Inherited change Banker

In a species of flowering plant grown in a research garden in Faisalabad, the allele for purple flowers, F, is dominant to the allele for white flowers, f.

1. (a) Explain what is meant by the terms 'dominant' and 'homozygous'. [2]
2. (b) A purple-flowered plant of unknown genotype is crossed with a white-flowered plant. Use genetic diagrams to show how the offspring would show the genotype of the purple-flowered plant. [4]

The ABO blood group gene has three alleles: Iᴬ and Iᴮ are codominant and both are dominant to Iᴼ. In a family in Peshawar, the father has blood group A and the mother has blood group B. Their first child has blood group O.

1. (a) State the genotypes of the father and the mother. Explain how you know. [2]
2. (b) Draw a genetic diagram to show the possible blood groups of their children and state the probability that their next child has blood group AB. [3]
3. (c) Explain why blood group AB shows codominance. [1]

In shorthorn cattle, the allele Cᴿ gives red hair and the allele Cᵂ gives white hair. Heterozygous cattle, CᴿCᵂ, are roan: they have a mixture of red hairs and white hairs.

1. (a) Explain why this is an example of codominance. [1]
2. (b) A farmer crossed roan bulls with roan cows and 24 calves were born. Draw a genetic diagram and state the expected numbers of red, roan and white calves. [4]

Haemophilia A is caused by a recessive allele, Xʰ, of a gene on the X chromosome. A woman whose father had haemophilia marries a man without haemophilia. Neither of them has haemophilia.

1. (a) State the genotypes of the woman and the man. [1]
2. (b) Draw a genetic diagram to show the possible genotypes and phenotypes of their children. [3]
3. (c) Calculate (i) the probability that a son of this couple has haemophilia, and (ii) the probability that any child of this couple will be a son with haemophilia. [2]
4. (d) Explain why haemophilia is much more common in males than in females. [1]

In a species of tomato, red fruit (R) is dominant to yellow fruit (r) and tall stem (T) is dominant to dwarf stem (t). The two genes are on different chromosomes. Two plants heterozygous for both genes were crossed and 320 offspring were produced.

1. (a) State the genotypes of the gametes that each parent can produce, and explain why there are four types. [2]
2. (b) Calculate the expected numbers of each phenotype among the 320 offspring. [3]

In summer squash, fruit colour is controlled by two genes on different chromosomes. A dominant allele W prevents any pigment forming, so the fruit is white. In ww plants, allele Y (dominant) gives yellow fruit and yy gives green fruit.

1. (a) Give the phenotypes of plants with genotypes WwYy, wwYy and wwyy. [1]
2. (b) Two WwYy plants are crossed and 160 plants are grown from the seeds. Calculate the expected number of white, yellow and green plants. Show your reasoning. [3]
3. (c) Explain how this cross differs from the 9 : 3 : 4 ratio seen in Labrador coat colour. [2]

Coat colour in rabbits is controlled by one gene with four alleles. In order of dominance: C (full colour) > cᶜʰ (chinchilla, silver-grey) > cʰ (Himalayan, white with dark extremities) > c (albino).

1. (a) A chinchilla rabbit with genotype cᶜʰc is crossed with a Himalayan rabbit with genotype cʰc. Draw a genetic diagram to show the expected genotypes and phenotypes of the offspring, and state the phenotypic ratio. [4]
2. (b) State the number of different genotypes possible for this gene in the rabbit population. [1]

In domestic cats, a gene on the X chromosome controls orange or black fur. Xᴼ gives orange fur and Xᴮ gives black fur. The alleles are codominant: a female XᴼXᴮ is tortoiseshell (patches of orange and black).

1. (a) A tortoiseshell female is mated with a black male. Draw a genetic diagram to show the phenotypes of the kittens and their sexes. [3]
2. (b) State the probability that a kitten from this cross is tortoiseshell. [1]
3. (c) Explain why tortoiseshell cats are almost always female. [2]

In a cereal plant, the allele for coloured grain (C) is dominant to colourless grain (c), and the allele for smooth grain (S) is dominant to wrinkled grain (s). A plant heterozygous for both genes, CcSs, which had received C and S from one parent and c and s from the other, was crossed with a plant with colourless, wrinkled grains. Offspring: coloured smooth 86; colourless wrinkled 82; coloured wrinkled 17; colourless smooth 15.

1. (a) State the ratio of phenotypes expected if the two genes assorted independently. [1]
2. (b) Explain why the results differ from this ratio. [3]
3. (c) Calculate the percentage of offspring that are recombinants. [1]

Coat colour in Labrador retriever dogs is controlled by two genes on different chromosomes. Gene B/b: B gives black pigment, b gives chocolate (brown) pigment. Gene E/e: at least one E allele allows pigment to be deposited in the hair; ee dogs are yellow whatever their B/b alleles.

