Question

A gene locus has two alleles aa. If the frequency of dominant allele a is 0.4, then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the population?

• A. 0.36(AA); 0.48(Aa); 0.16(aa)
• B. 0.16(AA); 0.24(Aa); 0.36(aa)
• C. 0.16(AA); 0.48(Aa); 0.36(aa)
• D. 0.16(AA); 0.36(Aa); 0.48(aa)

Answer 1

The correct option is C

0.16(AA); 0.48(Aa); 0.36(aa)

The correct option is C

Explanation of correct option

Step 1: Given data

Frequency of dominant allele (p) = 0.4

Step 2: Applying the formulae

Formulae used-

Since, (p+q)=1;

where p =dominant allele, q= recessive allele.

Thus, q=(1-p)

2.p² + 2pq + q² = 1

where, p² = homozygous dominant individual

2pq = heterozygous dominant individual

q² = homozygous recessive individual

Step 3: Solving equations to get required value

Frequency of recessive allele:

(q)=1- 0.4 =0.6

Frequency of homozygous dominant individuals:

(AA) = P² =( 0.4)( 0.4) =0.16

Frequency of heterozygous dominant individuals:

(Aa) = 2pq = 2(0.4)(0.6) = 0.48

Frequency of homozygous recessive individuals:

(aa) = q2 = (0.6) (0.6) = 0.36

Final answer:

The correct option is (c) i.e., 0.16(AA); 0.48(Aa); 0.36(aa)

Frequency of homozygous dominant individual = 0.16

Frequency of heterozygous dominant individual =0.48

Frequency of homozygous recessive individual=0.36.