Learning Genetics Can Be Fun - Solutions

1. Two black dogs could be homozygous black (BB) or heterozygous black (Bb). Yellow must be homozygous, therefore cannot be the same genotype as black.

2. P Cc x Cc

F₁ CC, Cc, Cc, cc       both parents are normal but “carry” the allele for CF. One in four children will inherit it.

3. a) P Rr x rr

b) R, r and r, r

F₁ Rr, Rr, rr, rr           1 round:1 wrinkled

F₂ RR, Rr, Rr, rr        3 round:1 wrinkled

4. P Ll x ll

F₁ Ll, Ll, ll, ll            1 long:1 short

5. P T_ x tt

F₁ 327 tall: 321 short - almost 1:1 therefore unknown parent must be heterozygous. Note: homozygous (LL) would give ALL tall plants in F₁.

6. The presence of all smooth in the offspring means smooth is dominant.

P SS x ss

F₁ Ss

F₂ 3:1

7. a) P Ss x Ss

F₁ SS, Ss, Ss, ss

b) P S_ x _ _

The female must be heterozygous as she produced non-spotted pups. The unknown male must be homozygous recessive (ss). If he were homozygous dominant, all pups would be spotted. If he were heterozygous, you would expect a 3:1 ratio in pups.

8. (i)    P T_ x tt                                             The male must be heterozygous (Tt) to be able to produce

            F₁ tt                                                     both trotters and pacers. If he were homozygous dominant

    (ii)    P T_ x tt                                              her would produce only trotters.

            F₁ Tt

    (iii)   P T_ x Tt

            F₁ tt

9. Normal woman Pp (must be heterozygous because father was albino)

Husband pp

Husband’s parents both Pp

Children Pp, Pp, pp

10. test cross W_ x ww

11. P Pp x Pp

F₁ PP, Pp, Pp, pp        chance of PKU is 1/4

12. P Bb x Bb

      F₁ BB, Bb, Bb, bb

a) 1/4              (b) 1/4             (c) 1/2

d) 1 homozygous brown:2 heterozygous brown:1 homozygous blonde

e) 3 brown:1 blonde

f) not possible because blonde (b) is recessive

g) C = curly; c = straight

h) P Cc x cc

F₁ Cc, Cc, cc, cc

i) C, c              (j) c, c             (k) 0                (l) 1/2              (m) 1/2

n) 1 heterozygous:1 homozygous recessive

o) 1 curly:1 straight

p) No. Straight hair is recessive so individual MUST be homozygous (cc).

13. B - black; b - white; S - short; s - long

a) P BBSs x bbss

F₁ BbSs, Bbss                         1 black, short:1 black, long

b) P BbSs x bbss

F₁ BbSs, Bbss, bbSs, bbss      1 black, short:1 black, long: 1 white, short:1 white, long

c) P BBss x BbSs

F₁ BBSs, BBss, BbSs, Bbss   1 black, short:1 black, long

d) i) (a) 1/2 (b) 1/4 (c) 1/2

ii) (a) 1/2 (b) 1/4 (c) 1/2

iii) (a) 0 (b) 1/4 (c) 0

14. B - black; b - white; S - solid; s- spotted

            male                female

a) P     B_S_               bbS_

F₁        2 BbS_ , 2 bbS_

Some white pups so the male must be Bb. The absence of any non-spotted pups suggests that female A is SS but we can’t say for sure.

b) P     BbSs               B_S_

F₁        bbss

the presence of white, non-spotted pups means that female B must be BbSs

c) P     BbSs               bbss

F₁        bbSs , bbss , BbSs , Bbss

The genotype of female C can be determined from her phenotype.

15. S^R - round; S^L - long

P S^RS^L x S^RS^L

F₂ S^RS^R, S^RS^L, S^RS^L, S^LS^L  (incomplete dominance)

16. P S^NS^M x S^NS^M

      F₁ S^NS^N, S^NS^M S^NS^M, S^MS^M 25% chance of having homozygous recessive child

17. C^RC^R - chestnut; C^MC^M - cremello; C^MC^R - palomino

P C^MC^R x C^MC^M

F₁ C^MC^M, C^RC^M       1 cremello:1 palomino

18. P F^RF^W x F^RF^W

      F₁ F^RF^R, F^RF^W, F^RF^W, F^WF^W

a) ½ pink

b) 1/4 red

c) 1/4 white

d) 1:2:1

e) 1:2:1

19. woman I^B_ x man I^A_

              F₁ ii is possible if mother and father were both heterozygous. The facts are inconclusive.

20. P ♂ ii x ♀ I^AI^B

F₁ I^Ai, I^Bi

AB ♀ could produce AB offspring if ♂ were type A, B, or AB; she could never produce type O in F₁ because she always donates either A or B.

21. a) P C^hC^a x C^aC^a

F₁ C^hC^a, C^aC^a            1 himalayan:1 albino

b) P CC^a x C^chC^a

F₁ 2 C_, C^ch_, C^aC^a

c) P C^chC^ch x C^chC^a

F₁ C^chC^ch, C^chC^a       1 chinchilla:1 light gray

d) P C^chC^h x C^aC^a      test cross

F₁ 5 C^hC^a, 5 C^chC^a

22. note: cc = no purple

            P Ppcc x PPCc

gametes Pc, pc PC, Pc

            F₁ PPCc, PPcc, PpCc, Ppcc

phenotypes 1 purple, curved: 1 white, straight: 1 purple, curved: 1 white, straight (1:1)

23. a)   P CCBB (black) x Ccbb (brown)

            F₁ CCBb (black), CcBb (black)

      b)   P ccBB (albino) x CcBb (black)

            F₁ CcBB (black), CcBb (black), ccBB (albino), ccBb (albino)

       c)   P CcBb (black) x ccbb (albino)

            F₁ CcBb (black), Ccbb (brown), ccBb (albino), ccbb (albino)

       d)  P CcBb (black) x CcBb (black)

            F₁ CCBB (black), CCBb (black), CcBB (black), CCbb (brown), CcBb (black), Ccbb (brown), ccBB (albino), ccBb (albino), ccbb (albino)

note: you would get the normal 9:3:3:1 as in any heterozygous dihybrid cross but ccBB, ccBb, and ccbb all combine to give 4 albino (a bit tricky, eh?)

24. a) 4 children

b) A is Dd, B is Dd

c) M is dd, N is dd

25.