Question

A certain gene occurs in the population in two allelic forms, a dominant A and a...

A certain gene occurs in the population in two allelic forms, a dominant A and a recessive a . The relative frequencies of these alleles in the population are 0.8 and 0,2 respectively.

a) Assuming a panmictic population, what is the probability that an individual will have a homozygous pair of alleles?

b) What is the probability that an individual will have a heterozygous pair of alleles?

c) Only the presence of homozygous "aa" is associated with an increased Type 2 Diabetes in people over 40. The joint probability that an individual is homozygous "aa" and develops Type 2 Diabetes after 40 is 0.03. The marginal probability that an individual will develop Type 2 Diabetes after 40 is 0.05. Construct a table showing the joint probabilities of the development or non-development Type 2 Diabetes and the presence or absence of homozygous "aa", and the marginal probabilities associated with the presence/absence of Type 2 Diabetes and the presence/absence of "aa".

d) Based on the table in (c), calculate the conditional probability that an individual who is homozygous "aa" will develop Type 2 Diabetes.

Homework Answers

Answer #1

Answer to part a)

Panmitic population implies that there is random pairing , there is no restriction on pairing

Homogenous implies: AA or aa

Probability = 0.8*0.8 + 0.2*0.2

Probability = 0.64 +0.04 = 0.68

.

Answer to part b)

Probability of Heterozygous pairs = P(Aa) + P(aA)

Probability = 0.8 *0.2 + 0.2*0.8

Probability = 0.16 +0.16 = 0.32

.

Answer to part c)

P(aa AND Type 2) = 0.03

P(Type 2) = 0.05

P(aa) = 0.2*0.2 = 0.04

Using these values we get the following table:

.

Answer to part d)

The conditional probability formula is:

P(A | B) = P(A and B) / P(B)

P(type II | aa) = P(type II and aa) / P( aa)

P( type II | aa) = 0.03 / 0.04 = 0.75

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