Genetics (non-Hersh) Flashcards
A couple is considering pregnancy and the male partner had bilateral retinoblastoma. What is the risk for a future child to develop retinoblastoma? Assume there is 80% penetrance.
40% risk. 0.5 x 0.8 = 0.40.
50% chance that child will inherit the gene for Rb
What is the chance that 2 carrier parents will have an unaffected child? What about an unaffected carrier? Chance of having an affected child?
affected child = 1/4
unaffected child = 3/4
unaffected carrier = 1/2
Unaffected sibling of an individual with an autosomal recessive condition. What is their risk of being a carrier?
2/3
1st degree relatives share how much of their genetic information?
1/2
How to calculate risk of having a child affected with an autosomal recessive disorder
mother’s carrier risk x father’s carrier risk x 1/4
Hardy-Weinberg equilibrium
p + q = 1. P and q represent proportions of the population genes (what proportion are common/wild v. disease?)
- p typically refers to more common or wild-type allel
- q typically refers to the disease allele
Hardy-Weinberg equilibrium in a 2 allele system.
p2 + 2pq + q2 = 1 Represents frequency of genotypes in a population. Equation that shows likelihood of inheriting the p or q allele from one parent AND the p or q allele from the other parent. p2 = unaffected homozygotes 2pq = carrier heterozygotes q2 = affected homozygotes
If 1 out of every 250,000 people have disease X, a non-lethal autosomal recessive disorder, what is the approximate carrier frequency of this disease? (Assume Hardy-Weinberg equilibrium)
q2 = incidence of the disease q2 = 1 in 250,000** q = 1/500 p = 1 - q = 499/500 = ~1 carrier frequency = 2pq = 2 x (1/500) x 1 = 1/250
*short cut to taking square root of big #s = square root of # in front and cut the number of zeros in half
What is 2pq represent in the Hardy-Weinberg equation? p2 and q2? What about just p and q?
p and q = frequency of alleles. Whereas p2 and q2 = frequency of genotypes:
p2 = incidence of normal (wild-type) homozygotes
q2 = incidence of disease
2pq = carrier frequencyp = frequency of normal (wild-type) allele
q = frequency of disease allele
For rare, recessive conditions. What can the carrier frequency be estimated to?
carrier frequency = 2q. Since the condition is rare, p (the frequency of the normal allele) can be estimated to ~1, thus 2pq (the carrier frequency) = 2q
For the ABO blood group, the allele frequencies are:
A = 0.30 (p)
B = 0.10 (q)
O = 0.60 (r)
What percent of the population should be blood group A?
Total genotype frequency = p2 + q2 + r2 + 2pq + 2pr + 2qr = 1
Phenotype A can be A/A (p2) or A/O (2pr), thus incidence of A blood grp =
p2 + 2pr = (0.3)^2 + 2(0.30)(0.60) = 0.45
Incidence of PKU is 1 in 10,000. If the father has a brother with PKU, what is the risk for the pregnancy to be affected?
Father’s carrier risk = 2/3. For mother (population)…
q2 = 1/10,000
q = 1/100
2pq = 1/50
So mother’s carrier risk = 1/50. Overall risk for pregnancy to be affected = 2/3 x 1/50 x 1/4 = 1/300
When doing problems of X-linked inheritance, what should you pay attention to in the question stem?
Note whether the question is asking for the risk to have an affected CHILD or the risk to have an affected SON. If son, need to multiply the risk of inheriting the allele by 1/2 (risk for child to be boy)
How does the Hardy-Weinberg equation differ between males and females when it comes to X-linked inheritance?
For males, you are only looking at p and q (since they only have 1 X): p = unaffected HEMIzygotes q = affected hemizygotes For females: p2 = unaffected homozygotes 2pq = carrier heterozygotes q2 = affected homozygotes
The incidence of an X-linked trait is 1/40. What is the frequency of female unaffected carriers? How many females will be affected?
incidence of trait in MALES = q = 1/40 (freq of disease gene)
frequency of female carriers = 2pq = 2(1/40)(39/40) =~1/20
freq of affected FEMALES = q2 = (1/40)^2 = 1/1600
