PLEASE HELP ASAP
Would really appreciate it but would understand if nobody replies.
Thanks in advance.

1.

Frequency of dominant allele = p=0.5196

Frequency of recessive allele= q=0.4804

% homozygote dominant= p^2=27%

% homozygote recessive= q^2=23%

% heterozygotes = 2pq= 50%

2.

p=0.7

q= 0.3

p^2=49%

q^2=9%

2pq= 42%

3.

p=0.409

q= 0.591

p^2=16.7%

q^2= 35%

2pq= 48.3%

4.

p=0.99

q= 0.01

p^2=98.01%

q^2=  0.01%

2pq= 1.98%

Explanation:

1. The percentage of homozygous dominant organism is expressed as p^2 and its value in this question is 27%. Using the information, we can find the frequency of the dominant allele (p)

p^2= 0.27

p= √0.27= 0.5196

After we get dominant allele frequency we can find recessive allele frequency

p+q=1

q= 1- 0.5196

q= 0.4804

With recessive allele frequency and dominant allele frequency, we can find the heterozygote and homozygote recessive.

homozygote recessive= q^2= (0.4804)^2= 0.2307= 23%

heterozygote =  2pq= 2(0.4804)(0.5196)=0.499= 50%

2.

Only an organism with homozygote recessive genotype will show symptoms and signs of sickle-cell disease, so 9% is the percentage of q^2.

q^2= 0.09

q= √0.09= 0.3

p+q=1

p=1-q

p=1-0.3= 0.7

p^2= (0.7)^2= 0.49= 49%

2pq= 2(0.7)(0.3)= 0.42 = 42%

3.

The 35% population of the moth is homozygote recessive (q^2). We can use the same step as problem 2 then

q^2= 0.35

q=√0.35= 0.591

p+q=1

p=1-q

p=1-0.591=  0.409

p^2= (0.409)^2=0.1672= 16.7%

2pq= 2(0.409)(0.591)= 0.4834 = 48.3%

p=0.409

q= 0.591

p2=16.7%

q2= 35%

2pq= 48.3%

4.

Cystic fibrosis is a homozygous recessive disease with a population rate of 1/10,000. That means the percentage of the population with q^2 gene will be: 1/10000= 0.01%

Then, the rest will use same step as question 2

q^2= 0.0001

q=√0.0001= 0.01

p+q=1

p=1-q

p=1-0.01=  0.99

p^2= (0.99)^2=0.9801= 98.01%

2pq= 2(0.99)(0.01)= 0.0198 = 1.98%

it would be the percentage of food eaten by the flock

i remember doing this lab

≅200-130 milions years ago pangea splited in 2 continents