Nonrandom Mating Historically the royal families of Europe were known to be very
ID: 169360 • Letter: N
Question
Nonrandom Mating Historically the royal families of Europe were known to be very inbred doe to generations of related marriages (often for strengthening political alliances). King Charles II (of Spain in 1665) likely suffered from two rare genetic disorders - combined pituitary hormone deficiency and distal renal tubular acidosis. He also appeared to be infertile (left no heir). Given his genealogy we can estimate his inbreeding coefficient (F). Here first estimate the frequency of homozygotes for a rare trait in a randomly mating population. Then estimate the frequency of homozygotes with inbreeding at the level of King Charles II. Keep 6 significant digits after the decimal. This rare trait occurs in a population at frequency of q = 0.04 Consider a random mating population of 15,000 individuals. If expression of a disease associated with this trait is only expressed in homozygous individuals (A2A2), how many will have this disease in this population? Now consider individuals as inbred as King Charles (F = 0.254). If the trait occurred In royal families at q = 0.04, using the inbreeding equation (page 278) in a population size of N = 15,000, determine how many would have this disease? A thought question - In a very large randomly mating population would you expect rare deleterious traits to be retained or not?Explanation / Answer
Inbreeding is the process of mating of closely related individuals and it tends to increase the number of homozygous individuals. Inbreeding is common in royal families in Europe. King Charles II suffered from two rare genetic disorders and his inbreeding coefficient was F = 0.254.
(3) (a) Here, we have to consider a random mating population.
Total number of individuals = 15,000
The frequency of the recessive allele = q = 0.04
The disease associated with the triat is only expressed in dominanthomozygous individuals (A2A2)
Let, p be the frequency of the dominant allele.
According to Hardy-Weinberg principle,
p+q = 1
p = 1-q
p = 1 - 0.04
p = 0.96
Frequency of homozygous dominant triat = p2 = (0.96 x 0.96) = 0.9216
Number of individuals with the disease = (0.9216 x 15,000) = 13,824 = 0.138240 x 105 (6 significant digits after the decimal).
(3) (b) Here, we have to consider inbreeding of individuals.
F = 0.254, q = 0.04, p = 0.96 , N = 15,000
f(A2A2) = ?
According to inbreeding equations,
f(A2A2) = (1-F) (p2) + (F) (p) (1) = p2 + Fp (1-p) = p2 + Fpq
f(A2A2) = (0.96 x 0.96) + (0.254) (0.96) (0.04) = 0.9216 + 0.0097536 = 0.931353
Number of individuals with the disease = ( 0.931353 x 15,000) = 13,970 = 0.139700 x 105 (6 significant figures after the decimal)
(3) (c) During random mating, the number of diseased individuals decreases as compared to the number of diseased individuals during inbreeding. The disease is expressed in dominant homozygous individuals. For this reason, in very large randomly mating population, rare deleterious traits would be retained but would not be expressed much because the number of diseased individuals would decrease.
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