QUESTION 1 A survey of a SNP site segregating for nucleotides C and G in a chimp

Question

QUESTION 1

A survey of a SNP site segregating for nucleotides C and G in a chimpanzee population found the following numbers for different genotypes:

CC 46

CG 68

GG 29

Assuming that the locus is in HWE, what is the expected number of CC genotypes in the next generation?

QUESTION 2

A survey of a SNP site segregating for nucleotides C and G in a chimpanzee population found the following numbers for different genotypes:

CC 46

CG 68

GG 29

Use a chi-square test to determine if this population in HWE.

What is the chi-square value?

QUESTION 3

If a specific gene has 9 different co-dominant alleles in a population, how many degrees of freedom are there when doing a chi-square test?

QUESTION 4

If H=.30, p=0.35 and q=0.65, calculate the fixation index (Fis).

QUESTION 5

You are investigating 5 different loci in 3 populations of polar bears. You run an exact test on your microsatellite data and get the following results.Which locus/loci are not in HWE in all 3 populations?

Population 1:

Locus 1: p=0.02

Locus 2: p=0.53

Locus 3: p=0.04

Locus 4: p=0.79

Locus 5: p=1.0

Population 2:

Locus 1: p=0.01

Locus 2: p=0.78

Locus 3: p=0.06

Locus 4: p=0.92

Locus 5: p=0.15

Population 3:

Locus 1: p=0.03

Locus 2: p=0.59

Locus 3: p=0.01

Locus 4: p=0.29

Locus 5: p=0.07

Locus 1

Locus 2, 4, and 5

Locus 1 and 3

Locus 5

QUESTION 6

If H=.30, p=0.35 and q=0.65, is there a deficiency or excess of heterozygotes in the population?

Excess

Deficiency

Can't tell with this information

QUESTION 7

If Fis=0.35, what sample size is needed to detect this effect at the p=0.05 level? Assume the locus is biallelic (2 alleles).

QUESTION 8

If p=0.32 and q=0.68 in a 2 allele system, calculate Nei's heterozygosity.

QUESTION 9

Locus 1

Locus 2, 4, and 5

Locus 1 and 3

Locus 5

Explanation / Answer

Question 1

GENOTYPE FREQUENCIES:
CC (p2) = 46/143 = 0.321
GC (2pq) = 68/143 = 0.475
GG (q2) = 29/143 = 0.202

ALLELE FREQUENCIES:
Freq of C = p = p2 + 1/2 (2pq) = 0.321 + 1/2 (0.475) = 0.321 + 0.238 = 0.559
Freq of G = q = 1-p = 1 - 0.559 = 0.441.

EXPECTED GENOTYPE FREQUENCIES (assuming Hardy-Weinberg):
CC (p2) = (0.559)2 = 0.312
CG (2pq) = 2 (0.559)( 0.441) = 0.493
GG (q2) = (0.441)2 = 0.194

EXPECTED NUMBER OF INDIVIDUALS of CC GENOTYPE is:
CC = 0.312 X 143 = 44
CG = .493 * 143 = 70
GG = .194 * 143 = 27

Question 2
chi-square value
X2 = ?(O - E)2 / E
X2 = (46-44)2 /44 + (68-70)2 /70 + (29-27)2 /27
4/44 + 4/70 + 4/27
.909 + .057 + .148 = 1.114

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