Population genetics example problem set. Please answer the following questions b
ID: 140431 • Letter: P
Question
Population genetics example problem set. Please answer the following questions before class. 1. Two alleles (81 and 82) are present in a population at a frequency of B1-0.6 (a) What is the expected frequency of heterozygous individuals in the population? (b) After further investigation, a third allele (B3) is discovered in the same population such that the new allele frequencies are B1-0.6 and B3-0.02. What is the new expected frequency of heterozygous individuals in the population? 2 One individual in 100 is observed to have the phenotype that is associated with the homozygous recessive genotype for a single locus trait in a population. (a) What is the frequency of the allele that confers this phenotype in the homozygous recessive condition?Explanation / Answer
1 (a). The sum of frequency of two alleles is 1
B1 + B2 =1
B2 = 1-0.6 = 0.4
frequency of B2 allele is 0.4
The frequency of heterozygotes is 2 pq ie 2 x frequency of B1 x frequency of B2 = 2 x 0.6 x 0.4 = 0.48
1 (b) If there are three allele B1, B2 and B3 . let p represent the frequency of B1, q represent the frequency of B2, and, r represent the frequency of B3. Thn p+q+r =1
p is 0.6 , r is 0..02
0.6 + q + 0.02 =1
q = 1 - 0.62 = 0.38
Frequency of B2 allele is 0.38
The possible heterozygotes can be B1B2, B1B3, B2B3
(p + q+ r)2 = 1
p2 + q2 + r2 + 2pq + 2pr + 2 qr =1
The frequency of B1B2 is 2pq = 2 x 0.6 x 0.38 = 0.456
The frequency of B1B3 is 2pr =2 x 0.6 x 0.02 = 0.024
The frequency of B2B3 is 2 qr = 2 x 0.38 x 0.02 = 0.0152
The expected frequencies of heterozygotes is 0.456, 0.024 and 0.0152
2. One out of 100 individuals have the homozygous recessive genotype. lets the recessive allele by the letter a. Humans are diploid i.e they have two allele for each gene. Therefore the total number of allele of a particular locus (gene) are 100x 2 =200.
The homozygous recessive individual contains 2 allele for the recessive trait.
Assuming no heterozygotes exist in the population
Frequency of allele a = 2 x number of individuals homozygous for allele a / 2 x total number of individual
Frequency of allele a = 2 x 1 / 2 x100 = 2/200 = 0.01
The frequency of allele which confers the phenotype in homozygous recessive condition is 0.01