inherited. Ex. Wither Height in Horses 5.) Compare and contrast heritability and inheritance. Heritability o A statistical term that describes the phenotypic variance due to genetic variance. Deals with genetics. Inheritance o What segregated from parents to progeny. Mendel’s Principles Meiosis Deals with genetics. 6.) What is the equation used to calculate repeatability? ???? ???????? +????2????????????+????2???????? Equation: ???? = 2 ???? ???? o BV= Breeding Value The value of the individual as a parent, (the sum of the independent genotypes). o P= Phenotypic Value The performance of an individual animal for a specific trait. GCV= Gene Combination Value An individual’s genotypic value of gene interaction Ep= Permanent Environmental Effect An environmental effect that permanently influences an individual’s performance for a repeated trait. Ex. Increase or decrease in performance.

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