Question

Stubble bristles (Sb) is dominant to wildtype bristles (Sb⁺) in Drosophila melanogaster. In a population where there are 137 flies with wildtype bristles and 836 flies with stubble bristles, what are the allele frequencies for the Sb and Sb⁺ alleles (assuming the population is in Hardy-Weinberg equilibrium)? Please calculated your answers to 3 decimal places.

f(Sb)=

f(Sb⁺)=

If the population is in Hardy-Weinberg equilibrium, what proportion of the population would you expect to be heterozygous?

Answer 1

Here, the wildtype bristles are the dominant trait.
So, 'sb+' is the dominant allele responsible for the wildtype bristles and 'sb' is the corresponding recessive allele responsible for the Stubble bristles.

Then, the following are the possible genotypes and corresponding phenotypes -
sb+ sb+ = wildtype bristles
sb+ sb = wildtype bristles
sb sb = stubble bristles
Now, let 'p' be the frequency of the dominant allele and 'q' be the frequency of the recessive allele,
Also, 'p2' is the frequency of the homozygous dominant individuals (sb+ sb+)
'q2' is the frequency of the homozygous recessive individuals (sb sb)
and '2pq' is the frequency of heterozygous individuals (sb+ sb).

Here, p2 + 2pq = 137/973
and q2 = 836/973 = 0.859
So, q = square root of 0.859 = 0.927

Now, according to the hardy weinberg equilibrium, we know that - p + q = 1.
So, p = 1 -q = 1 - 0.927 = 0.073

Hence, p = 0.927 and q = 0.073
So, the frequency of the sb+ allele = 0.073
Frequency of the sb alele = 0.927

Now, the frequency of the heterozygous individuals having the genotype sb+ sb =
= 2*p*q
= 2 * 0.927 * 0.073
= 0.135

So, the numbers of the heterozygous individuals expected = 0.135*973 = 132 (approx.)

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