1. GENETIC
PROBLEMS

2. Question #1
• How many different kinds of gametes
could the following individuals produce?
1. aaBb
2. CCDdee
3. AABbCcDD
4. MmNnOoPpQq
5. UUVVWWXXYYZz

3. Question #1
• Remember the formula 2n
• Where n = # of heterozygous
1. aaBb = 2
2. CCDdee = 2
3. AABbCcDD = 4
4. MmNnOoPpQq = 32
5. UUVVWWXXYYZz = 2

4. Question #2
• In dogs, wire-haired is due to a
dominant gene (W), smooth-haired is
due to its recessive allele (w).
• WW, Ww = wire haired
• ww = smooth haired

5. Question #2A
• If a homozygous wire-haired dog is mated
with a smooth-haired dog, what type of
offspring could be produced.
W W
w
w

6. Question #2A
W W
w Ww Ww
fg F1 generation
w Ww Ww all heterozygous

7. Question #2B
• What type(s) of offspring could be produced
in the F2 generation?
• Must breed the F1 generation to get the F2.
• Results of F1 Cross: Ww x Ww

8. Question #2B
W w
W WW Ww F2 generation
w Ww ww
genotype: 1:2:1 ratio
phenotype: 3:1 ratio

9. Question #2C
• Two wire-haired dogs are mated. Among the
offspring of their first litter is a smooth-
haired pup.
• If these, two wire-haired dogs mate again,
what are the chances that they will produce
another smooth-haired pup?
• What are the chances that the pup will wire-
haired pup?

10. Question #2C
W w
W WW Ww F2 generation
w Ww ww
- 1/4 or 25% chance for smooth-haired
- 3/4 or 75% chance for wire-haired

11. Question #2D
• A wire-haired male is mated with a
smooth-haired female. The mother of
the wire-haired male was smooth-haired.
• What are the phenotypes and genotypes
of the pups they could produce?
• Show the results of crossing: Ww x ww

12. Question #2D
W w
w Ww ww
w Ww ww
phenotypes: 1:1 ratio
genotypes: 1:1 ratio

13. Question #3
• In snapdragons, red flower (R) color is
incompletely dominant over white flower
(r) color.
• The heterozygous (Rr) plants have pink
flowers.
RR - red flowers
Rr - pink flowers
rr - white flowers

14. Question #3A
• If a red-flowered plant is crossed
with a white-flowered plant, what are
the genotypes and phenotypes of the
plants F1 generation?
• RR x rr

15. Question #3A
R R
r Rr Rr F1 generation
r Rr Rr
phenotypes: 100% pink
genotypes: 100% heterozygous

16. Question #3B
• What genotypes and phenotypes will
be produced in the F2 generation?
• Rr x Rr

17. Question #3B
R r
R RR Rr F2 generation
r Rr rr
phenotypes: 1:2:1 ratio
genotypes: 1:2:1 ratio

18. Question #3C
• What kinds of offspring can be
produced if a red-flowered plant is
crossed with a pink-flowered plant?
• RR x Rr

19. Question #3C
R R
R RR RR
r Rr Rr
50%: red flowered
50%: pink flowered

20. Question #3D
• What kind of offspring is/are produced
if a pink-flowered plant is crossed with
a white-flowered plant?
• Rr x rr

21. Question #3D
R r
r Rr rr
r Rr rr
50%: white flowered
50%: pink flowered

22. Question #4
• In humans, colorblindness (cc) is a
recessive sex-linked trait.
• Remember: XX - female
XY - male

23. Question #4A
• Two normal people have a colorblind son.
• What are the genotypes of the parents?
• XCX_? x XCY
• What are the genotypes and phenotypes
possible among their other children?

24. Question #4A
XC Y ← parents
XC XCXC XCY
Xc XCXc XcY
50%: female (one normal, one a carrier)
50%: male (one normal, one colorblind)

25. Question #4B
• A couple has a colorblind daughter.
• What are the possible genotypes and
phenotypes of the parents and the
daughter?

