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

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 = = 0.59 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.

easy

explanation: