Answers to the Three Dozen Genetics Problems

(if you think I have coded in a mistake to the questions, please let me know)

1a. M 

1b. Z, z 

1c. AB, aB 

1d. RT, Rt, rT, rt (assuming no linkage) 

1e. AbCDEf, AbcDEf

2a. 1

2b. 1

2c. 2

2d. 2

2e. 6

3a. dom

3b. dom

3c. dom, dom

3d. dom, dom

3e. A, C, D, E are dom; B and F are rec

4a. 3 yellow : 1 green  or 75% yellow  25% green

4b. 100% yellow

4c. 100% yellow

5a. Ww

5b. Ww (black) or ww (white)

5c. 1 black : 1 white  or 50% black  50% white

6a. 25% chance  or 1/4 probability

6b. 50% chance or 1/2 probability

6c. assuming that the female is homozygous dom, there is 0% chance of an albino offspring - but, you can't be sure the female is homozygous dom, even if the trait hasn't shown up for three generations.

7. Aa x Aa  (both heterozygous)   25% probability that the next child will be albino.

8. Ff (heterozygous)

9. Yy x yy  (the yellow parent is heterozygous, the green is homozygous recessive)

10. Note:  widow's peak is dominant.  50% prbability that any one child will have a widow's peak.  There is a 1/2 x 1/2 = 1/4 or 25% chance that BOTH will have widow's peak.

11. 50% brown or blue

12. You can't be certain that he is homozygous dominant, but the probability is good that he is.  You could do a statistical test, like chi square to "prove" our conclusion.

13. F  (we could both be heterozygous, therefore, there is 25% chance that one received both recessive alleles)

14. F  since the trait is dominant, I must be Rr or RR.

15. F

16. F (if Linda is homo dom then there is 100% probability Erin can taste it)

17. T (since Im homo rec, she must receive the allele from me)

18. T (although genetic surprises do happen)

19. T

20. F (they could have O if Linda is heterozygous)

21. F

22. F (at least one parent must have a dominant allele - could be homozygous dom or heterozygous. The other parent must have a recessive allele - could be heterozygous or homo rec) 23. T (they receive it from Linda)

24. F (he could be if Evelyn is heterozyg)

25. F (if Evelyn is heterozy then it is true, but if Evelyn is homo dom then there is no chance)

26. T

27. F (they could both be heterozygous, then yield 1/4 or 25% with FH)

28. T

29. F (each parent could be heterozygous)

30. F

31. T (at least three)

32. F (since it is a 3:1 ratio both parents must be heterozygous)

33. F (50% would be expected.  The man must be homoz rec and the woman heteroz since her rather was homo rec)

34. T

35. T

36. F (one parent must have freckles)

37. F (4/52).  The odds of picking an Ace of Spades is 1/52.

38. T

39. F (1/4 or 25%.  The probability of two independent events occurring is the product of the individual probabilities.  Thus 1/2 times 1/2 equals 1/4.  As an example, the probability of any one child being a female is 1/2.  Even if a family has 10 children, all boys, the probability of the next child being a girl is 1/2.  However, the probability of ALL 10 children being boys is 1/2 times 1/2 times 1/2.....(ten times).

Last updated: July 14, 2009     � Copyright by SG Saupe