Mendelian Genetics in Populations I: Selection and Mutation

Question 1

population genetics

Answer 1

• integrates theory of evolution + mendelian genetics
• changes in relative abundance of traits in population -> tied to changes in relative abundance of alleles that influence them
• provides theoretical foundation for much of modern understanding of evolution

Question 2

how can we measure genetic variation?

Answer 2

requires determination of genotypes at particular loci for individuals

(1) infer genotype based on phenotype
(2) examine proteins encoded by alleles
(3) examine the DNA itself

Question 3

ex. heterozygosity - measure of genetic variation

Answer 3

heterozygotes / # sampled individuals
AA = 25, AB = 50, BB = 25
Heterozygosity (H) = 500/100 = 0.5

Question 4

infer genotype based on phenotype
ex. intestinal schistosomiasis

Answer 4

susceptibility influenced by 1 gene
- AA-susceptible
- AB-moderate susceptibility
- BB-resistant

by determining susceptibility, one can infer genotype

Question 5

examine proteins encoded by alleles

Answer 5

measuring variation w electrophoresis

Question 6

protein electrophoresis

Answer 6

first method to determine an individual’s genotype and measure diversity of population at a locus

heterozygosity in natural populations - protein electrophoresis in 60s and 70s revealed substantial genetic variation

Question 7

examine DNA itself

Answer 7

PCR - polymerase chain reaction

polyacrylamide gel electrophoresis

Question 8

PCR

Answer 8

step 1: denaturation
- 1 minute 94 degrees C
step 2: annealing (forward/reverse primers)
- 45 sec 54 degrees C
step 3: extension (only dNTPs)
- 2 minutes 72 degrees C

exponential amplification
2^(n+1) = # copies, where n is # cycle

Question 9

polyacrylamide gel electrophoresis

Answer 9

cathod (-)
anode (+)

DNA moves through matrix (gel) vertially
DNA (-ive) runs from cathode -> anode
Speed factors: DNA fragment size (1) & charge (2)

Question 10

how to tell if it’s an allele (polyacrylamide gel)?

Answer 10

if the row has bands for the population
- heterozygote if 2 bands
- homozygote if 1 thick band

Question 11

allele frequencies from polyacrylamide gel

Answer 11

count up individuals
count up alleles

calculate:
ex. allele 1: homo 2x2 + hetero 7x1 = 11
formula: allele y: homo _x2 + hetero _x1 = ?

total of all allele COPIES should add up to # individual x 2 (diploid)

calculate (#allele copies)/(#total copies) = ex. allele 1 = 0.393

Question 12

calculate allele frequency if:
204 individuals
3 alleles, 6 genotypes

Genotypes:
AA 20
BB 30
CC 22
AB 49
AC 44
BC 39

Answer 12

F(A) = (2AA + AB + AC)/(2N)
= 2(20) + 49 + 44 / 2(204)
= 0.326

F(B) = (2BB + AB + BC)/(2N)
= 2(3) + 49 + 39 / 2(204)
= 0.363

F(C) = 1-0.326-0.363 = 0.311

Question 13

Hardy-Weinberg Principle

Answer 13

null principle
derived by G. Hardy and W. Weinberg
applies to all diploid populations
assumes idealized/panmictic population

Question 14

panmictic meaning

Answer 14

random mating
can be thought of as a pool of gametes/alleles

Question 15

Hardy-Weinberg assumptions

Answer 15

no selection
no mutation
no gene flow/dispersal/migration
no gene drift (large population size)
individuals choose mates randomly

Question 16

Hardy-Weinberg vs Evolution

Answer 16

H-W equilibrium
- no selection
- no mutation
- no gene flow
- large population size
- random mate slection

evolution
- selection
- mutation
- gene flow
- genetic drift
- non-random mate selection

Question 17

under assumptions, H-W principle can predict _____ ____ from ____ ____

Answer 17

genotype frequencies from allele frequencies

Question 18

H-W formula for genotype frequencies

Answer 18

p^2 + 2pq + q^2 = 1

Question 19

calculate expected genotype frequencies

diploid alleles, A(0.65) & B(0.35)

Answer 19

p = frequency A
q = frequency B

P(AA) = p^2 = 0.65^2 = 0.4225
P(AB) = 2pq = 2(0.65)(0.35) = 0.4550
P(BB) = q^2 = 0.35^2 = 0.1225

check = 0.4225 + 0.4550 + 0.1225 = 1

Question 20

is population evolving?

given: genotypes

Answer 20

calculate allele frequencies from genotype frequencies

then calculate expected genotype frequencies from actual allele frequencies

Question 21

does H-W hold true?

100 individuals
AA = 40, AB = 47, BB = 13

Answer 21

Step 1: calc allele frequencies
F(A) = 240 + 147 / 200 = 0.635
F(B) = 213 + 147 / 200 = 0.365

Step 2: calc expected genotype
p^2 + 2pq + q^2 OR A^2 + 2AB + B^2 = 1
AA = A^2 = 0.635^2 = 0.4032
AB = 2AB = 2(0.635)(0.365) = 0.4635
BB = B^2 = 0.365^2 = 0.1332

H-W holds true! (ignore rounding error)

Question 22

does H-W hold true?

100 individuals
AA = 37, AB = 19, BB = 44

Answer 22

Step 1: calc allele frequencies
F(A) = 237 + 119 / 200 = 0.465
F(B) = 244 + 119 / 200 = 0.535

Step 2: calc expected genotype
p^2 + 2pq + q^2 OR A^2 + 2AB + B^2 = 1
AA = A^2 = 0.465^2 = 0.216
AB = 2AB = 2(0.465)(0.535) = 0.498
BB = B^2 = 0.535^2 = 0.285

H-W does not holds true!