Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Pad 5:02 PM Assignment 10: Hardy-Weinberg Detail Grade Assignment 10: Hardy-Wein

ID: 3165121 • Letter: P

Question

Pad 5:02 PM Assignment 10: Hardy-Weinberg Detail Grade Assignment 10: Hardy-Weinberg Due: Apr 15, 2018 at 11:59 PM Watch the Lightboard lecture, examine the figures, and answer the following questions Let us determine the genetic makeup of a population for PTC tasting identical to that of another laboratory section. To determine the makeup of the population you must first deterinine the frequency of M and m in your lab section. You were supplied with a small piece of filter paper impregnated with PTC that you placed on your tongue; if you do not taste a very bitter taste you are a non-taster. With your instructor, you tallied the number of individuals in your lab that are non- tasters (mm) Next, you determined what fraction of the entire class is non-taster. This fraction is the frequency of the genotype mm, or q2. q would then equal the square root of this fraction. p would equal one minus this square root, because p +q. By substituting the values of p and q into the Hardy-Weinberg equation you can determine the probable distribution of genotypes in your lab section: p2+2pq + Compute the Alelle and Genotype frequencies of a Sample of Individuals Number of non-tasters 15 Total number of individuals 20 Fraction of students who are non- asters q (frequency of mm) Dashboard Calendar To Do Inbox

Explanation / Answer

Fraction of non-taster = 15 /20

                                           = 3 / 4

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

Frequency of non-taster (mm) (q2) = 0.75

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

If q2 = 0.75, then q = 0.866 (Frequency of m allele)

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

p + q = 1

p + 0.866 = 1

p = 0.134 (Frequency of M allele)

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

p2 = 0.017 (Frequency of MM genotype)

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

p = 0.134, q = 0.866,

Then, 2pq = 0.232 (Frequency of Mm genotype)

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

P2 + 2pq + q2 = 1

0.017 + 0.232 + 0.75 = 0.999

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