What is the contour interval or C.I.?
2 answers:
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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 = = 0.59 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.
Answer:
1/4
Explanation:
Cystic fibrosis is a recessive disorder and therefore, is expressed only in homozygous recessive genotype. Let's assume that the allele "c" is responsible for the disease in the homozygous state. Since the couple already has a child with cystic fibrosis (cc), they both are heterozygous carriers for the disease.
The genotype of male and female each is "Cc". A cross between Cc x Cc gives progeny in the following phenotype ratio=1/4 diseased: 1/2 normal carriers: 1/4 normal. Therefore, the probability of getting a diseased child for them is 1/4.
