1 p QUESTION 5 The remaining questions for the quiz will be based off this examp

Question

1 p QUESTION 5 The remaining questions for the quiz will be based off this example. In order to answer everything work out the problem on paper first, then answer the following the observed total population (n-2000) have these genotypes. Determine the expected frequencies of each genotype using Hardy-Weinberg. Round to 4 decimal places, do not round up Total 1000 1000 AS 275 . 25 700 587 Children 319 94 Adults Fill in the blanks. Adult's p- 587 1 pa QUESTION 6 (Round to 4 decimal Fillin the blanks. Children's p 7o0 places, do not round up)

Explanation / Answer

Adults: AA = 587

P2= 587/1000 = 0.587

Therefore P = 0.766

Expected frequency =

                AS = 319

SS = 94

q2 = 94/1000 = 0.094

q =0.306

2pq = 2*0.766*0.306*1000 = 468.792

So the expected frequency of AA = 1000*0.766 = 766

Expected frequency of SS = 0.306*1000 = 306

Expected AS = 2pq = 468.792

Children:

AA = 700

P2= 700/1000 = 0.7

Therefore P = 0.836

Expected frequency of AA = 0.836*1000 = 836

q2= 25/1000= 0.025

q = 0.158

Expected frequency of SS = 0.158*1000 = 158

Expected AS = 2pq = 2*0.836*0.158 = 264.176

Q7. The expected frequency of adult AS = 468.792

Q8. Expected no. of SS children = 158

Q9. Children with AA genotype = 836

Q10. Since children show deviation and adults do not, this indicates that selection is not present and this population adheres to Hardy Weinberg principles is the correct option as seen from above calculations.

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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