Answers to Genetic 04

Q1.Tt X Tt

X	T	t
T	TT	Tt
t	Tt	tt

a. 75 % Taster and 25% non-tasters

b. As above

c. 75%

d. 0.25 x 0.25 x0.25 = 0.015 of 1.5%

Q2. Trees : Allele Key B = Smooth and b = wrinkled

Heterozygote X Heterozygote

Phenotype: Smooth X Smooth

Genotype: Bb X Bb

X	B	b
B	BB	Bb
b	Bb	Bb

Offspring Genotype: BB :2Bb : bb

Phenotype Ratio: 3 Smooth: 1 Wrinkled

100 offspring : 75 Smooth : 25 Wrinkled

Q3.

1. Rooster feather colour is a codominant condition. In which the grey is the heterozygote.
2. Offspring 50% Black and 50% Grey feather

Q4. Allele Key T1 = Long Tail T2= Short Tail

a. Long Tailed Bloomer T1 T1 x Short tailed Bloomer T2 T2

X	T2	T2
T1	T1T2	T1T2
T1	T1T2	T1T2

b. All Offspring T1 T2 = Medium tailed

Q5. Red dominant over White R = Red r = white

a. Heterozygote X Heterozygote

Rr X Rr

X	R	r
R	RR	Rr
r	Rr	rr

Offspring Genotype: RR : 2 Rr ; rr

Offspring Phenotype; 25% Homozygous Dominant : 50% Heterozygous: 25% Homozygous recessive

b. ¾ Red : ¼ White

Q6.

a) Homozygous Wired x Smooth

WW x ww

x	w	w
W	Ww	Ww
W	Ww	Ww

Genotype of Offspring : All Ww

Phenotype of Offspring: All Wired

b) F2 = F1 x F1

Genotype Ww x Ww

Cross

x	W	w
W	Ww	Ww
w	Ww	ww

Offspring genotype: WW: 2 Ww: ww

Offspring Phenotype: ¾ Wired : ¼ Smooth

c) Wired x Wired

Genotypes W_ x W_

Offspring some ww

Parent types must be Ww and Ww

Chance of Another Smooth Pup = 25%

Chance of another wired = 75%

d) Mother ww

Wired Haired male x Smooth Haired female

Genotype Ww x ww

Cross

x	w	w
W	Ww	Ww
w	ww	ww

50% Wired : 50 % Smooth

Q7. Snapdragon

Red (R) incompletely dominant to White ( R) Heterozygotes are (Rr)= pink

a) Red Flower X White flower

Rr X rr

x	R	R
r	Rr	Rr
r	Rr	Rr

All Rr

All Pink

b.

F2 = F1 X F1

= Rr x Rr

x	R	r
R	RR	Rr
r	Rr	rr

25% RR , 50% Rr, 25% rr

25% Red; 50% Pink ; 25% White

c. Red Plant x Pink Plant

RR x Rr

x	R	R
R	RR	RR
r	Rr	Rr

50% Red (RR) and 50% Pink (Rr)

d. Pink Flower X White Flower

Rr x rr

x	R	r
r	Rr	Rr
r	Rr	rr

¾ Pink and ¼ White

Q8. Roan (Rr) Red (RR) white (rr)

Roan x Roan = 1 red ; 2 roan; 1 white

a) Red (RR) x White (rr) = all Roan (rr)

b) Roan (Rr) x Roan (Rr) = 25% Red; 50 % Roan; 25% White

c) White(rr) x White (rr) = White (rr)

d) Red (RR) x Roan (Rr) = 50% Red (RR) ; 50 % Roan (Rr)

Q9. A pure breeding roan herd cannot be produced since roan is a heterozygous condition. Such heterozygotes will always produce some homozygotes.

Q10. Black X^b Orange X^B Calico X^b X^B

Orange male X^B Y x Black Female X^b X^b

x	X^b	X^b
X^B	X^BX^b	X^BX^b
Y	Y X^b	Y X^b

Female Cats all Calico (X^b X^B )

Male Cats all Black (Y X^b )

Q11. a) Woman X^H X^h and the man is YX^H

b) P =0.25

c) Carrier P= 0.5, Haemophiliac P = 0.5

d) Haemophiliac son P =0.25

Q12. Woman Red-green Colourblind X^r X^{r x} Male Normal Vision Y X^R

x	X^r	X^r
X^R	X^RX^r	X^RX^r
Y	Y X^r	Y X^r

All males have red green colourblindness (Y X^r)

All females are normal vision but are carriers of the colourblind allele

Q13. Cystic fibrosis is found on the autosomal chromosomes. It is a recessive disorder in which the allele key is dominant C^Fand C^f for the recessive allele.

a. C^F C^f X C^F C^f

x	C^F	C^f
C^F	C^F C^F	C^F C^F
C^f	C^F C^f	C^f C^f

b. 75%