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PS 2 answers, probs 6+7+8

# PS 2 answers, probs 6+7+8 - R and C W in this population C...

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Problem 6. a.) What is the allele frequency of the M allele in the sailor population? (450+450+45)/1000= 0.945 b.) 10 years later 1000 children have been born on the coastline. (All unions were consensual!) If the population of 1000 young people on the coast mate randomly, how many of the 1000 children would you expect to have MN bloodtype? 469.9 ~ 470. (Be careful with rounding too early!) Tunisian Population: 45 MM 210 MN 245NN Tunisian population+ Sailor Population New Genotype Frequency: 495 MM 255 MN 250NN New Allele Frequency M=0.6225 N= 0.3775 2pq= 2 (0.6225)(0.3775)= 469.9 c.) In fact, 50 children have MM bloodtype, 850 have MN bloodtype and 100 have NN blood type. What is the observed allele frequency of the N allele among the children? Frequency of N (850+100+100)/ 2000= 0.525 = 52.5% Problem 7. a. What are the genotype frequencies in the population? C R C R = 100/1000=0.10 C R C W = 740/1000= 0.74 C W C W = 160/1000= 0.16 b. What are the allele frequencies of C
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Unformatted text preview: R and C W in this population? C R = 200+740 / 2000= 0.47 C W = 1- 0.47 = 0.53 c. what are the expected frequencies of the genotypes if the population is at Hardy- Weinberg equilibrium? P2 = (0.47) 2 = 0.22 C R C R Q2= (0.53) 2 = 0.28 C W C W 2PQ= 2 (0.47)(0.53) = 0.50 C R C W Problem 8. Assuming that the population is at Hardy-Weinberg equilibrium, what proportion of the males in this population will also have the condition? X s = dominant allele for condition. X= recessive, doesn’t have the condition. Females with X s _ have the condition and 36% are X s X s and X s X genotypes. Thus, 64% is XX. Since 64% is XX (p 2 ), the X (p) allele frequency is 0.8. Using p + q =1, X s (q) allele frequency is 0.2 For male population: Since male has only one X, the genotype frequencies are equal to the allele frequencies. . Thus males with X s Y have the condition and there are 20% since the X s allele frequency is 0.2....
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