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 wrinkled
  
3. 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 one
6. 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 CchCa
  
7. 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, Bbss
  
10. 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 , Bbss
  
11. 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 albino
  
13.	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 dominance
  
2.	P Ll x ll 
	F1 Ll, ll
  
3.  both are heterozygous
  
4. (1)	P Tt x tt 
	F1 tt 

   (2)	P Tt x tt 
	F1 Tt 

   (3)	P Tt x Tt 

	F1 tt  
5. codominance - 1:2:1 indicates codominance 

multiple alleles - would give ratios other than 1:2:1 because of dominance hierarchy
  
6. 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 HB
  
2. 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  
3. 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 normal
  
2. 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)