Answer:
d. pxp +2pq
Explanation:
The formula for genotype frequency for a population in Hardy-Weinburg equilibrium is as under:
p² + 2pq + q² = 1
where, p = dominant allele
q = recessive allele
Here,
p² represents frequency of homozygous dominant genotype
2pq represents frequency of heterozygous genotype
q² represents frequency of homozygous recessive genotype
Also, although the genotypes p² & 2pq are different from each other yet phenotypically they both will collectively produce dominant trait i.e. free ear lobes not attached earlobes. So the term "p² + 2pq or pxp + 2pq" represents the frequency of the individuals who show the dominant phenotype in this particular population. Dominant phenotype will comprise 75% of the population.
Which of these is an example of an achieved status? d. husband
I think the answer should be 16 because 1×2^4=16
Answer:
.
Explanation:
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