46. In a group of 50 students, 18 determined no flavor in the PTC paper (non-tas

Question

46. In a group of 50 students, 18 determined no flavor in the PTC paper (non-tasters) and 32 were tasters. The dominant (T) trait is tasting and the recessive (t) is non-tasting. If we assume the population is in Hardy-Weinberg equilibrium, solve for the following: What is 2pq, the frequency of the heterozygous individuals?
A. 50 B. 18 C. .48 D. .16 E. .36 I picked C .48

47.One Central American baby out of every twenty thousand is born with a recessive condition that is lethal. How many people, out of 1,000 individuals in this population, would carry the recessive allele? A. 5 b. 10 c.12 d. 14 e. 22 I picked A 5

48.When looking at snapdragons, the dominant flowers are red (R) and the recessive flowers are white (r). This plant displays incomplete dominance since heterozygotes have flowers that are pink. If we look at population of 100 snapdragons with 50 having have pink flowers and 25 having white flowers, what is the frequency of homozygous dominant individuals in the population?
a. 0.5

b. 0.30

c. 1

d. 0.75

e. 0.25 I picked E .25

49. You have calculated that a certain population has a homozygous recessive genotype (aa) at 36%. Use that 36% to determine the frequency of the "A" allele.

a. 40%
b. 36%
c. 60%
d. 48%
e. 16%
I picked E 16%

Explanation / Answer

Alright so I don't really know if you want me to show my work or not or just agree with you but...
#46 is correct. 2pq= 2(.6)(.4)= .48
#47 ain't. So they tell ya that 1 baby out of 20,000 is born w/ the condition.

So q2=1/20000= 5 x 10-5 and 5x10-5 = 0.00707 leaving p to be 1-q=.99293

Carriers are gunna be the population that is 2pq or (2)(.00707)(.99293)=.014 or 1.4% of a population

Therefore in 1000 people population 1000(.014)= 14 peeps making d) the right answer

#48 is right. p=.5 so p2=.52= .25(100)=25 flowers that are homozygous dominant

#49 isn't right because you got confused on what they were asking for. It's all good they make these problems to try and get ya like that.

So you figured everything out right just answered the wrong question. You got q=.6 and p=.4. But the question you answered was how many are homozygous dominant which is 16%. However they are asking for the allele frequency, not a genomic frequency. Therefore they asked for the A allele frequency = p = .4 x 100 = 40%. So a) is the right answer.

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

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