Frequency problem practice

1. Frequency problem practice

2. For the following problems: • Copy the answer on your notebook and explain the steps labeled with a number: why do we need to do that? How do we come up with this step? • 把答案抄在你的笔记本上并且解释有数字标记的每一步为什么要这样做，这一步是怎么得出来的。

3. Problem 1 • Tay-Sachs disease is caused by a recessive allele. The frequency of this allele is 0.1 in a population of 3,600 people. What is the frequency of the dominant allele, and how many of the 3600 people will be heterozygous for the condition?

4. Answer: • Ans: since the frequency is recessive, then: q2= 0.1 q = 0.3162 (1) Because p + q = 1: (2) • p = 1-q = 1 – 0.3162 = 0.6838 • 2pq = 2x0.6838x0.3162 = 0.4324 (3) • The frequency of heterozygous carrier will be: 0.4324 • 3600 x 0.4324 = 1557 (4) • So: 1557 people may carry heterozygous alleles. • Frequency of the dominant allele will be: 1 – 0.1 = 0.9 (5)

5. Problem 2 • In a population of 2,000 earthworms, there is a condition governed by a recessive allele, where the worms do not have setae. These are tiny hair-like structures needed by the worm to move through the ground. 500 worms were found not having setae. What percent of the population were heterozygous for the setae? What was the actual number of earthworms containing the setae?

6. Answer: • Ans: Because 500 don’t have setae, the frequency will be: 500/2000 = 0.25 (1) • So: q2 = 0.25 (2) • So: q = 0.5 (3) • So: p = 1-q = 1- 0.5 = 0.5 (4) • So: 2pq = 2 x 0.5 x 0.5 = 0.5 (5) • So: 50% of the population carry heterozygous allele. • The actual number of the population contains the dominant trait will be 2000 – 500 = 1500 (6)

7. Problem 3 • In Drosophilia, the allele for normal length wings is dominant over the allele for vestigial wings. In a population of 100,000 individuals, 3,600 show the recessive phenotype. How many individuals would you expect to be homozygous dominant and heterozygous for this trait?

8. Answer: • Ans: Because 3600 show recessive phenotype: • So: the recessive frequency: q2 = 3600/100000 • q = 0.1897 (1) • So: p = 1 – q = 1 – 0.1897 = 0.8103 (2) • So: 2pq = 2 x 0.8103 x 0.1897 = 0.3074 • So: 100000 x 0.3074 = 30740 (3) • So: 30740 individuals will carry the heterozygous allele. • So: 100000 – 30740 – 3600 = 65660 (4) • So: 65660 will carry homozygous dominant allele.

9. Problem 4 • The allele for the ability to roll one's tongue is dominant over the allele for the lack of this ability. In a population of 500,000 individuals, 25% show the recessive phenotype. How many individuals would you expect to be homozygous dominant and heterozygous for this trait?

10. Answer: • Do this problem yourself and discuss the answer with your group.

11. Problem 5 • The allele for the hair pattern called "widow's peak" is dominant over the allele for no "widow's peak." In a population of 300,000 individuals, 1.7% carry heterozygous allele. How many individuals would you expect of each of the possible three genotypes for this trait?

12. Answer: • Ans: Because 2pq = 1.7% = 0.017 • Also: p + q = 1 (1) • So: 2q (1-q) = 0.017 (2) • So: 2q – 2q2 = 0.017 (3) • Multiply (-1) both side: 2q2 – 2q = -0.017 • Add 0.017 both side: 2q2 – 2q + 0.017 = 0 (4) • Use formula: q1 = (2 + 1.9657)/4 = 0.9914 (too big) • q2 = (2 – 1.9657)/4 = 0.008575 • q2 = 0.00007353 • So: 300000 x 0.00007353 = 23 (5) • So: 23 people carry recessive phenotype. • Because 300000 x 0.017 = 5100 (6) • So: 300000 – 5100 – 23 = 294877 will carry homozygous dominant trait • 5100 will carry heterozygous dominant trait

13. Problem 6 • In a bird population, birds with black head are recessive to white head. 360,000 of the 800,000 birds carry heterozygous alleles. How many birds in this population have black head?

14. Answer: • Do this problem yourself and discuss the answer with your group.