Question

Determine the genotypes and corresponding phenotypes for the traits listed below.

• Example: In a plant, purple flowers are dominant to white flowers.
• Answer:    PP = homozygous dominant = purple flowers;  Pp = heterozygous = purple flowers; pp = homozygous recessive = white flowers

Part A questions:

1. In a plant, red flowers are dominant to white flowers.

2. In cats, normal tail length is dominant to bobbed tails.

3. In pea plants, yellow peas are dominant to green peas.

   Part B: Use Punnett Squares to answer the questions. Example: In humans, brown eyes are dominant to blue eyes. A man with brown eyes marries a woman with blue eyes. Together they have a son with blue eyes. What are the genotypes of the parents? Answer: The parents are: Bb (brown eyes) x bb (blue eyes)

4. In humans, having dimples is dominant. A man with dimples marries a woman with dimples; they reproduce and have child with no dimples. What are the genotypes of the child's parents?

5. In humans, lactose intolerance (inability to digest lactose) is recessive. Brian is lactose intolerant, and Carolyn is not (she can digest lactose). If both Brian and Carolyn are homozygous for this trait, what are the possible genotypes, and corresponding phenotypes, of their children?

6. In a species of birds, feather color follows incomplete dominance. Birds homozygous for one allele are black, and birds homozygous for the other allele are white. If a black bird and a white bird mate, what are the possible genotypes, and corresponding phenotypes, of their offspring?

7. In humans, hemophilia is an X-linked recessive disorder. Is it possible for a healthy (no hemophilia) man and a woman who is a carrier to have a son with hemophilia?

8. Shown above is the Punnett Square for the example question.

Answer 1

Answer part A : 1) In a plant, red flowers are dominant to white flowers.

Answer: RR = homozygous dominant = Red flowers; Rr = heterozygous = Red flowers; rr = homozygous recessive = white flowers.

2) In cats, normal tail length is dominant to bobbed tails.

Answer: NN = homozygous dominant = normal tail length: Nn = heterozygous = normal tail length; nn = homozygous recessive = bobbed tails.

3) In pea plants, yellow peas are dominant to green peas.

Answer: YY = homozygous dominant = Yellow peas; Yy = heterozygous = yellow peas; yy = homozygous recessive = green peas.

Answer part B:

1. In humans, having dimples is dominant. A man with dimples marries a woman with dimples; they reproduce and have child with no dimples. What are the genotypes of the child's parents?

Answer: The genotype of the parents are: Dd (with dimples) x Dd (with dimples)

(Note none of the parents can be homozygous, in that case all the children will be having dimples.)

Punnett square

gamates	D	d
D	DD With dimples	Dd With dimples
d	Dd With dimples	dd No dimples

2. In humans, lactose intolerance (inability to digest lactose) is recessive. Brian is lactose intolerant, and Carolyn is not (she can digest lactose). If both Brian and Carolyn are homozygous for this trait, what are the possible genotypes, and corresponding phenotypes, of their children?

Answer: The genotype of the parents are: ll (Brian, lactose intolerant) x LL (Carolyn, lactose tolerant)

Punnett square

gamates	l	l
L	Ll lactose tolerant	Ll lactose tolerant
L	Ll lactose tolerant	Ll lactose tolerant

3. In a species of birds, feather color follows incomplete dominance. Birds homozygous for one allele are black, and birds homozygous for the other allele are white. If a black bird and a white bird mate, what are the possible genotypes, and corresponding phenotypes, of their offspring?

Answer: The genotype of the parents are: BB(black) x WW (white)

Punnett square

gamates	B	B
W	BW Grey	BW Grey
W	BW Grey	BW Grey

As the character of feather color is showing incomplete dominance, hence both the characters will be expressed partially in the birds with genotype BW and they will show the phenotype of grey color.

4. In humans, hemophilia is an X-linked recessive disorder. Is it possible for a healthy (no hemophilia) man and a woman who is a carrier to have a son with hemophilia?

Answer: The genotype of the man will be XY ( a man is having single X chromosome, hence if he is having even if a single allele, he will be haemophilic, in this case since the man is without haemophilia hence his genotype is XY)

The genotype of the woman will be XX^h ( a woman can be a carrier of the disease with the single allele, The possibility of a women having hemophila is less,since they need to have the defective allele on both the X chromosomes)

Punnett square

gamates	X	Y
X	XX Normal daughter	XY Normal son
Xh	XXh Carrier daughter	XhY Haemophilic son

Hence for a healthy (no hemophilia) man and a woman who is a carrier to have a son with haemophilia is there, there are 50% (1/2) chances of the son being haemophilic.