Question

1. You are studying a population of sandblossoms (Linanthus parryae) that has individuals with blue and white flowers. The allele for white flowers (A) is dominant to the allele for blue flowers (a). In the population you survey, 91 out of 100 individuals have white flowers. Based on this information:

a. Calculate the frequency of the A and a alleles.

b. Calculate the numbers of each genotype.

2. A population of snapdragons (Antirrhinum hispanicum) has two additive alleles for flower color: R (red) and r (white). In a sample of 200 individuals you find 100 red individuals, 25 pink individuals, and 75 white individuals. In terms of the Hardy-Weinberg theorem, the 25 pink individuals in this population represents:

- p

- q

- p^2

- 2pq

- 2pqn

- q^2

3. At the SMH locus in a population of Drosophila melanogaster you find two alleles, S and S-, with a frequency of 0.516 and 0.484, respectively in a sample of 128 individuals.

a. Calculate the expected numbers of each genotype using Hardy-Weinberg.

b. After genotyping all individuals at the SMH locus, you find the following number of individuals with each genotype: 42 SS, 48 SS-, and 38 S-S-. Is this population in Hardy-Weinberg equilibrium (give the X² test statistic you calculate)?

c. Name a specific mechanism that could produce the discrepancy between expected and observed genotype numbers you calculated in b.

4. In a study of peppered moths (Biston betularia) with two alleles for color (B = black and b = white, with B dominant to b), 87% of individuals were black.

a. What is the frequency of the bb genotype?

b. What is the frequency of the b allele (assuming Hardy-Weinberg equilibrium)?

c. What is the frequency of the B allele (assuming HWE)?

5. Why is it so hard for natural selection to fix an advantageous allele with complete dominance?

6. A survey through 23andMe.com found that, of 4737 individuals of European ancestry, 3002 said they could smell asparagus in their urine and 1735 said they could not. If the A allele for odor detection is dominant over the G allele for lack of odor detection, and 1027 individuals are heterozygotes:

a. What are the allele frequencies for this locus assuming random mating?

b. What are the genotype frequencies?

c. Is the population in Hardy-Weinberg equilibrium?

7. Why do dominant alleles tend to be advantageous and recessive alleles tend to be disadvantageous in most populations?

8. You are studying a population of parrots and you observe that there are two alleles for wing color. Allele 1 codes for green wings and is dominant over allele 2, which codes for red wings. Allele 2 is at a frequency of 70% of the population, what should be the frequency of allele 1 homozygotes?

9. You are studying a population of chupacabras that has two alleles for spine length (L for long and S for short, with L dominant to S). In a population you surveyed, you find 25% of individuals with the short-spine phenotype. If you were to consider this information in the context of the Hardy-Weinberg theorm, the 25% you have been given represents:

-q

-2pq

-q^2

-2qn

-q^2*n

10. Name one form of selection that can generate genotypic diversity.

Answer 1

Answer-

According to the given question-

Question 1-

We have a population of sandblossoms also called Linanthus parryae having blue and white flowers.

where white flowers is dominant over blue flowers. we have total 100 population out of which 91 individuals have white flowers-

we have to calculate frequency of the A and a alleles as well as numbers of each genotype.

we know that p2 + 2pq + q2 = 1

and we also know that p + q = 1

where p = allele frequency of dominant trait

q = allele frequency of recessive trait

p2 = number of homozygous dominant

q2 = number of homozygous Recessive

2pq = number of heterozygous population

so considering question -

91 are white out of 100 are white

Frequency of individuals = individuals / Total population= 91 / 100 = 0.91

so p2 = 0.91

so p  = 0.95 and we know that p + q = 1

so 0.95 + q = 1

q = 1 - 0.95 = 0.05

hare Frequency of dominant allele (A) i.e. p = 0.95 and Frequency of recessive allele (a) i.e. q = 0.05

numbers of AA genotype= p2 * total number of individuals= (0.95)2   * 100 = 0.9025 * 100 = 90

while number of aa genotype = q2 * total number of individuals= (0.05)2 * 100 = 0.0025 * 100 = 0.25

Question 2-

in this question a population of snapdragons also called Antirrhinum hispanicum has two additive alleles for flower color such as

:red is denoted by R

white ius denoted by r

out of  200 individuals there are 100 red individuals, 25 pink individuals, while 75 white individuals. Hardy-Weinberg theorem, the 25 pink individuals in this population represents:

so here Red is dominant over the white which is recessive

Considering Hardy-Weinberg theorem

we know that p2 + 2pq + q2 = 1

p2 = number of homozygous dominant

q2 = number of homozygous Recessive

2pq = number of heterozygous population

so according to the given information -

number of homozygous dominant = p2 = 100

number of homozygous Recessive = q2 = 75 so rest will be heterozygous i.e. 2pq which is = 25

so here the pink are heterozygous and denoted by 2pq = 25

Question 3-

at SMH locus in a population of Drosophila melanogaster we have fing two alleles, which is S and S- with frequency of 0.516 and 0.484  respectively out of total 128 individuals.

(a) we have to calculate the numbers of each genotype

here frequency of p = 0.516 and frequency of q = 0.484 and total population = 128

numbers of SS genotype = p2  * total number of population

= (0.516)2 * 128 = 0.266 * 128 = 34

numbers of SS genotype = 34

numbers of S-S- genotype = q2 * total number of population = (0.484)2 *128 = 0.234 *128 = 30

Number os S-S- genotype = 30 and numbers of SS genotype = 34

(b). here we have onserved fequency of individuals such as -

number of SS genotype = 42

number of SS- genotype = 48

number of  S-S- genotype = 38 .

we have to find X2 test statistic in  Hardy-Weinberg equilibrium -

Expected number of SS genotype = 34

Expected number of S - S- genotype = 30

Expected number of SS- genotype i.e. heterozygous will be = 2pq * total number of population

= 2 * 0.516 * 0.484 * 128 = 64

Expected number of SS- genotype = 64

Genotype	Observed frequency (O)	Expected frequency (E)	O- E	(O-E)2	(O-E)2/ E
Homozygous dominant SS	42	34	42 - 34 = 12	144	144 / 34 = 4.2
Homozygous recessive S-S-	38	30	38 - 30 = 8	64	64 / 30 =2.13
Heterozygous SS-	48	64	48 - 64 = - 16	256	256 / 64= 4
	Total = 128	Total = 128			X2 = 4.2 +2.13 +4 = 10.33

Chi square statics X2 = 10.33

(c) here we have 3 categories but have estimated one parameter so degree of freedom = 1

We know that for 1 degree of freedom the critical valve of X2  at probability 0.5 is 3.84 . so our calculated Chi square statics X2  is larger than the critical valve of X2  i.e. 10.33 > 3.84 so we reject the null hypothesis and we can say that the population is in Hardy-Weinberg equilibrium .

Question 4-

peppered moths also called Biston betularia have two alleles for color

B = black and b = white, where the black colour is dominant over white and 87% of individuals were black.

here 87% of individuals were black which menas that 87 out of total 100 are black .

Frequency of individuals = individuals / Total population= 87 / 100 = 0.87

so frequency of black p2 = 0.93 or 93 %

so p  = 0.93 and we know that p + q = 1

so 0.93 + q = 1

q = 1 - 0.93 = 0.07 or 7 %

frequency of the bb genotype = 0.07 *0.07 = 0.0049

frequency of the b allele = 0.07 or 7 %

frequency of the B allele = 0.93 or 93 %  

According to the Chegg guidelines i am only suppose to answer first four question . so please post the remaining question separately if you want the answer. thanks.