Thanks II. After the chart is completely.filled out, for each trait, I would lik

Question

Thanks II. After the chart is completely.filled out, for each trait, I would like you to use the Hardy-Weinberg Equilibrium equation to figure out the p value, q value, and 2pq (carrier frequency) value for all of these traits. First, for each trait, you will tally up how many people showed the dominant phenotype and how many people showed the recessive phenotype. Based on the recessive phenotype () value, you can now solve for p. Using that number, you can solve for 2pq (heterozygotes). Use the lecture, the video on Hardy-Weinberg, and the book

Explanation / Answer

THEORY:

The Hardy-Weinberg equilibrium is a mathematical relationship of the alleles and genotypes in a population.
The relationships are as follow:

Alleles: p + q = 1

where p is frequency of dominant allele
and q is frequency of recessive allele

Genotypes: p2 + 2pq + q2 = 1

where p2 is frequency of homozygous dominant genotype
q2 is frequency of homozygous recessive genotype
2pq is frequency of heterozygous recessive genotype

---------------------------------------------------------------------------------------------------------------------------------------

Now let take first trait, tongue rolling

10 out of 10 person have the homozygous dominant genotype so by Hardy-Weinberg equation:

p2 = 10/10 = 1

p = 1

and q2 = 0/10 = 0

q = 0

and 2pq = 0

Result- frequency of dominant allele (p) will be 100%

no recessive allele and no carrier will be there in tongue case.

-------------------------------------------------------------------------------------------------------------------------------------

Now second trait is widow's peak

2 out of 10 persons have homozygous recessive genotype and 8 persons have homozygous dominant genotype so by Hardy-Weinberg equation:

q2 = 2/10 = 0.2

Take the square root of both sides

q = 0.44

This means that 44% of the alleles in the system are recessive alleles

for p use equation p + q = 1

p + 0.44 = 1

p = 0.56

This means that 56% of the alleles in the system are dominant alleles

now calculate frequency of carriers i.e. 2pq

2pq = 2 (0.56) (0.44 )

2pq = 0.49

Converting this into a percent, we see that 49% of the population is heterozygous.

----------------------------------------------------------------------------------------------------------------------------

rest all traits have no variation in type so-

By following method all kind of frequencies can be calculated.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.