26. Question #4B
Xc Y
XC XCXc XCY
Xc XcXc XcY
parents: XcY and XCXc or XcXc
father colorblind
mother carrier or colorblind
daughter: XcXc - colorblind

27. Question #5
• In humans, the presence of freckles is
due to a dominant gene (F) and the
non-freckled condition is due to its
recessive allele (f).
• Dimpled cheeks (D) are dominant to
non-dimpled cheeks (d).

28. Question #5A
• Two persons with freckles and dimpled cheeks
have two children: one has freckles but no
dimples and one has dimples but no freckles.
• What are the genotypes of the parents?
Parents: F__D__ x F__D__
Children: F__dd x ffD__

29. Question #5B
• What are the possible phenotypes and
genotypes of the children that they
could produce?
• Cross: FfDd x FfDd
• This is a dihybrid cross

30. Question #5B
• Possible gametes for both: FD Fd fD fd
FD Fd fD fd
FD FFDD FFDd FfDD FfDd
Fd FFDd FFdd FfDd Ffdd
fD FfDD FfDd ffDD ffDd
fd FfDd Ffdd ffDd ffdd

31. Question #5B
Phenotype : Freckles/Dimples: 9
Freckles/no dimples: 3
no freckles/Dimples: 3
no freckles/no dimples: 1
Phenotypic ratio will always been 9:3:3:1
for all F1 dihybrid crosses.

32. Question #5B
Genotypic ratio: FFDD - 1
FFDd - 2
FFdd - 1
FfDD - 2
FfDd - 4
Ffdd - 2
ffDD - 1
ffDd - 2
ffdd - 1

33. Question #5C
• What are the chances that they
would have a child whom lacks both
freckles and dimples?
• This child will have a genotype of
ffdd
• Answer: 1/16

34. Question #5D
• A person with freckles and dimples whose
mother lacked both freckles and dimples
marries a person with freckles but not
dimples whose father did not have freckles
or dimples.
• Cross: FfDd x Ffdd
• Possible gametes:
FD Fd fD fd x Fd fd

35. Question #5D
• What are the chances that they would have
a child whom lacks both freckles and
dimples?
FD Fd fD fd
Fd FFDd FFdd FfDd Ffdd
fd FfDd Ffdd ffDd ffdd
Answer: 1/8

36. Question #6
• Sixteen percent of the human population is
known to be able to wiggle their ears.
• This trait is determined to be a recessive
gene.
• These is a population genetics question.
• Use the following equation: 1 = p2 + 2pq + q2

37. Question #6A
• What of the population is homozygous
dominant for this trait?
• q2 = 16% or .16: √ q2 = √ .16
q = .4
• then use : 1 = p + q
1 = p + .4
1- .4 = p
p = .6
• Now use p2 for answer: .62 = .36 or 36%

38. Question #6B
• What of the population is heterozygous for
this trait?
• We know that q = .4 and p = .6
• Now use 2pq for answer: 2(.6)(.4) = .48 or
48%

39. Question #7
• In dogs, the inheritance of hair color
involves a gene B for black hair and gene b
for brown hair b.
• A dominant C is also involved. It must be
present for the color to be synthesized.
• If this gene is not present, a blond condition
results.
BB, Bb - black hair CC, Cc - color
bb - brown hair cc - blond

40. Question #7A
• A brown haired male, whose father was a
blond, is mated with a black haired female,
whose mother was brown haired and her
father was blond.
Male: bbCc (gametes: bC bc)
Female: BbCc (gametes: BC Bc bC bc)
• What is the expected ratios of their
offspring?

41. Question #7A
BC Bc bC bc
bC BbCC BbCc bbCC bbCc
bc BbCc Bbcc bbCc bbcc
Offspring ratios:
Black: 3/8
Brown: 3/8
Blond: 2/8 or 1/4

42. Question #8
• Henry Anonymous, a film star, was involved in
a paternity case. The woman bringing suit had
two children, on whose blood type was A and
the other whose blood type was B.
• Her blood type was O, the same as Henry’s!
• The judge in the case awarded damages to the
woman damages to the woman, saying that
Henry had to be the father of at least one of
the children.

43. Question #8A
• Obviously, the judge should be sentenced to
Biology. For Henry to have been the father of
both children, his blood type would have had to
be what?
IA IB ← Answer
i IAi IBi
i IAi IBi