** ANSWERS — POPULATION GENETICS PROBLEMS**

1) A study on blood types in a population found the following genotypic distribution among the people sampled: 1101 were MM, 1496 were MN and 503 were NN. Calculate the allele frequencies of M and N, the expected numbers of the three genotypic classes (assuming random mating). Using X2, determine whether or not this population is in Hardy-Weinberg equilibrium.

**GENOTYPE FREQUENCIES**:

MM (p2) = 1101/3100 = 0.356

MN (2pq) = 1496/3100 = 0.482

NN (q2) = 503/3100 = 0.162

**ALLELE FREQUENCIES**:

Freq of M = p = p2 + 1/2 (2pq) = 0.356 + 1/2 (0.482) = 0.356 + 0.241 = 0.597

Freq of N = q = 1-p = 1 – 0.597 = 0.403.

**EXPECTED GENOTYPE FREQUENCIES (assuming Hardy-Weinberg):**

MM (p2) = (0.597)2 = 0.357

MN (2pq) = 2 (0.597)(0.403) = 0.481

NN (q2) = (0.403)2 = 0.162

**EXPECTED NUMBER OF INDIVIDUALS of EACH GENOTYPE:**

# MM = 0.357 X 3100 = 1107

# MN = 0.481 X 3100 = 1491

# NN = 0.162 X 3100 =502

**CHI – SQUARE (X2):**

X2 = Σ(O – E)2 / E

X2 = (1101-1107)2 /1107 + (1496-1491)2 /1491 + (502-503)2 /503

= (-6)2 /1107 + (5)2 /1491 + (-1)2 /503

= 0.0325 + 0.0168 + 0.002

= 0.0513

X2 (calculated) < X2 (table) [3.841, 1 df, 0.05 ls].

Therefore, conclude that there is no statistically significant difference between what you observed and what you expected under Hardy-Weinberg. That is, you fail to reject the null hypothesis and conclude that the population is in HWE.

2) A scientist has studied the amount of polymorphism in the alleles controlling the enzyme Lactate Dehydrogenase (LDH) in a species of minnow. From one population, 1000 individuals were sampled. The scientist found the following fequencies of genotypes: AA = .080, Aa = .280; aa = .640. From these data calculate the allele frequencies of the “A” and “a” alleles in this population. Use the appropriate statistical test to help you decide whether or not this population was in Hardy-Weinberg equilibrium.

**Genotype frequencies:**

Freq AA = p2 = 0.08

Freq Aa = 2pq = 0.28

Freq aa =q2 = 0.64

**Allele frequencies:**

p = Freq A = 0.08 + 1/2 (0.28) = 0.08 + 0.14 = 0.22

q = Freq a = 1 – 0.22 = 0.78

** IF population is in HWE, then you’d expect the following frequencies:**

p2 = (0.22)2 = 0.0484

2pq = 2(0.22)(0.78) = 0.3432

q2 = (0.78)2 = 0.6084

Genotype |
Expected Numbers |
Observed Numbers |

AA | 0.0484 X 1000 = 48.4 | 0.080 X 1000 = 80 |

Aa | 0.3432 X 1000 = 343.2 | 0.280 X 1000 = 280 |

aa | 0.6084 X 1000 = 608.4 | 0.640 X 1000 = 640 |

X2 = [(80 – 48.4)2/ 48.4] + [(280 – 343.2)2 / 343.2] + [(640 – 608.4)2/ 608.4]

= 20.63 +11.64+1.64

= 33.91

X2 (Calculated) > X2 (table), therefore reject null hypothesis. Not in HWE.

3)The compound phenylthiocarbamide(PTC)tastes very bitter to most persons. The inability to taste PTC is controlled by a single recessive gene. In the American white population, about 70% can taste PTC while 30% cannot (are non-tasters). Estimate the frequencies of the Taster (T) and nontaster (t) alleles in this population as well as the frequencies of the diploid genotypes.

Estimated Freq t = q =square root of q2=square root of 0.30 = 0.5477

Freq T = p = 1 – q = 1 – 0.5477 = 0.4523

Expect under HWE:

TT = p2 = (0.4523)2 = 0.2045

Tt = 2pq = 2(0.4523)(0.5477) = 0.4956

tt = q2 = (0.5477)2 = 0.2999

4) In another study of human blood groups, it was found that among a population of 400 individuals,230 were Rh+ and 170 were Rh-.. Assuming that this trait (i.e., being Rh+) is controlled by a dominant allele (D), calculate the allele frequencies of D and d. How many of the Rh+ individuals would be expected to be heterozygous?

