Question 1 (10 marks). The normal color of snapdragons is red.
Some pure lines showing variations of flower colour have been found .
When these lines were crossed they gave the following results.

What is the mode of inheritance: dihybrid with one allelic series plus epistastic modifier

Write the genotypes of the parents, the F1 and the F2 in the places provided.
 
 

	Parents	F1	F2
Phenotypes: Genotypes:	Orange X Yellow yoyo X yy	Orange yoy	3 orange: 1 yellow yo_ : yoyo
Phenotypes: Genotypes:	Red X Orange y+y+ X yoyo	Red y+yo	3 red: 1 orange y+_: yoyo
Phenotypes: Genotypes:	Red X Yellow y+y+ X yy	Red y+y	3 red: 1 yellow y+_: yy
Phenotypes: Genotypes:	Red X White w+w+ X ww	Red w+w	3 red: 1 white w+_: ww
Phenotypes: Genotypes:	Yellow X White w+w+ yy X ww y+y+	Red w+w y+y	9 red: 3 orange: 4 white w+_ y+_: w+_ yy: ww _ _
Phenotypes: Genotypes:	Orange X White w+w+ yoyo X ww y+y+	Red w+w y+yo	9 red: 3 orange: 4 white w+_ y+_: w+_ yoyo: ww _ _
Phenotypes: Genotypes:	Red X White w+w+ y+y+ X ww yy	Red w+w y+y	9 red: 3 yellow: 4 white w+_ y+_: w+_ yy: ww _ _

Question 1b (10 marks).  Some Himalayan rabbit breeds have either black or red-furred extremities.
The following is the result of a set of carefully done experiments.
Write the genotypes of the parents, the F1 and the F2 in the places provided.

	Parents	F1	F2
Phenotypes: Genotypes:	Himalayan black X albino r+r+ chch  X r+r+ cc	Himalayan black r+r+ chc	3 Himalayan black: 1 albino r+r+ ch__ : r+r+ cc
Phenotypes: Genotypes:	black X Himalayan black r+r+ c+c+ X r+r+ chch	black r+r+ c+ch	3 black: 1 Himalayan black r+r+ c+__: r+r+ chch
Phenotypes: Genotypes:	black X albino r+r+ c+c+ X r+r+ cc	black r+r+ c+c	3 black: 1 albino r+r+ c+__: r+r+ cc
Phenotypes: Genotypes:	Himalayan red X albino rr chch X rr cc	Himalayan red rr chc	3 Himalayan red: 1 albino rr ch__: rr cc
Phenotypes: Genotypes:	red X Himalayan red rr c+c+ X rr chch	red rr c+ch	3 red: 1 Himalayan red rr c+__: rr chch
Phenotypes: Genotypes:	red X albino rr c+c+ X rr cc	red rr c+c	3 red: 1 albino ______ : rr cc
Phenotypes: Genotypes:	black X red r+r+ c+c+ X rr c+c+	black r+r c+c+	3 black: 1 red r+__ c+c+: rr c+c+
Phenotypes: Genotypes:	red X Himalayan black rr c+c+X r+r+ chch	black r+r c+ch	9 black: 3 Himalayan black: 3 red: 1 Himalayan red r+__ c+__: r+__ chch: rr c+__: rr chch
Phenotypes: Genotypes:	red X albino rr c+c+X r+r+ cc	black r+r c+c	9 black: 3 red: 4 albino r+__ c+__: rr c+__: _____ cc
Phenotypes: Genotypes:	Himalayan black X Himalayan red r+r+ chch X rr chch	Himalayan black r+r chch	3 Himalayan black: 1 Himalayan red r+__chch: rr chch

The best way to describe the relationship between the Himalayan condition and coat colour is separate pathways.
The best way to describe the relationship between the albino condition and coat colour is epistasis.
The best way to describe the relationship between the Himalayan and albino traits is allelic.
 
 

Question 2 (10 marks). A mutant allele in mice causes a kinked tail when compared to the normal straight tail.
Six pairs of mice were crossed to investigate the genetic basis of the kinked tail phenotype.
The phenotypes of the parents and progeny are listed below.
 

Cross	female parent	male parent	female progeny	males progeny
1  Genotypes:	normal tail t+ t+	kinked tail tK Y	All kinked tail tK t+	All normal tail t+ Y
2  Genotypes:	kinked tail tK t+	normal tail t+ Y	half kinked half normal tK t+; t+ t+	half kinked half normal tK Y; t+ Y
3  Genotypes:	kinked tail tK tK	normal tail t+ Y	All kinked tail tK t+	All kinked tail tK Y
4  Genotypes:	normal tail t+ t+	normal tail t+ Y	All normal tail t+ t+	All normal tail t+ Y
5  Genotypes:	kinked tail tK tK	kinked tail  tK Y	All kinked tail  tK tK	All kinked tail tK Y
6  Genotypes:	kinked tail tK t+	kinked tail  tK Y	All kinked tail tK tK; tK t+	half kinked half normal tK Y; t+ Y

What is the mode of inheritance: X-linked dominant.
Give the Genotypes of Parents and Progeny in the places available in the chart.
 

Question 3 (10 marks). This pedigree deals with a very rare trait.

What is the mode of inheritance: autosomal recessive.

In this pedigree, the probability that indicated individual is a carrier is Ö
 
 

Individual Designation	Probability of being a carrier
II.5	P=1.0 (1/1)
II.8	P=1.0 (1/1)
III.5	P=0.5 (1/2)
IV.5	P=0.25 (1/4)
IV.11	P=0.6667 (2/3)