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In a population of plants, the allele for long stems (S) is completely dominant over the allele for short stems (s). If 35% of the population has short stems, calculate the percentage of the population that is expected to be heterozygous (Ss). 48 percent 59 percent 41 percent 17 percent

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Answer:

The correct answer would be 48 percent.

It can be solved by using the Hardy-Weinberg equation.

p² + 2pq+ q² = 1

p + q = 1

where, p² represents homozygous dominant's frequency that is, SS in this case,

q² represents homozygous recessive's frequency that is, ss in this case,

2pq represents heterozygous dominant's frequency that is, Ss in this case.

p represents dominant allele's frequency and q represents the recessive allele's frequency.

Now, the percentage of homozygous recessive or short stems is 35%. Therefore, the frequency of homozygous recessive would be 0.35.

Thus, the value of q² = 0.35

So, q =
√(0.35) = 0.59 which is approximately equals to 0.6.

As p + q = 1 so, q = 1 - 0.6 = 0.4

The frequency of heterozygous (Ss) = 2pq = 2(0.6)(0.4) = 0.48

Hence, the frequency of heterozygous would be 48 percent.

User Dave Olson
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