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# ANEQ 328 Week 12 Class Notes ANEQ328-001

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This 6 page Class Notes was uploaded by Destinee on Sunday April 17, 2016. The Class Notes belongs to ANEQ328-001 at Colorado State University taught by Milton Thomas in Spring 2016. Since its upload, it has received 19 views. For similar materials see Foundation in Animal Genetics in Animal Science at Colorado State University.

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Date Created: 04/17/16

ANEQ 328 Foundations In Animal Genetics Week 12 Class Notes (4/5/16-4/7/16) Allele and Genotypic Frequency and Hardy Weinberg Equilibrium Foundational Information Needed To Understand Inheritance o The genotypes of the parents. o How genes interact. o Mendel’s Principles o Genotype and Allele Frequency Vocab Terms o Population A breeding group of individuals. o Artificial Selection When humans make the selection in the breeding process of animals based on desirable traits. o Natural Selection When breeding occurs naturally without humans interfering or “hand-selecting” for desirable traits. Ex. Inter-se mating and random mating. o Allele Frequency 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. o Genotypic Frequency 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. q equals the frequency of the homozygous recessive individuals. o Allele Fixation The point at which a particular allele becomes the only allele at its locus in a population—the frequency of the allele becomes 1. Example of Solving For Allele Frequency o When solving for allele frequency the following equation is used p+q=1 p equals the frequency of the dominant alleles. q equals the frequency of the recessive alleles. o 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. Where the genotype NN signifies the horse is normal and doesn’t have the condition. Where the genotype N/HRD signifies the horse is normal but is a carrier of the condition. Where the genotype HRD/HRD signifies the horse has HERDA. ANEQ 328 Foundations In Animal Genetics Week 12 Class Notes (4/5/16-4/7/16) 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 : o Use the equation p+q=1 to find the frequency of those that are Normal within the population. 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. o Use the equation p+q=1 to find the frequency of those that have HERDA within the population. 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%. Example of Solving For Genotypic Frequency ANEQ 328 Foundations In Animal Genetics Week 12 Class Notes (4/5/16-4/7/16) o Genetic frequency can be solved by using the following equation 2 2 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. o 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. Where the genotype NN signifies the horse is normal and doesn’t have the condition. Where the genotype N/HRD signifies the horse is normal but is a carrier of the condition. 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 (Has2 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 Disea2e) (q ) Total: 10 2 2 o Use the equation p +2pq+q =1 to find the frequency of those that are Normal (homozygous individuals) within the population. 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 2 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 2 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. ANEQ 328 Foundations In Animal Genetics Week 12 Class Notes (4/5/16-4/7/16) 2 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 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 equals q . 2 We can double check out work by plugging in our values that we got for p , 2pq, and q into the equation p +2pq+q =1. 2 p =.70 2pq=.20 2 q =.10 .70+.20+.10 does indeed equal 1. Why be concerned with allele and genotypic frequency? o Gene frequencies can be used to predict the gene frequency of offspring. o Genetic change can allow us to measure change in gene frequency. 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. o There are 5 conditions in which 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 o The equation p +2pq+q =1 can be used to calculate Hardy-Weinberg Equilibrium Deriving Hardy Weinberg Equilibrium With Punnett Squares p +2pq+q =12 Solving For Allele and Genotypic Frequency Using Hardy-Weinberg Equilibrium ANEQ 328 Foundations In Animal Genetics Week 12 Class Notes (4/5/16-4/7/16) o Using the equation 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. 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 . 2 2 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 2 which is q . 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 2 that we found an2 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 valu2s we just found. Since p equals the frequency of the homozygous dominant individuals which is 2 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. ANEQ 328 Foundations In Animal Genetics Week 12 Class Notes (4/5/16-4/7/16) 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.

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