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2 answers:
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To calculate the frequency of the heterozygote genotype (Pq) for this gene we must use the Hardy-Weinberg equation ( p2 + 2pq + q2 = 1 ). This equation relies on the Hardy-Weinberg principle, a model in population genetics that states that the frequency of the alleles in a population is never changing, only the combinations (the genotypes) are changing.
If there are only two alleles (variations) of this gene in a population, then their frequencies should add up to 1 (100%). From this, we can calculate the frequency of the q allele.
p +q=1
0,3 +q=1
q= 1-0,3
q= 0,7
Now hat we have the frequency of the q allele we can use the HW equation to calculate the frequency of the heterozygotes.
0,09 + 2pq +0.49= 1
2pq +0,58= 1
2pq= 1-0.58
2pq=0,42
The freqency of the heterozygotes in this population is 0.42
Answer:
1/2
Explanation:
This question involves a single gene coding for feather color in chickens. The black allele (B) is incompletely dominant over the white allele (W). This means that the allele B will not mask the phenotypic expression of allele W, hence, a third intermediate blue phenotype (BW) will be produced.
According to the question, if two blue chickens are crossed i.e. BW × BW, the gametes B and W will be produced by each parent. Using these gametes in a punnet square (see attached image), the following offsprings will be produced:
BB, BW and WW in the ratio 1:2:1
BB is black, BW is blue, WW is white. Hence, the probability of producing a blue chicken from this cross is 1/2.
