<span>Polydactyly in its most common form, is an autosomal dominant mutation. Let's represent the mutation as P and the normal state as +. Thus, if both parents are heterozygous (P+), both would have the mutation (and thus have six fingered). They have 3/4 chances of having a child with the extra digit, 1/4 normal. </span>
Answer:
1. Allele frequency of b = 0.09 (or 9%)
2. Allele frequency of B = 0.91 (0.91%)
3. Genotype frequency of BB = 0.8281 (or 82.81%)
4. Genotype frequency of Bb = 0.1638 (or 16.38%)
Explanation:
Given that:
p = the frequency of the dominant allele (represented here by B) = 0.91
q = the frequency of the recessive allele (represented here by b) = 0.09
For a population in genetic equilibrium:
p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)
(p + q)^2 = 1
Therefore:
p^2 + 2pq + q^2 = 1
in which:
p^2 = frequency of BB (homozygous dominant)
2pq = frequency of Bb (heterozygous)
q^2 = frequency of bb (homozygous recessive)
p^2 = 0.91^2 = 0.8281
2pq = 2(0.91)(0.9) = 0.1638
Answer:
X linked inheritance is one of the mechanisms of the inheritance which is associated with the mutation of the gene present on the X chromosome.
The characteristics of the X linked recessive inheritance that must be observed in the pedigree analysis will be-
1. X linked trait is always passed from the mother to the sons (females to males)
2. The males (XY) are usually affected by the gene as only one copy of the recessive gene in males can show its effect.
3. Females are the carrier of the disease but they are not affected by the condition.
