1-38 Population Genetics

Question 1

population

Answer 1

in the context of genetics, a breeding group or gene pool

Question 2

gene frequency

Answer 2

how often an allele for a gene appears in population.

an array of frequencies describes the genetic constitution of the population.

Question 3

genotype

Answer 3

the allele combo for a locus that yields a particular characteristic.

all genotype frequencies for a particular locus must add up to 1.

Question 4

all genotype frequencies for a single locus must

Answer 4

add up to 1

Question 5

hardy weinberg law and definition of symbols

Answer 5

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

Question 6

hardy weinberg assumptions

Answer 6

1. population large
2. mating is random with respect to the locus in question
3. allele frequencies remain constant over time (mutation rate low, immigrants have same genetic makeup)

Question 7

Heterozygote frequency From Homozygote Frequency

Answer 7

can be calculated from p+q=1

Question 8

Balance Between Mutation Rate (μ) and Selection (s)

Answer 8

while mutation rate allows for steady introduction of new alleles, selection simultaneously removes disadvantageous existing alleles from the system

Question 9

natural selection has a harder time removing disease causing alleles as..

Answer 9

the disease becomes rarer. the frequency of carriers will decline, but the ratio of carriers to those affected will increase.

Question 10

Reasons for an increased autosomal recessive disease frequency? (4)

Answer 10

• founder effect and drift
• heterozygous advantage
• elevated mutation rate
• genome instability

Question 11

Spouse has a sister with CF, how likely is it that your children will have CF?

Answer 11

her risk of carrying = 2/3
your risk of carrying= population risk

x .25

Question 12

genotypes must add up to

Answer 12

1

P+H+Q = 1

Question 13

Q =

Answer 13

q^2, people with 2 disease causing alleles

Question 14

hardy weinberg assumption frequently made

Answer 14

2pq»»>q^2

for disease causing alleles, expect far more heterozygotes than homozygotes (more carriers than afflicted)

we also assume p is about 1

Question 15

frequency of CF in caucasions is one in 2,500 births. What is a person’s carrier risk?

Answer 15

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

Question 16

wife has a brother with CF. You are from a population birth incidence is 1/360,000

chance baby will have CF?

Answer 16

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

Question 17

we assume p

Answer 17

is about 1

Question 18

the ratio of carriers to affected inviduals will ______ as the frequency of the disease decreases

Answer 18

increases

Question 19

assuming hardy weinberg for an X-linked recessive trait. derive q directly from frequency of male births.

Answer 19

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

Question 20

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?

Answer 20

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

Question 21

in small populations the allele frequency..

Answer 21

becomes fixed, never in HW equilibrium

Question 22

fitness definition and equation

Answer 22

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

Question 23

dominant mutations tend to be…

Answer 23

milder because the alleles cannot hide and are faced with selection

Question 24

tay sachs may be more resistant to

Answer 24

TB

Question 25

equation for new mutation rate

Answer 25

new mutation rate = n/2N

n = number of affected patients born to unaffected parents

N = total number of births

Question 26

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

Answer 26

m = n/2N

m = 20/2(1,000,000) = .00001
m = 10^-5

Question 27

Mutation rate changes will alter allele frequencies. (TF)

Answer 27

True

Question 28

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.

Answer 28

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.

Question 29

What is the new mutation rate for an autosomal dominant disorder (genetic lethal) that results in 40 new cases amongst 2,000,000 newborns?

Answer 29

1x10^-5

Question 30

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?

Answer 30

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

Question 31

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?

Answer 31

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

Question 32

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?

Answer 32

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.

Question 33

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?

Answer 33

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

Question 34

your first cousin has a rare autosomal recessive mutation. What are the chances you carry it?

Answer 34

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