Number of dd individuals = 170, therefore the frequency of the genotype dd (q2) is 170/400 = 0.425. From this, we can estimate q as:

q = square root of q2 = square root of 0.425 = 0.652.

The allele frequency of D is:

p = 1-q = 1- 0.6519 = 0.348.

**Assuming HWE, genotype frequencies are as follows:**

DD = p2 = 0.121

Dd = 2pq = 0.454

dd = q2 = 0.425

Using the expected genotype frequencies, the number Dd among the Rh+ individuals is:

= 2pq (400)

0.454 (400) = 182

5) Phenylketonuria is a severe form of mental retardation due to a rare autosomal recessive allele. About 1 in 10,000 newborn Caucasians are affected with the disease. Calculate the frequency of carriers (i.e., heterozygotes).

Given the above, estimate q from q2

q = square root of q2 =square root of 1/10,000 = square root of 0.0001 = 0.01

Therefore, p = 1 – q = 1 – 0.01 = 0.99

** Using Hardy-Weinberg Law, calculate the expected number of individuals of each genotype as:**

DD = p2 = 0.9810

Dd = 2pq = 0.0198

dd = q2 = 0.0001

Therefore, 1.98% of the population is expected to be carriers.

6) For a human blood, there are two alleles (called S and s) and three distinct phenotypes that can be identified by means of the appropriate reagents. The following data was taken from people in Britain. Among the 1000 people sampled, the following genotype frequencies were observed SS = 99, Ss = 418 and ss = 483. Calculate the frequency of S and s in this population and carry out a X2 test. Is there any reason to reject the hypothesis of Hardy-Weinberg proportions in this population?

** Observed Genotype frequencies:**

SS = p2 = 99/1000 = 0.099

Ss = 2pq = 418/1000 = 0.418

ss = q2 = 483/1000 = 0.483

** Allele frequencies**:

Frequency of S = p = p2 + 1/2 (2pq) = 0.099 + 1/2 (0.418) = 0.308

Frequency of s = q = 1 – p = 1 – 0.308 = 0.692.

** Expected Genotype frequencies:**

SS = p2 = (0.308)2 = 0.095

Ss = 2pq = 2 (0.308)(0.692) = 0.426

ss = q2 = (0.692)2 = 0.479

** Expected number of individuals:**

SS = 0.095 (1000) = 95

Ss = 0.426 (1000) = 426

ss = 0.479 (1000) = 479

X2 = (99-95)2 /95 + (418-426)2 /426 + (483-479)2 /479

= 16/95 + 64/426 + 16/479

= 0.168 + 0.150 + 0.033

= 0.351

X2 (calculated) < X2 (table) [3.841, 1 df at 0.05 ls).

Therefore, fail to reject null hypothesis and conclude that the population is in HWE.

7) A botanist is investigating a population of plants whose petal color is controlled by a single gene whose two alleles (B & B1) are codominant. She finds 170 plants that are homozygous brown, 340 plants that are homozygous purple and 21 plants whose petals are purple-brown. Is this population in HWE (don’t forget to do the proper statistical test)? Calculate “F” (inbreeding coefficient) and explain what is happening in this population.

** Genotype frequencies:**

Freq. of brown (BB) = p2 = 170/531 = 0.32

Freq. of purple-brown (B1B) = 2pq = 21/531 = 0.04

Freq. of purple (B1 B1) = q2 = 340/531 = 0.64.

** Allele frequencies:**

Freq of B = p = p2 + 1/2 (2pq) = 0.32 + 1/2(0.04)

= 0.32 + 0.02

= 0.34.

Freq of B1 = q = 1- p = 1 – 0.34

= 0.66

** Expected Genotype Frequencies:**

BB = p2 = (0.34) 2 = 0.1156

B1B = 2 pq = 2 (0.34)(0.66) = 0.4488

B1 B1 = q2 = (0.66) 2 = 0.4356

** Expected Numbers: **

BB = 0.1156 X 531 = 61.4

B1B = 0.4488 X 531 = 238.3

B1 B1 = 0.4356 X 531 = 231.3

X2 = (170-61.4)2 /61.4 + (21-238.3)2 /238.3 + (340-231.3)2 /231.3

= 441.3185

X2 (Calculated) > X2 (table), therefore reject null hypothesis. Not in HWE.

F = (Ho – H)/ Ho

= (0.4488 – 0.04)/ 0.4488

= 0.91

Inbreeding.