Learning Genetics Can Be Fun - Solutions
1. 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
2. P TTYy x ttYy
F1 TtYY, TtYy, TtYy, Ttyy
600 tall, yellow: 200 tall, green
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. 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
4. 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.
5. (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 he would produce only trotters.
F1 Tt
(iii) P T_ x Tt
F1 tt
6. 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).
7. Agnes Bb
Agnes’ mother bb. This tells us that Agnes is heterozygous.
Ralph Bb
Ralph’s father bb. This tells us Ralph is heterozygous.
Agnes x Ralph
P Bb x Bb
F1 BB, Bb, Bb, bb chance of a blond child is 1/4
8. P TmTn x TmTn
F1 TmTm, 2TmTn, TnTn (1 severe : 2 mild : 1 normal)
9. B - black; b - white
S - short; s - long
P bbSS x BBss
F1 BbSs black, short
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 )
10, 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.
11. P iiRh-Rh- x IA_Rh+_
gametes iRh- IARh+, IARh-, iRh+, iRh-
F1 IAiRh+Rh-, IAiRh-Rh-, iiRh+Rh-, iiRh-Rh-
The genotype of the male is unknown so we have to consider all possibilities.
12. P B_ x A_
F1 ii the man could be father but this is not proof
13. 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
14. 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
15. SR - round; SL - long
P SRSR x SLSL
F1 SRSL
F2 SRSR, SRSL, SRSL, SLSL (incomplete dominance)
16. CRCR - chestnut; CMCM - cremello; CMCR - palomino
P CMCR x CMCM
F1 CMCM, CRCM 1 cremello:1 palomino
17. 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?)
18. 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.
19. HB - baldness - dominant in ♂; recessive in ♀
Hn - normal - dominant in ♀, recessive in ♂
P ♂ HnHn x ♀ HBHn
F1 HnHB
- a bald girl is impossible because the father cannot donate HB
20. a) 4 children
b) A is dd, B is dd
c) M is dd, N is dd
21.