Biology (Genetics Inheritance) Homework: Coursework

1. (8 points) Below is a pedigree of a family with a high frequency of Tay Sachs Disease. What is the probability that the “?” individual will have the disease? Show your work.

2. (8 points) Please note that this is a very similar question to your homework but it is not identical.

            A three-point testcross was made in corn. The results and a recombination analysis are shown in the display below, which is typical of three-point testcrosses (p = purple leaves, + = green; v = virus-resistant seedlings, + = sensitive; b = brown midriff to seed, + = plain). Study the display and answer parts a through c.

P                              +/+ · +/+ · +/+  ×  p/p · v/v · b/b

Gametes + · + · +                   p · v · b

a.     (2 points) Fill in the “Recombinant for” columns.

b.  (4 points) Draw a map that shows distances in map units.

b.     (2 points) Calculate interference I.

3. (8 points) Below is a table of 17 SNPs from dogs with black coats and dogs with white coats.

a. (2 points) Which of these SNPs is completely associated with coat color?

b. (2 points) Which four SNPs are completely unassociated with coat color?

c. (1 point) What type of statistical test would you need to do in order to show that a SNP is significantly associated with a phenotype or disease? You do not need to actually do the test.

d. (3 points) What are the three limitations of GWAS?

                i.

                ii.

                iii. 

4. (8 points) Patients with Xeroderma pigmentosum have mutations in genes for nucleotide excision repair. This is an autosomal recessive disease.

a. (5 points) Below is hypothetical complementation data for mutations in patients with XP. How many genes are represented in this data and which mutations belong in which group?

b. (3 points) Two parents affected with Xeroderma pigmentosum have all unaffected offspring. How is this possible?

5. (5 points) For each of the double mutants below, select the correct interpretation.

a. The mutant phenotype of the double mutant is the same as each single mutant.

i.   These genes are in the same pathway

ii.  Gene B is a suppressor

iii  These genes are in two different pathways that control the same phenotype.

iv. These genes are redundant.

v.  Gene A is epistatic to gene B.

vi. These genes are synthetically lethal.

b. The mutant phenotype of the double mutant is worse than each single mutant.

i.   These genes are in the same pathway

ii.  Gene B is a suppressor

iii  These genes are in two different pathways that control the same phenotype.

iv. These genes are redundant.

v.  Gene A is epistatic to gene B.

vi. These genes are synthetically lethal.

c. There is no mutant phenotype in the single mutants, but there is a mutant phenotype in the double mutants.

i.   These genes are in the same pathway

ii.  Gene B is a suppressor

iii  These genes are in two different pathways that control the same phenotype.

iv. These genes are redundant.

v.  Gene A is epistatic to gene B.

vi. These genes are synthetically lethal.

d. There is a mutant phenotype when Gene A is mutated, but the phenotype is normal in a double mutant with Gene A and Gene B.

i.   These genes are in the same pathway

ii.  Gene B is a suppressor

iii  These genes are in two different pathways that control the same phenotype.

iv. These genes are redundant.

v.  Gene A is epistatic to gene B.

vi. These genes are synthetically lethal.

e. Single mutations in Gene A and Gene B live, but the double mutants die.

i.   These genes are in the same pathway

ii.  Gene B is a suppressor

iii  These genes are in two different pathways that control the same phenotype.

iv. These genes are redundant.

v.  Gene A is epistatic to gene B.

vi. These genes are synthetically lethal.

6. (8 points) The length of wings in adult fruit flies is a complex trait that depends on genetic and environmental factors. You have isolated 8 inbred lines and grown them in 4 environmental conditions. Average wing lengths (in mm) for each experiment is shown in the table below.

a. (3 points) What is the mean, variance and standard deviation of the entire data set?

b. (2.5 points) What is the genetic variance under the condition of 21°C and 40% relative humidity? Please show your work for this part in order to receive full credit.

c. (2.5 points) If the total genetic variance is 0.036, what is the Broad-sense heritability (H²) of wing length?

7. (5 points) In a population of beetles, the total variance of body weight is 140 mg². It is estimated that the environmental variance is 40 mg², and dominance genetic variance is 35 mg². Calculate the additive genetic variance and the narrow-sense heritability (h²) of body weight in these beetles.