Inheritance: Biology 9700 (Cambridge International AS & A Level)

The lac operon in the bacterium Escherichia coli controls the production of enzymes that allow the bacterium to digest the sugar lactose. The operon includes a regulatory gene that produces a repressor protein, and an operator sequence that this repressor protein can bind to. When no lactose is present, the repressor protein binds tightly to the operator, physically blocking RNA polymerase from transcribing the structural genes lacZ, lacY and lacA.

Which statement correctly describes what changes when lactose becomes available in the bacterium's environment?

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]

The ABO blood group system in humans is controlled by a single gene with three alleles: IA and IB, which are codominant with each other, and i, which is recessive to both IA and IB. A person's blood group depends on their genotype as follows: genotype IAIA or IAi gives blood group A; genotype IBIB or IBi gives blood group B; genotype IAIB gives blood group AB; and genotype ii gives blood group O.

A man with blood group A has the genotype IAi. A woman with blood group B has the genotype IBi. They have children together.

(a) State which two alleles of this gene show a codominant relationship with each other, and which allele is recessive to both of the others. [2] (b) Construct a genetic diagram for a cross between this man and this woman, and use it to determine the possible blood groups of their children and the expected ratio between them. [4] (c) Explain how it is possible for this couple, who have blood groups A and B themselves, to have a child with blood group O. [2]

Haemophilia A is a blood clotting disorder caused by a recessive allele of the F8 gene, which is carried on the X chromosome. The F8 gene normally codes for clotting factor VIII; a person with two copies of the recessive allele (or, in a male, one copy) cannot produce functional factor VIII and does not clot blood normally.

A woman does not have haemophilia herself, but her father does have haemophilia. This woman has children with a man who does not have haemophilia.

(a) Using the symbols X^H (dominant allele, normal clotting) and X^h (recessive allele, haemophilia), state the genotype of the woman's father, and explain how the woman must have inherited an X^h allele from him even though she does not have haemophilia herself. [2] (b) Construct a genetic diagram for a cross between this woman and her partner, and use it to determine the expected genotypes and phenotypes of their children with respect to haemophilia, showing daughters and sons separately. [4] (c) Explain, in terms of the number of X chromosomes each sex inherits, why sons are more likely than daughters to show the haemophilia phenotype in crosses such as this one. [2]

A gardener is investigating two genes in a species of ornamental plant called the moonflower. Petal colour is controlled by gene P, with allele P (purple, dominant) and allele p (white, recessive). Leaf shape is controlled by a separate gene, gene L, with allele L (broad leaves, dominant) and allele l (narrow leaves, recessive).

The gardener crosses two plants that are both heterozygous at both loci (genotype PpLl). If the two genes assort completely independently, a dihybrid cross of this kind is expected to produce offspring in the ratio "9 purple, broad-leaved : 3 purple, narrow-leaved : 3 white, broad-leaved : 1 white, narrow-leaved".

The gardener grows 160 offspring from this cross and records the following numbers in each phenotype category:

• purple, broad leaves: 100
• purple, narrow leaves: 15
• white, broad leaves: 15
• white, narrow leaves: 30

The gardener carries out a chi-squared test to compare these observed numbers with the numbers expected if the two genes assort independently.

(a) State a suitable null hypothesis for this chi-squared test. [1] (b) (i) Calculate the expected number of offspring in each of the four phenotype categories, assuming a "9:3:3:1" ratio for a total of 160 offspring. [1] (b) (ii) Using the formula "chi-squared = the sum of (observed minus expected) squared, divided by expected", calculate the value of chi-squared for the gardener's data. Show your working. [3] (c) The critical value of chi-squared at the p = 0.05 significance level, for the appropriate number of degrees of freedom, is 7.815. Compare this critical value with your calculated value, and state what you conclude about the two genes. [2] (d) In some pairs of genes, the alleles of one gene mask or alter the phenotypic effect of alleles at a completely different gene locus; this is called epistasis. State one way in which epistasis differs from the type of gene interaction identified in part (c). [2]

During meiosis in a diploid cell, two separate events contribute to genetic variation before fertilisation even occurs. In prophase I, non-sister chromatids of each pair of homologous chromosomes can exchange sections of DNA at a chiasma. Later, in metaphase I, each pair of homologous chromosomes lines up at the equator of the cell independently of every other pair, so that the maternal and paternal chromosomes are distributed to the two poles of the cell in random combinations.

Which statement correctly distinguishes the genetic variation produced by these two events?

A species of nematode worm has a diploid chromosome number of 8, meaning its body cells contain 4 pairs of homologous chromosomes. Gametes are produced by meiosis.

(a) State what happens to the pairs of homologous chromosomes during metaphase I and anaphase I that allows independent assortment to generate genetic variation. [2] (b) Assuming no crossing over takes place, calculate the number of genetically different combinations of whole chromosomes that a single individual of this species can produce in its gametes through independent assortment alone. Show your working. [2] (c) Explain why crossing over during prophase I means that the actual number of genetically distinct gametes this worm can produce is far greater than the number calculated in part (b). [2] (d) Explain how random fertilisation provides a further source of genetic variation in the offspring of this species, in addition to the variation already generated during meiosis. [2]

In a species of fruit fly, body colour is controlled by gene G, with allele G (grey body, dominant) and allele g (ebony body, recessive). Wing length is controlled by a separate gene, gene W, located on a different chromosome, with allele W (normal wings, dominant) and allele w (vestigial wings, recessive).

A fly that is heterozygous for both genes (genotype GgWw, grey body with normal wings) is crossed with another fly that is also heterozygous for both genes (genotype GgWw).

(a) State the gametes that each parent fly can produce with respect to these two genes. [2] (b) Construct a genetic diagram for this cross and use it to determine the expected phenotypic ratio of the offspring with respect to body colour and wing length. [4] (c) Explain, with reference to meiosis, why the alleles of gene G and the alleles of gene W assort independently of each other in this cross. [2]

In a species of leaf beetle, body colour is controlled by two genes on different chromosomes. Gene E codes for an enzyme needed to make any pigment at all: allele E (functional enzyme, dominant) allows pigment to be produced, while allele e (non-functional enzyme, recessive) means no pigment can be made, regardless of the genotype at the second gene. Gene B then determines which pigment is made, but only in beetles that can produce pigment at all: allele B (black pigment, dominant) is dominant to allele b (brown pigment, recessive).

A beetle that is heterozygous at both genes (genotype EeBb, black) is crossed with another beetle that is also heterozygous at both genes (genotype EeBb).

(a) Define the term epistasis, referring to genes E and B in your answer. [2] (b) Construct a genetic diagram for this cross and use it to determine the expected phenotypic ratio of the offspring, showing how the genotypes are grouped into phenotypes. [4] (c) A beetle has the genotype eeBB. Explain why this beetle is not black, even though it carries two copies of the dominant allele B for black pigment. [2]

In a species of guinea pig, coat colour is controlled by a single gene, with allele C (black, dominant) and allele c (cream, recessive). A breeder crosses two black guinea pigs that are both heterozygous (Cc), expecting offspring in the ratio 3 black : 1 cream if the C and c alleles behave as simple dominant and recessive alleles.

Out of 320 offspring from this cross, the breeder counts 228 black guinea pigs and 92 cream guinea pigs.

Using the formula chi-squared equals the sum of (observed minus expected) squared, divided by expected, which of the following is the calculated value of chi-squared for these results, to 1 decimal place?