A2 Biology 9700: Monohybrid Cross for Albinism and a Test Cross

Albinism in humans is caused by a non-functional version of the enzyme tyrosinase, encoded by the TYR gene. Tyrosinase normally catalyses an early step in the production of the pigment melanin; without functional tyrosinase, no melanin can be made, and a person with albinism has unpigmented skin, hair and eyes. The allele for albinism is recessive to the allele for normal pigmentation.

A man and a woman, both with normal skin pigmentation, have a child who has albinism.

(a) Using the symbols T (dominant allele, functional tyrosinase) and t (recessive allele, non-functional tyrosinase), state the genotype of each parent, and explain how you arrived at your answer. [2] (b) Construct a genetic diagram for a cross between the two parents, and use it to determine the expected phenotypic ratio of their children with respect to skin pigmentation. [3] (c) If the couple have another child, state the probability that this child will have albinism, giving your answer as both a fraction and a percentage. [1] (d) A separate individual with normal skin pigmentation is thought to be either homozygous dominant or heterozygous for this gene. Explain how a test cross could be used to determine which of these two genotypes the individual has. [2]

Part (a): Deducing the parents’ genotypes

The child has albinism, so their genotype must be homozygous recessive, tt (this is the only genotype that gives the albino phenotype, since T is dominant to t). For the child to inherit a t allele from each parent, both parents must each carry at least one t allele.

Both parents have normal skin pigmentation, so neither of them can be tt (that genotype would give albinism). The only way a parent can show the dominant phenotype while still carrying a t allele is to be heterozygous.

So both parents have the genotype Tt.

Part (b): Genetic diagram and phenotypic ratio

Parents: Tt (father) x Tt (mother)

Gametes from father: T, t Gametes from mother: T, t

	T (from father)	t (from father)
T (from mother)	TT	Tt
t (from mother)	Tt	tt

Offspring genotypes: 1 TT : 2 Tt : 1 tt.

TT and Tt both have at least one functional T allele, so both produce enough functional tyrosinase for normal pigmentation. Only tt individuals lack functional tyrosinase entirely and show albinism.

Grouping the genotypes by phenotype gives 3 normally pigmented (TT, Tt, Tt) : 1 albino (tt), so the expected phenotypic ratio is 3 : 1 (normal pigmentation : albinism).

Part (c): Probability for the next child

Each fertilisation is an independent event, and the Punnett square in part (b) shows that 1 out of the 4 equally likely genotype combinations is tt (albino). This is true for every pregnancy this couple has, regardless of the phenotype of any earlier children.

So the probability that the next child has albinism is 1/4, which is 25%.

Part (d): Using a test cross to identify an unknown genotype

An individual showing the dominant phenotype (normal pigmentation) could have genotype TT or Tt, these cannot be told apart just by looking at the individual. A test cross resolves this by crossing the unknown individual with a homozygous recessive individual, genotype tt (someone with albinism), because this is the only cross where the recessive allele shows up in the offspring whenever the unknown parent carries it.

If the unknown individual is TT: cross TT x tt gives only Tt offspring. Every child shows normal pigmentation, and no child ever shows albinism.
If the unknown individual is Tt: cross Tt x tt gives offspring in the ratio 1 Tt : 1 tt. Approximately half of the children are expected to show albinism.

So, by observing whether any children with albinism appear among a reasonable number of offspring, the unknown individual’s genotype can be determined: the appearance of an albino child confirms the individual is heterozygous (Tt), while consistently normal-pigmented children over several offspring supports (though does not absolutely prove, given the small sample sizes typical of human families) a homozygous dominant genotype (TT).

Final answers

(a) Both parents have genotype Tt, because the albino child (tt) must have received a t allele from each unaffected parent.
(b) The cross Tt x Tt gives offspring in the phenotypic ratio 3 normally pigmented : 1 albino.
(c) The probability the next child has albinism is 1/4 (25%).
(d) A test cross against a homozygous recessive individual (tt) distinguishes the genotypes: no albino offspring indicates TT; the appearance of albino offspring indicates Tt.