Python Scripts for Generating Genetics Homework/Quiz problems
Created by gh-md-toc
What will the results of a blood test look like for a person with O- blood type?
- A.
- B.
- For the ABO blood group in humans, the iA and iB alleles are codominant and the i allele is recessive. If a female ♀ with type A blood marries a male ♂ with type O blood, which of the following blood types could their children possibly have? Check all that apply.
- A. Type O blood
- B. Type A blood
- C. Type B blood
- D. Type AB blood
- For the ABO blood group in humans, the iA and iB alleles are codominant and the i allele is recessive. A father ♂ who has blood type AB has a son ♂ who has blood type A, which of the following blood types could the mother ♀ possibly have? Check all that apply.
- A. Type O blood
- B. Type A blood
- C. Type B blood
- D. Type AB blood
- E. None of the above are possible
- Download list of 2 gene questions in blackboard upload format
- Download list of 3 gene questions in blackboard upload format
- A mother has a HLA genotype of A2,B5,C6 on one chromosome and A1,B1,C3 on the other. The father has a HLA genotype of A7,B9,C2 on one chromosome and A8,B3,C5 on the other. Which one of the following is a possible genotype for one of their offspring?
- A. A2,A8,B1,B9,C2,C3
- B. A1,A2,B1,B5,C3,C6
- C. A2,A7,B5,B9,C2,C6
- D. A7,A8,B3,B9,C2,C5
- E. A1,A7,B1,B3,C5,C6
- How many unique gametes could be produced through independent assortment by an individual with the genotype AA Bb cc dd Ee Ff Gg ?
- A. 22 = 4
- B. 23 = 8
- C. 24 = 16
- D. 25 = 32
- E. 26 = 64
- In a specific cross, F2 progeny exhibit a modified dihybrid ratio of 15:1 (instead of 9:3:3:1 ). What phenotypic ratio would be expected from a test-cross with an individual from the F1 progeny?
- A. 1:4
- B. 2:2 or 1:1
- C. 4:1
- D. 3:1
- E. 2:1
- F. 1:3
- An F1 heterozygote individual from dihybrid cross is used for a test-cross. The progeny from the test-cross exhibited a modified ratio of 3:1 (instead of 1:1:1:1). What phenotypic ratio would be expected in the F2 progeny if the dihybrid cross is continued?
- A. 13:3
- B. 12:4
- C. 11:5
- D. 10:6
- E. 9:7
1. The white-eyed phenotype is an X-linked recessive disorder in fruit flies. The red allele, +, is dominant to the white allele, w. The offspring of size 400 from the mating of a single female and a single male are shown in the table below:
| phenotype | female ♀ | male ♂ |
|---|---|---|
| red-eyed (wildtype) | 0 | 0 |
| white-eyed (mutant) | 185 | 215 |
What are the genotypes of the parents in this cross?
- A. homozygous wildtype female (++) and male of unknown genotype
- B. heterozygous female (+w) and wildtype male (+–)
- C. heterozygous female (+w) and mutant male (w–)
- D. homozygous mutant female (ww) and wildtype male (+–)
- E. homozygous mutant female (ww) and mutant male (w–)
| Table of Chi-Squared (χ2) Critical Values | ||||||||
|---|---|---|---|---|---|---|---|---|
| Degrees of Freedom | Probability | |||||||
| 0.95 | 0.90 | 0.75 | 0.50 | 0.25 | 0.10 | 0.05 | 0.01 | |
| 1 | 0.00 | 0.02 | 0.10 | 0.45 | 1.32 | 2.71 | 3.84 | 6.63 |
| 2 | 0.10 | 0.21 | 0.58 | 1.39 | 2.77 | 4.61 | 5.99 | 9.21 |
| 3 | 0.35 | 0.58 | 1.21 | 2.37 | 4.11 | 6.25 | 7.81 | 11.34 |
| 4 | 0.71 | 1.06 | 1.92 | 3.36 | 5.39 | 7.78 | 9.49 | 13.28 |
1. The final result gives the chi-squared (χ2) test value of 3.58 with 3 degrees of freedom. Using the Table of χ2 Critical Values and a level of significance α=0.50, we get a critical value of 2.37. Since the chi-squared (χ2) test value of 3.58 is greater than the critical value of 2.37, the null hypothesis was rejected.
| Phenotype | Expected | Observed | Calculation | Statistic |
|---|---|---|---|---|
| Yellow Round (Y–R–) | 90 | 94 | (94-90)2⁄ 90 | 0.178 |
| Yellow Wrinkled (Y–rr) | 30 | 21 | (21-30)2⁄ 30 | 2.700 |
| Green Round (yyR–) | 30 | 33 | (33-30)2⁄ 30 | 0.300 |
| Green Wrinkled (yyrr) | 10 | 12 | (12-10)2⁄ 10 | 0.400 |
| (sum) χ2 = | 3.578 | |||
Your lab partner has done a chi-squared (χ2) test for your lab data (above), for the F2 generation in a standard dihybid cross. They wanted to know if the results confirm the expected phenotype ratios, but as usual they did something wrong. Where did they do wrong?
- A. the expected progeny for the null hypothesis is incorrect
- B. the degrees of freedom is wrong
- C. the numbers in the calculation have to be squared
- D. the wrong numbers in the calculation were used for division
- E. the wrong rejection criteria was used
A women has nine (9) children, what is the probability that she has exactly five (5) boys and four (4) girls?
