Two alleles called T1 and T2 of gene T are surveyed using PC

Two alleles called T1 and T2 of gene ‘T’ are surveyed using PCR and gel electrophoresis of blood samples from turtles. The results show 16 turtles are T1T2, 28 are T1T1, and 7 are T2T2.

a. What are the observed genotype frequencies?

b. What are the allele frequencies?

c. What are the H-W expected genotype frequencies?

d. What number of turtles of each genotype would have been found if the population showed a very good fit to the H-W model?

Solution

Answer:

When two alleles called T1 and T2 of gene ‘T’ are surveyed using PCR and gel electrophoresis of blood samples from turtles, the results show 16 turtles are T1T2, 28 are T1T1, and 7 are T2T2. For that case:

a. Frequency of genotype (f) = Number of a particular genotype/Total number of genotypes

f(T1T1) = 28/51= 0.549

f(T1T2) = 16/51= 0.314

f(T2T2) = 7/51= 0.137

**f(T1T1) + f(T1T2) + f(T2T2) = 0.549 + 0.314 + 0.137 = 1

b. Frequency of the allele = Number of particular allele/total number of alleles

= (2 x number of particular allele homozygote) + (number of particular allele heterozygote)/(2 x total number of individuals)

f(T1) OR p = [2(28) + 16] / 2(51)

= 72/102 = 0.706

f(T2) OR q = [2(7) + 16] / 2(51)

= 30/102 = 0.294

**p+q = 0.706 + 0.294 = 1

c. According to the Hardy-Weinberg principle,

The expected frequency of T1T2 genotype is pq+pq = 2pq = 2(0.208) = 0.416

The expected frequency of T1T1 genotype is : p² = 0.498

The expected frequency of T2T2 genotype is : q² = 0.086

And p² + 2pq + q² = 0.416 + 0.498 + 0.086 = 1

d. If the population showed a very good fit to the H-W model, the number of turtles of each genotype would be calculated as:

Number of genotype (T1T1) = 0.498 x 51 = 25 about

Number of genotype (T1T2) = 0.416 x 51 = 21 about

Number of genotype (T2T2) = 0.086 x 51 = 5 about