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. a) P Rr x rr
b) R, r and r, r
c) F1 Rr, Rr, rr, rr 1 round:1 wrinkled
d) F2 RR, Rr, Rr, rr 3 round:1 wrinkled
3. P L_ x ll
F1 327 tall: 321 short - almost 1:1 therefore unknown parent must be heterozygous. Note: homozygous (LL) would give ALL tall plants in F1.
4. The presence of all smooth in the offspring means smooth is dominant.
P SS x ss
F1 Ss
F2 3:1
5. P WW x ww
F1 Ww all white
F2 WW, Ww, Ww, ww 3 white:1 yellow
c) P Ww x WW
F1 Ww, WW all white
6. 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.
7. (i) P T_ x tt
F1 tt
(ii) P T_ x tt
F1 Tt
(iii) P T_ x Tt
F1 tt
The male must be heterozygous (Tt) to be able to produce both trotters and pacers. If he were homozygous dominant her would produce only trotters.
8. 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).
9. P TmTn x TmTn
F1 TmTm, 2TmTn, TnTn (1 severe : 2 mild : 1 normal)
10. The trick here is that the parents produce all yellow or all green seeds. That means they muts be
homozygous, otherwise they would produce a mixture.
P TTYY x ttyy
F1 TtYy
F2 9:3:3:1
11. 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
12. B - black; b - white
S - short; s - long
P bbSS x BBss
F1 BbSs
F2 1 BBSS, 2 BBSs, 2 BbSS, 2 Bbss, 4 BbSs, BBss, bbSS, 2 bbSs, bbss (typical heterozygous dihybrid
cross phenotypic ratio of 9:3:3:1 )
13. 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.
14. P TTBb x Ttbb
F1 TTBb, TtBb, TTbb, Ttbb
a) 1 tall, brown:1 tall, blonde
b) 1/4
c) 1/4
15. P C_aa x ccA_
F1 ccaa this means the parents must both be heterozygous (Ccaa and ccAa)
16. P Pp x Pp
a) P(P_) x P(P_) x P(P_) = 3/4 x 3/4 x 3/4 = 27/64
b) 1 - 27/64 = 37/64
c) P(all disease) = P(pp) x P(pp) x P(pp) = 1/4 x 1/4 x 1/4 = 1/64
d) 1 - 1/64 = 63/64
17. R - rose; r - single
F - feather; f - clean
P A x C
F1 all feather, mostly rose (F_R_ and F_rr)
P A x D
F1 feathered and clean but all rose (F_R_ and ffR_)
P B x C
F1 feathered and clean, most rose but some single (F_R_, F_rr,ffR_, ffrr )
A FfRr
B FfRr
C FfRr
D FfRR (?)
18. P B_ x A_
F1 ii the man could be father but this is not proof
19. P D1D3 x D2D3
F1 2D1_, 1 D2D3, 1 D3D3 the presence of two recessive alleles in one offspring means each parent must
have donated one
20. a) P ChCa x CaCa
F1 ChCa, CaCa 1 himalayan:1 albino
b) P CCa x CchCa
F1 2 C_, Cch_, CaCa
c) P CchCch x CchCa
F1 CchCch, CchCa 1 chinchilla:1 light gray
d) P CchCh x CaCa test cross
F1 5 ChCa, 5 CchCa
21. SR - round; SL - long
P SRSL x SRSL
F1 SRSR, SRSL, SRSL, SLSL (incomplete dominance)
22. CRCR - chestnut; CMCM - cremello; CMCR - palomino
P CMCR x CMCM
F1 CMCM, CRCM 1 cremello:1 palomino
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. P Ccbb x CcB_
A test cross would be necessary to determine the genotype of the black parent. Cross it with a CCbb
mouse. If any tan offspring are present in F1, the black mouse is heterozygous. If all black, the black
mouse is homozygous.
25. a) P rrPP (pea) x RRpp (rose)
F1 RrPp (walnut)
b) P RrPp (walnut) x RRPP (walnut)
F1 RRPP (walnut), RRPp (walnut), RrPP (walnut), RrPp (walnut)
c) P RrPP (walnut) x rrPP (pea)
F1 RrPP (walnut), rrPP (pea)
d) P RrPp (walnut) x RrPp (walnut)
F1 RRPP (walnut), 2RRPp (walnut), RRpp (rose), 2RrPP (walnut), 4RrPp (walnut),
2 Rrpp (rose), rrPP (pea), 2 rrPp (pea), rrpp (single)
26. Codominance gives a typical 1:2:1 ratio. Multiple alleles would give ratios other than 1:2:1 because of any dominance hierarchy.
27. P ii x IAIB
F1 IAi, IBi
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.
28. C - color; c - albino
B - black; b - brown
P CcBb x CcBb
F1 CCBB, CCBb, CcBB, CcBb, CCbb, Ccbb, ccBB, ccBb, ccbb
- the expected ratio of 9:3:3:1 becomes 9:3:4 due to epistasis.
29. HB - baldness - dominant in males; recessive in females
Hn - normal - dominant in females, recessive in males
P HnHn x HBHn
F1 HnHB
- a bald girl is impossible because the father cannot donate HB
30. a) 4 (DUH ! Gee, challenging)
b) D - normal; d - recessive
P Dd x Dd
c) P (F) dd x (G) dd
F1 (M) dd, (N) dd