Using the provided information about dominant and recessive alleles for finger length and hairline shape, answer the following: State the genotype at Z, give the genotype of the parents, determine the

Here are the solutions to Questions 8, 9, and 10.

Question 8

• Dominant alleles: F (short fingers), H (widow's peak)
• Recessive alleles: f (long fingers), h (continuous hairline)

8.1. State the genotype at Z.
The genotype at Z is formed by combining the gamete 'fH' from the top row and 'fh' from the left column.

$$\text{fH} + \text{fh} = \text{ffHh}$$

The genotype at Z is ffHh.

8.2. Give the:

a) genotype of the parents
The problem states that both parents are heterozygous for both characteristics.
The genotype of the parents is FfHh.

b) number of genotypes that could result in offspring with short fingers and a continuous hairline

• Short fingers: F_ (FF or Ff)
• Continuous hairline: hh
• Possible genotypes: FFhh, Ffhh The number of genotypes is 2.

c) allele for a continuous hairline
Continuous hairline is the recessive trait.
The allele for a continuous hairline is h.

d) phenotype of a child who is homozygous recessive for both characteristics
Homozygous recessive for both characteristics means the genotype is ffhh.

• ff: long fingers
• hh: continuous hairline The phenotype is long fingers and continuous hairline.

Question 9

• Dominant alleles: T (tall stems), R (red fruit)
• Recessive alleles: t (short stems), r (yellow fruit)

9.1. Name this type of genetic cross.
This cross involves two different characteristics (stem height and fruit colour).
This is a dihybrid cross.

9.2. Give the genotype of a homozygous tall, yellow tomato plant.

• Homozygous tall: TT
• Yellow fruit (recessive): rr The genotype is TTrr.

9.3. List the genotypes of ALL the possible gametes for a plant that is heterozygous for both characteristics.
A plant heterozygous for both characteristics has the genotype TtRr.
The possible gametes are formed by combining one allele for stem height and one allele for fruit colour.
The genotypes of the possible gametes are TR, Tr, tR, tr.

Question 10

Haemophilia is a genetic disorder caused by a recessive allele on the X chromosome. A haemophiliac female marries a normal male. Explain why all their sons will be haemophiliacs.

Let X$^H$ be the normal allele and X$^h$ be the haemophilia allele.

• A haemophiliac female has the genotype X$^h$X$^h$. She can only pass on the X$^h$ allele to her offspring.
• A normal male has the genotype X$^H$Y. He passes on the X$^H$ allele to his daughters and the Y chromosome to his sons.
• When they have sons, the mother contributes an X$^h$ chromosome and the father contributes a Y chromosome.
• Therefore, all sons will inherit the genotype X$^h$Y, which means they will be haemophiliacs.