1-38 Population Genetics Flashcards
population
in the context of genetics, a breeding group or gene pool
gene frequency
how often an allele for a gene appears in population.
an array of frequencies describes the genetic constitution of the population.
genotype
the allele combo for a locus that yields a particular characteristic.
all genotype frequencies for a particular locus must add up to 1.
all genotype frequencies for a single locus must
add up to 1
hardy weinberg law and definition of symbols
p^2 + 2pq + q^2 = 1
p = frequency of recessive allele q = frequency of dominant allele p^2 = frequency of aa q^2 = frequency of AA 2pq = remainder of punnet square, the frequency of carriers
hardy weinberg assumptions
- population large
- mating is random with respect to the locus in question
- allele frequencies remain constant over time (mutation rate low, immigrants have same genetic makeup)
Heterozygote frequency From Homozygote Frequency
can be calculated from p+q=1
Balance Between Mutation Rate (μ) and Selection (s)
while mutation rate allows for steady introduction of new alleles, selection simultaneously removes disadvantageous existing alleles from the system
natural selection has a harder time removing disease causing alleles as..
the disease becomes rarer. the frequency of carriers will decline, but the ratio of carriers to those affected will increase.
Reasons for an increased autosomal recessive disease frequency? (4)
- founder effect and drift
- heterozygous advantage
- elevated mutation rate
- genome instability
Spouse has a sister with CF, how likely is it that your children will have CF?
her risk of carrying = 2/3
your risk of carrying= population risk
x .25
genotypes must add up to
1
P+H+Q = 1
Q =
q^2, people with 2 disease causing alleles
hardy weinberg assumption frequently made
2pq»»>q^2
for disease causing alleles, expect far more heterozygotes than homozygotes (more carriers than afflicted)
we also assume p is about 1
frequency of CF in caucasions is one in 2,500 births. What is a person’s carrier risk?
q^2 = 1/2,500
q=sqrt(1/2500)=1/50
p = 1-q = 49/50
H = 2pq = 2 (1/50)(49/50) = about 1/25
**49/50 is about 1
wife has a brother with CF. You are from a population birth incidence is 1/360,000
chance baby will have CF?
her risk of carrying is 2/3 (unaffected sibling of affected)
your risk of carrying is the population risk =
q^2 = 1/360,000
q = sqrt(1/360,000) = 1/600
p is about 1
2pq = 21(1/600) = 2/600 = 1/300
2/3 * 2pq x 1/4
2/3 *1,300 * 1/4 = 1/1,800
we assume p
is about 1
the ratio of carriers to affected inviduals will ______ as the frequency of the disease decreases
increases
assuming hardy weinberg for an X-linked recessive trait. derive q directly from frequency of male births.
1-q is p
carrier females will be 2pq
females have 2 X’s, only effected when both X’s are bad (q^2)
males only have a single x, so the disease incidence of x-linked recessives in males is q
q = recessive allele frequency = male birth frequency
x-linked recessive traits. derive female heterzyogus frequency from the male birth rate (1/5,000)
derive affected female frequency
ratio of carrier emales to affected males?
ratio of affected males to affected females?
because x-linked recessive
q= male birth rates = 1/5,000
2pq = carrier females
=assume p is 1 (again)
21(1,500) = 1/2500 for carrier female
q^2 would be affected females
(1/2500)^2 = 1/25,000,000
2:1
5,000:1
in small populations the allele frequency..
becomes fixed, never in HW equilibrium
fitness definition and equation
f = (1-s) s= coefficient of selection
fitness = the probability of transmitting enes to the next gen and of the survival in that generation to be pased on to the next in relation to the average probabilty for the population
dominant mutations tend to be…
milder because the alleles cannot hide and are faced with selection
tay sachs may be more resistant to
TB
equation for new mutation rate
new mutation rate = n/2N
n = number of affected patients born to unaffected parents
N = total number of births
20 children with an autosomal dominant lethal disease were born in a series of 1,000,000 newborn children. the new mutation rate would be
m = n/2N
m = 20/2(1,000,000) = .00001 m = 10^-5
Mutation rate changes will alter allele frequencies. (TF)
True
A LOD score (Z) of -3 at a recombination distance (θ) of 5 cM is good evidence that the the gene and marker are linked.
False
If the LOD score is high (Positive LOD), it means that the traits are closely linked, and therefore usually inherited together. Low scores (Negative LOD), on the other hand, indicate a low linkage.
What is the new mutation rate for an autosomal dominant disorder (genetic lethal) that results in 40 new cases amongst 2,000,000 newborns?
1x10^-5
Your pregnant patient, Diana, is from an island in the Atlantic Ocean that has been relatively isolated for many generations and has a population in Hardy-Weinberg equilibrium for an autosomal recessive trait, “stiff-upper-lip” (SUL) which has a birth incidence of 1/1600. Neither Diana or her husband, Chuck, have SUL. Chuck is from a different population with an incidence of 1/400 births for SUL. What is the probability that their child will be affected with SUL?
q2 = 1/1600, q = 1/40, 2pq ~ 1/20; q2 = 1/400, q = 1/20, 2pq ~ 1/10; 1/20 x 1/10 x 1 / 4 (Chance both pass it on) = 1/800
Your patient, Jane, is from a population in which an autosomal recessive disorder is in Hardy-Weinberg equilibrium and has a birth incidence of 1/16,000,000. Like Jane, her husband Roger, is phenotypically normal, but on taking a family history you find that Roger’s maternal uncle was affected with this disorder. Since Jane is pregnant, what do you tell her is the risk to the fetus?
q2 = 1/16,000,000, q = 1/ 4,000, 2pq ~ 1/ 2,000 (Jane’s carrier risk); Roger’s mother’s risk of being a carrier is 2/3 (because she is the unaffected sibling of an affected individual), so Roger’s carrier risk is 1/3; Risk to fetus is 1/3 x 1/2,000 x 1/ 4 = 1/ 24,000
Your patient, John, has a brother, Peter, who has an autosomal recessive disorder seen in 1/ 40,000 births. John’s wife, Sophie, has a first cousin with this disorder. How likely is it that their first child would also be affected?
John’s carrier risk is 2/3, his wife’s carrier risk is 1/ 4 and there is a 1/ 4 risk they would both pass on their mutations: 1/24.
Now assume that John and Sophie had an affected child. John remarries, and his second wife, Rachel, is from a population with a birth incidence of this disorder of 1/ 400. How likely is it that their first child will be affected?
We now know John is a carrier. Rachel has a population carrier risk. q squared = 1/400; q = 1/20; 2pq ~ 1/10; 1 x 1/10 x 1/ 4 = 1/40
your first cousin has a rare autosomal recessive mutation. What are the chances you carry it?
1/4.
One of your grandparents carried it
your mother had a 1/2 chance of carrying
you have a 1/4 chance of carrying
