Question

1. The occurrence of the NN blood group genotype in the US population is 1 in 400, consider NN as the homozygous recessive genotype in this population.

You sample 1,000 individuals from a large population for the MN blood group, which can easily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). They are typed accordingly:

BLOOD TYPE	GENOTYPE	NUMBER OF INDIVIDUALS	RESULTING FREQUENCY
M	MM	490	0.49
MN	MN	420	0.42
N	NN	90	0.09

2. Using the data provide above, calculate the following:
   1. The frequency of each allele in the population.
   2. Based on HWE what are the expected frequencies for the three genotypes
   3. Are the observed frequencies different than the expected frequencies?

_____________________________________________________________________

3. Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following.
   1. The frequency of the recessive allele in the population.
   2. The frequency of the dominant allele in the population.
   3. The percentage of heterozygous individuals (carriers) in the population.

4. A lethal recessive condition is responsible for the death of 1 out of every 20,000 babies born in South America.

a. What is the frequency of the recessive allele in the population?

b. What is the frequency of the dominant allele in the population?

c. What is the frequency of homozygous recessive individuals in the population?

d. How many people, out of 1,000 individuals in this population, would carry the recessive allele?

5.         In snapdragons, R is dominant for red flowers and r is recessive for white flowers. Snapdragons exhibit incomplete dominance. In a population of 100 snapdragons, 50 have pink flowers and 25 have white flowers.

a. What is the frequency of the dominant allele in the population?

b. What is the frequency of the recessive allele in the population?

c. What is the frequency of heterozygous individuals?

d. What is the frequency of homozygous recessive individuals in the population?

e. What is the frequency of homozygous dominant individuals in the population?

            ______________________________________________________________________

6. Andallusizian chickens have beautiful plumage that is governed by a single gene with two alleles, which we’ll call S and s. SS individuals are black. Individuals with ss are white, and heterozygous individuals show blue plumage.

A sample of a population of these moths recovered the following numbers of each phenotype: black (SS) = 1020; blue (Ss) = 300; white (ss) = 6.

Calculate the frequencies of each allele and each genotype

Use the allele frequencies you calculated above and the Hardy-Weinberg equations to calculate the expected genotype frequencies assuming Hardy-Weinberg equilibrium.

Use the genotype frequencies from the first part of this example to calculate the expected numbers of individuals for each genotype.

Use the expected numbers for each genotype to calculate a chi square for this problem. How do the expected numbers of individuals compare to what was actually observed?

Phenotype	Observed (o)	Expected (e)	(o-e)	(o-e)2	(o-e)2      e
Black
Blue
White
∑ (o-e)2       e

Answer 1

Answer:

1).

A).

Numbe of M alleles = 490*2+1/2(420*2) = 1400

Numbe of N alleles = 90*2+1/2(420*2) = 600

Total alleles = 2000

Frequency of M allele = 1400/2000= 0.7

Frequency of N (recessive) allele = 600/2000 = 0.3

B).

Expected genotype frequencies:

MM = 0.7*0.7 = 0.49

MN = 2*0.7*0.3 = 0.42

NN = 0.3*0.3 = 0.09

C).

Observed genotype frequencies:

MM = 490/1000 = 0.49

MN = 420/1000 = 0.42

NN = 90/1000 = 0.09

The observed and expected genotype frequencies are similar but not different.

2).

A).

Recessive phenotype frequency = 1/2500 = 0.0004

Recessive allele frequency = SQRT of 0.0004 = 0.02

B). As p+q=1

The frequency of dominant allele = 1 – 0.02 = 0.98

C).

The frequency of heterozygous individuals = 2*0.98*0.02 * 2500 = 98

3).

a).

Recessive phenotype frequency = 1/20000 = 0.00005

Recessive allele frequency = SQRT of 0.00005 = 0.0071

b). As p+q=1

The frequency of dominant allele = 1 – 0.0071 = 0.9929

c).

The frequency of homozygous recessive individuals in the population = 1/20000 = 0.00005

d).

The number of people, out of 1,000 individuals in this population, would carry the recessive allele = 0.0071 * 0.0071 * 1000 =0.0504

5).

Based on this information, red = 15, pink=50; white = 25

The frequency of recessive (white) phenotype = 25/100 = 0.25

The frequency of recessive allele = SQRT of 0.25 = 0.5

The frequency of dominant allele = 1-0.5 = 0.5

a). The frequency of the dominant allele in the population = 0.5

b).The frequency of the recessive allele in the population = 0.5

c). The frequency of heterozygous individuals = 2 * 0.5 * 0.5 * 100 = 50

d). The frequency of homozygous recessive individuals in the population = 0.5 * 0.5 * 100 = 25

e).The frequency of homozygous dominant individuals in the population= 0.5 * 0.5 * 100 = 25