In humans, red blood cells have a number of proteins embedded in the cell membrane. One type of protein, the Rh factor, is controlled by a single gene and is either present or missing from the red blood cells. If present, the individual has the Rh+ phenotype. If missing, the individual has the Rh− phenotype. Rh+ is the dominant to Rh−. Suppose that, in the Yoruban population, the frequency of the Rh− phenotype is 0.01.0.01. Using the Hardy–Weinberg equations, calculate the frequency of the Rh+ allele to at least two decimal places.

the answer would be that the frequency of the allele Rh + = 0.6

Explanation:

the frequency of the Rh value is incorrect, the value must be expressed in an integer that can have decimals but not a meaningless numerical sequence(0.01.0.01?), however, an exercise is performed taking the frequency value 0.16 as an example for the development of subsequent

we can start from the formula taking into account that the population is in Hardy-Weinberg equilibrium

 p2 + 2pq + q2 = 1.0

the above is based on the fact that

p2 is considered as the dominant homozygous ratio

q2 as the proportion of recessive homozygotes

2pq as the proportion of heterozygous individuals

knowing that the Rh phenotype is recessive and has a value of 0.16, we replace obtaining

q2 = 0.16 or q = 0.4

where solving the equation

p + q = 1

p = 1- 0.4 = 0.6

the answer would be that the frequency of the allele Rh + = 0.6

from the original order (top to bottom) the new order will go

1 (founder)

4 (hawk)

2 (mutate)

3 (next generation)