Genetics-Population Studies Flashcards
Risk that II-4 is affected?
01/04/14
Risk that III-1 is affected? III-2?
III-1: 1/2. III-2: low (spontaneous mutation rate)
Risk that III-3 is affected in this X-linked disorder?
Since II-4 is an obligate heterozygote, there is 50% chance of transmitting X chromosome and 50% chance of being male. 1/2 * 1/2 = 1/4.
What is population genetics?
Study of allele frequency in a population and factors that influence the frequency
Hardy-Weinberg equilibrium “p”.
For an autosomal locus with two alleles (A and a), it is the frequency of allele A.
Hardy-Weinberg equilibrium “q”.
For an autosomal locus with two alleles (A and a), it is the frequency of allele a in the population.
Hardy-Weinberg equilibrium
p + q = 1
Hardy-Weinberg equation fo an autosomal locus with two alleles A and a?
p^2 + 2pq + q^2 = 1.
Cystic fibrosis (CF) is an autosomal recessive disorder. It affects 1/2,500 whites of northern and central European origin. What is the frequency of the mutant CF allele? What is the carrier frequency?
Mutant allele (aa) frequency: q^2 = 1/2,500. So sqrt(1/2,500) = 1/50. Carrier frequency (Aa = 2pq): p = 1-q = 49/50. 2pq = 2(49/50)(1/50) = 1/26
II-3 has phenylketonuria (PKU), an autosomal recessive disorder present in 1/10,000 children. What is the risk for III-1?
Risk of II-4 of being a carrier is 2/3. Risk of II-5 being a carrier is 2pq. q^2 = 1/10,000. q=sqrt1/10,000 = 1/100. p = 1 - q = 1 - 1/100 = 99/100. 2pq = 2(99/100)(1/100) = 1/50. The risk of both II-4 and II-5 being carriers is 2/3 * 1/50. The risk of III-1 getting the homozygous trait (2/3)(1/50)(1/4) = 1/300.
Alterations of Hardey-Weinberg calculations if autosomal dominant? If X-linked recessive?
AD = Homozygotes are rare and are excluded = 2pq. X-linked recessive = in affected males the disease frequency is = allele frequency and = q. In carrier females = 2pq.
What is the ratio of affected males to affected females in X-linked recessive disorders?
(q/q^2) = 1/q
Hemophilia A is an X-linked recessive disorder found in 1/5,000 males. What are the frequencies of carrier and affected females?
q = 1/5000, p = 1 - 1/5000. Carrier frequency = 2q = 2/5000 = 1/2500. Affected females = a^2 = (1/5,000)^2 = 1/25,000,000.
Hemophilia A is an X-linked recessive disorder found in 1/5,000 males. What is the ratio of affected males to affected females?
q/q^2 = 1/q = 5000.
Assumptions of Hardy-Weinberg?
Large population (in small populations you have founder effect & genetic drift), random mating (assertive mating and consanguinity), and no mutations (10^-5), migration (genetically links 2 separate populations) or selection (affects frequency of alleles).