p. 576 1. homozygous black - BB; heterozygous black - Bb yellow must be homozygous, therefore cannot be same genotype as black
2. a) P Rr x rr
F1 Rr, Rr, rr, rr 1 round:1 wrinkled
F2 RR, Rr, Rr, rr 3 round:1 wrinkled3. a) P Ss x Ss
F1 SS, Ss, Ss, ss
b) Unknown male must be homozygous recessive (ss). If he were homozygous dominant, all pups would be spotted. If he were heterozygous, would expect 3:1 ratio in pups.4. Ratio of 3:1 observed; conclude parents are both heterozygous
p. 580 5. 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 one6. a) P ChCa x CaCa
F1 ChCa, CaCa1 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 CchCa7. SR - round; SL - long
P SRSL x SRSL
F1 SRSR, SRSL, SRSL, SLSL
(incomplete dominance)8. CRCR - chestnut; CMCM - cremello; CMCR - palomino
P CMCR x CMCM
F1 CMCM, CRCM 1 cremello:1 palomino
p. 585 9. B - black; b - white; S - short; s - long
a) P BBSs x bbss
F1 BbSs, Bbss
b) P BbSs x bbss
F1 BbSs, Bbss, bbSs, bbss
c) P BBss x BbSs
F1 BBSs, BBss, BbSs, Bbss10. B - black; b - white; S - solid; s- spotted
male female
a) P BbS_ bbSS
F1 2 BbS_ , 2 bbS_
b) P BbSs BbSs
F1 bbss
c) P BbSs bbss
F1 bbSs , bbss , BbSs , Bbss11. Female OORh-Rh-
Male AORh+Rh- AORh+Rh- , AORh-Rh- , OORh+Rh- , OORh-Rh-
AORh+Rh+ AORh+Rh- , OORh+Rh-
AARh+Rh- AORh+Rh- , AORh-Rh-
AARh+Rh+ AORh+Rh-
p. 591
12. 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: would get normal 9:3:3:1 as in any heterozygous dihybrid cross but ccBB, ccBb, ccBb,
and cbb all combine to give 4 albino13. P Ccbb x CcB_
to find other parent must perform test cross with CCbb individual.14. 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)
P. 594
1. all behaved as simple, complete dominance2. P Ll x ll
F1 Ll, ll3. both are heterozygous
4. (1) P Tt x tt
F1 tt
(2) P Tt x tt
F1 Tt
(3) P Tt x Tt
F1 tt5. codominance - 1:2:1 indicates codominance
multiple alleles - would give ratios other than 1:2:1 because of dominance hierarchy6. 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.7. P TmTn x TmTn
F1 TmTm, 2TmTn, TnTn (1 severe : 2 mild : 1 normal)8. 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 )9. R - rose; r - single
F - feather; f - clean
A FfRr
B FfRr
C FfRr
D FfRR (?)10. C - color; c - albino
B - black; b - brown
P CcBb x CcBb
F1 CCBB, CCBb, CcBB, CcBb, CCbb, Ccbb, ccBB, ccBb, ccbb
- expected 9:3:3:1 becomes 9:3:4 due to epistasis.1. HB - baldness - dominant in male; recessive in female
Hn - normal - dominant in female, recessive in male
P HnHn x HBHn
F1 HnHB
- bald girl impossible because father cannot donate HB2. 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) dd3. The presence of recessive alleles in heterozygous individuals is masked by the presence of dominant alleles. In matings between close relatives (who may carry the same recessive alleles) the chance of producing homozygous recessive offspring increases.
p. 602
1. a) P XHXh x XHy
F1 Xhy
b) no because father is normal2. a) P XnXN x Xny
F1 XnXn, Xny, XNXn, XNy
b) XNXN individual would have to receive XN from father who would be dead
c) unlikely that XN would live long enough to breed, also heterozygous individuals would be
less common in nature.
3. b - brown; v - vermillion; BV - wt; bv - white
a) P BBVV x bbvy
F1 BbVv, BbVy (all wt)
b) P BbVv x BbVy
F1 BBVV, BBVy, BbVV, BbVy, BBVv, BBVy, BbVv, Bbvy, bbVV, bbVy, bbVv, bbvy (9 wt:3 brown:3 vermillion:1 white)
c) P bbVV x Bbvy
F1 BbVv, BbVy, bbVv, bbVy (2 wt:2 brown)