In population genetics we often calculate the probability of

In population genetics we often calculate the probability of a zygote’s genotype, by treating alleles contributed by each parent as random draws from the same pool of alleles (this is the random mating assumption). A particular locus has a common allele (T) at frequency of 90% and one other allele (t). Assume that the organism is diploid and the gene is autosomal (in other words, a zygote’s genotype consists of two copies of the gene).
a) What is the probability that a zygote will have at least on T allele ?
b) What is the probability that mother and father contribute the same allele to the zygote (in other words, what is the chance that the zygote will be homozygous at this locus)?

Solution

We have frequency of T = 90%

i.e. p = 0.90

Similarly frequency of \'t\' = 100 - 90 = 10%

Thus, q = 0.10

All possible combinations of genotypes are TT, Tt, tT, tt.

Where frequency of TT = p² = 0.90² = 0.81

Frequency of tt = q² = 0.10² = 0.01

Frequency of Tt or tT = 2pq = 2*0.90*0.10 = 0.18

Thus, frequency of Tt = frequency of tT = 0.18/2 = 0.09

Following is the probability distribution:

a) probability that a zygote will have atleast one T allelle.

P(T>=1) = P(T = 1) + P(T = 2)

P(T = 1) = frequency of Tt + frequency of tT = 0.09 + 0.09 = 0.18

P(T = 2) = frequency of TT = 0.81

Thus,

P(T>=1) = 0.18 + 0.81 = 0.99

Therefore, probability that a zygote will have atleast one T allelle is 0.99

b) probability that a zygote will have both allelle of same type.

Required probability = P(TT) + P(tt)

                                = frequency of TT + frequency of tt = 0.81 + 0.01 = 0.82

Thus probability that a zygote will be homozygous is 0.82