All answers must start with the equation that was used to answer the question, a

Question

All answers must start with the equation that was used to answer the question, and all work must be shown.

6. In a diploid population where the homozygous recessive phenotype occurs at a frequency of 0.04 and has a relative fitness 90% that of the dominant phenotype, what is the mean fitness?

6b. How does the frequency of the recessive allele change as a function of purifying selection?

7. In a sample of 48 bats, 18 were observed to be heterozygous. If allele frequencies are 0.65 and 0.35, is the population in HWE?          

Explanation / Answer

Answer 6.

The homozygous recessive phenotype is same as the homozygous recessive genotype.Therefore,the genotype of the homozygous recessive(aa) will be 0.04 that is q2. = 0.04, q=0.02

now we know, p+q=1(p=allele frequency of dominant allele,(A) and q=allele frequency of recessive allele(a)

So, p=1-q = 1-0.02 = 0.08

Therefore p2= 0.64 that is p2 = the genotype frequency of the dominant homozygote(AA).

genotype frequency of the dominant heterozygote(Aa)= 2pq= 0.32

homozygous recessive phenotype has a relative fitness 90% that of the dominant phenotype.(given)

Relative fitness of homozygous recessive phenotype,Waa=0.9

Relative fitness of homozygous dominant phenotype,WAA=0.1

Relative fitness of heterozygous dominant phenotype,WAa=0.1

Therefore mean fitness Wmean = (p2WAA)+ (2pqWAa)+ (q2Waa)

Wmean=(0.64 X 0.1)+(0.32 X 0.1)+(0.04 X 0.9) =0.132.

Answer 6b.

Let the frequency of the recessive allele (a) after one gen. of selection be q'
q'= q2 Waa/Wmean + pqWAa/Wmean

q' = (0.04 X 0.9/0.132 )+(0.16 X 0.1/0.132) =0.278+0.121=0.399

Answer 7.

Total population of bats=48

Number of heterozygote=18

heterozygote genotype frequency(Aa) 2pq=18/48

Therefore number of homozygotes = 48-18 =30

homozygote genotype frequency (AA + aa) p2+q2 = 30/48

If the population is in HWE, then p2+2pq+q2 =1

Therefore, p2+q2 +2pq=(18/48) +(30/48) =48/48 = 1

You can also show the same by using p2 =0.65 and q2 = 0.35,then also the above equation will be equal to 1.

So the population is in HWE.

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