2) In a certain human population, the percentage of individuals that have sickle

Question

2) In a certain human population, the percentage of individuals that have sickle cell anemia, a homozygous recessive disorder that increases the curvature of red blood cells, is 8%. Given a population of 420 individuals from this population, calculate the following:

a) The expected number of individuals that have sickle cell anemia.

b) The expected allelic frequency of the normal blood cell curvature trait.

c) The expected number of individuals that are homozygous dominant for normal blood cell curvature.

d) The expected number of individuals that are heterozygous in this population.

e) Suppose the population has 35 people that are homozygous recessive for the sickle cell trait, 69 people that are heterozygous for the trait, and 316 people that are homozygous dominant for normal blood cell curvature. Is this population in Hardy-Weinberg Equilibrium? Use a Chi-Square Test to confirm.

Explanation / Answer

Total Population = 420
% population having sickle cell anemia = 8%
a)
Expected number of individuals that have sickle cell anemia = 0.08 x 420
= 33.6
b)
Since it's a homozygous recessive disorder, the person affected by it has identical alleles on both homologous chromosomes i.e xx
Those with XX, Xx, xX, do not have this trait.
Expected allelic frequency of the normal blood cell curvature triat :
3/4 = 0.75 or 75%
c)
Probability of random person to be homozygous dominant: 1/4 = 0.25 (Since only XX is homozygous dominant out of XX, xX, Xx, xx)
Expected number of individual that are homozygous dominant for normal blood cell curvature = 0.25 x 420 = 105
d)
Probability of random individual being heterozygous = 2/4 = 0.5 (xX and Xx)
Expected number of individuals that are heterozygous in this population =
- 0.5 x 420 = 210

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