Population Genetics and Personalized Medicine Flashcards
given the follow genotype frequencies, what is the allele frequency of a?
AA = 0.41
Aa = 0.51
aa = 0.08
assume population of 100
AA = 41
Aa = 51
aa = 8
total allele # = 200
allele a # = (51 x 1) + (8 x 2) = 67
allele frequency of a = 67 / 200
*don’t forget to find the proportion out of total alleles after calculating total count of allele of interest
what are the 4 criteria for Hardy-Weinberg Law to apply?
- large population
- random mating
- allele frequencies same in M/F
- negligible mutation, selection, migration
what are the two Hardy-Weinberg equations
p = A (common allele)
q = a (minor allele, disease-producing)
p + q = 1
(p + q)^2 = p^2 +2pq + q^2
Cystic fibrosis is an AR disease with a 1/2500 incidence among a population of Europeans. What is the predicted incident of het. carriers?
1/2500 = q^2 (genotypic frequency)
so q = 1/50 = 0.02
[stop and actually think about this math!]
p + 0.02 = 1, so p = 0.98
(p + q)^2 = p^2 +2pq + q^2
het. carriers = 2pq = 2(0.98)(0.02) = 0.04 = 4/100 = 1/25
the incidence of phenylketonuria is 1/10,000 in the US. What is the predicted incidence of het. carriers?
q^2 = 1/10,000
sqrt. of 10,000 is 100
so q = 1/100 = 0.01
p then = 0.99
(p + q)^2 = p^2 +2pq + q^2
het. carriers = 2pq = 2(0.99)(0.01)
use these clues to determine the probability of a cystic fibrosis affected child:
-grandparents are both carriers
-mother is unaffected
-father is unaffected
-father is Italian. frequency of CF in Italian-Americans is 1/2500
-child is of unknown sex
- probability that mother is a carrier is 2/3 (AR disorder, she is unaffected, both parents are carriers)
1b. probability mother passes mutant allele IF she is a carrier = 1/2 - use H-W to determine probability that father is a carrier:
q^2 = 1/2500
so q = 1/50 = 0.02
so p essentially = 1
so 2pq = (2)(0.02)(1) = 0.04 = 1/25
2b. probability that father passes mutant allele IF he is a carrier = 1/2
- total probability =
(2/3)(1/2) x (1/25)(1/2) = 2/300 = 1/150
[child sex doesn’t matter for autosomal disorders]
using these clues, determine probability of affected child:
-frequency of AR disease in population is 1%
-heterozygous carrier male mates with unaffected woman
- probability male passes mutant allele is 1/2
- use H-W to find woman’s probability of being a carrier
q^2 = 0.01, so q = 0.1
p then = 0.9
2pq = 2(0.1)(0.9) = 0.18
- probability woman passes mutant allele IF she is a carrier = 1/2
- total probability =
(0.5) x (0.18)(0.5)
use these clues to determine probability of affected child:
-frequency of AR disease in general population is 1%
-het. carrier male mates with woman from general population
*if disease freq. is high, risk of disease needs to be ADDED in
- prob. male passes mutant allele = 0.5
- prob. of woman HAVING disease = 0.01
2b. if woman has disease (AR), prob. of passing variant allele = 1 - use H-W to find prob. woman is a carrier
q^2 = 0.01, so q = 0.1
p = 0.9
2pq = 2(0.9)(0.1) = 0.18
- prob. of women passing variant allele IF she is a CARRIER = 0.5
- total prob. =
risk if woman is a carrier PLUS risk if she has disease
[(0.5) x (0.18)(0.5)] + [(0.01)(1) x (0.5)]
remember to include father’s 50% chance of passing allele in BOTH scenarios
given an AD disorder with a frequency of 1/500 in the US, what is the mutant allele freq. in this population?
- with AD disorders, you assume affected individuals are heterozygotes (phenotype of homozygous mutation is usually very severe)
so 2pq = 1/500
- also assume that p^2 (two normal alleles) is very close to 1, and that there aren’t really any q^2
so p^2 = 499/500 ~ 1
- fill in the variable
2pq = 2(1)(q) = 1/500
q = 1/1000 = 0.001
what are the Hardy-Weinberg rules for X-linked recessive disorders? (how will genotype frequencies differ in sex)
M (single X) -
disease frequency = allele frequency
[genotypes or p or q]
F -
allele frequency in males = allele frequency in females
you assume females are all heterozygous**
[genotypes are p2, 2pq, q2]
if the frequency of an X-linked recessive disorder is 1/10 in males, what is the allele frequency in males, and in females?
in males, disease freq. = allele freq. —> allele freq. of mutation is 1/10 in males
females:
allele freq = q^2 = (1/10)^2 = 1/100
how is consanguinity denoted in pedigrees
double line
(mating between individuals who are second cousins or closer —> increases risk of recessive disorders)
how does genetic drift affect allelic frequencies?
alternations due to random events, independent of genotype
for example, when an AD disorder is passed to more (or less) than 50% of children -> overtime instances like this can increase/decrease allele freq. in a population (esp. if population is small)
Founder effect
a subgroup of a population breaks off and starts their own population. Introduction of new mutation/individual carrying variant allele into isolated population can result in high incidence of a disorder that is rare elsewhere
what is the equation for heritability
h^2 = (variance in DZ pairs - variance in MZ pairs)
———————————————————-
variance in DZ pairs
DZ = dizygotic, MZ = monozygotic (identical)
0 = trait NOT due to heritability (numerator = 0)
1 = trait COMPLETELY due to heritability (variance in MZ = 0)