1. (a) Name this type of gene interaction and explain why. [2]
2. (b) Two black dogs, both BbEe, are mated many times. Use a genetic diagram to show the expected phenotypic ratio of their puppies. [4]
3. (c) Over several litters the dogs had 64 puppies. Calculate the expected number of yellow puppies. [1]

In the fruit fly Drosophila melanogaster, white eyes are caused by a recessive allele, Xʷ, on the X chromosome (red eyes Xᵂ). Vestigial wings are caused by a recessive allele, v, of a gene on an autosome (normal wings V). A red-eyed female with normal wings, genotype XᵂXʷ Vv, is crossed with a white-eyed male with normal wings, genotype XʷY Vv.

1. (a) Calculate the probability that an offspring fly will be white-eyed with vestigial wings. Show your working. [3]
2. (b) 400 offspring were counted. Calculate the expected number of white-eyed flies with vestigial wings, and the expected number of these that are female. [2]
3. (c) Explain why a male fly cannot pass the Xʷ allele to his sons. [1]

1. (a) The lac operon of Escherichia coli contains a promoter, an operator and three structural genes. State what is meant by a structural gene and a regulatory gene. [2]
2. (b) Describe how the lac operon is kept switched off when lactose is absent. [3]
3. (c) Name the enzyme encoded by lacZ. [1]

1. (a) Explain what is meant by a transcription factor in a eukaryotic cell. [2]
2. (b) Suggest why different cell types in the same plant, such as guard cells and root hair cells, make different proteins although they contain the same genes. [2]

Mutant strains of E. coli were grown in a medium with no lactose and then in a medium with lactose. The amount of β-galactosidase was measured.

1. (a) Strain 1 has a mutation in the regulatory gene so that the repressor protein cannot bind to the operator. Predict and explain the amount of β-galactosidase made in each medium. [3]
2. (b) Strain 2 has a mutation in the regulatory gene so that the repressor still binds to the operator but cannot bind lactose (allolactose). Predict and explain the amount of β-galactosidase made in each medium. [3]

When a barley grain absorbs water, the embryo releases gibberellin (GA). GA causes cells of the aleurone layer to make amylase, which hydrolyses starch in the endosperm.

1. (a) Outline how gibberellin causes the gene for amylase to be transcribed in an aleurone cell. [4]
2. (b) Explain why the production of amylase is important for germination. [2]

A bacterium makes the amino acid tryptophan using five enzymes coded by genes in one operon. When tryptophan was added to the growth medium, the concentration of these enzymes fell to almost zero within 20 minutes.

1. (a) State whether the tryptophan-synthesising enzymes are repressible or inducible, and explain your answer. [2]
2. (b) Suggest how tryptophan could switch off transcription of this operon. [2]
3. (c) Explain the advantage to the bacterium of controlling these genes in this way. [2]

In eukaryotes, genes are not organised into operons. Each gene has its own promoter, and transcription starts only when the correct transcription factors and RNA polymerase have bound.

1. (a) Explain why the three structural genes of the lac operon are always transcribed together, while three genes coding for enzymes of one pathway in a eukaryote may be switched on separately. [2]
2. (b) A mutation changes the base sequence of the promoter of a eukaryotic gene so that a transcription factor that activates the gene can no longer bind. Suggest the effect on the cell. [2]
3. (c) State one way in which a transcription factor may be activated in response to a signal reaching a cell. [1]

A culture of E. coli growing without lactose was transferred at time 0 to a medium containing lactose and no glucose. β-galactosidase activity (arbitrary units): 0 min – 2; 2 min – 2; 5 min – 15; 10 min – 50; 20 min – 120; 30 min – 190.

1. (a) Calculate the mean rate of increase in β-galactosidase activity between 10 and 20 minutes, in arbitrary units per minute. [1]
2. (b) Explain the increase in β-galactosidase activity after lactose was added. [4]
3. (c) Suggest why there was no increase in activity in the first two minutes. [1]
4. (d) Explain why β-galactosidase is described as an inducible enzyme. [1]

1. (a) Semi-dwarf wheat varieties grown widely in Pakistan since the 1960s carry mutant alleles of the Rht genes, which code for DELLA proteins. The mutant DELLA proteins are not broken down when gibberellin binds to its receptor. Explain why these plants have short stems even though they make normal amounts of gibberellin. [3]
2. (b) In peas, tall plants have the dominant allele Le, which codes for an enzyme that converts an inactive gibberellin into active GA₁. The recessive allele le has a single base substitution. Suggest why lele plants are short. [3]
3. (c) Suggest one advantage to farmers of growing semi-dwarf wheat. [1]

1. (a) Explain what is meant by a homologous pair of chromosomes. [2]
2. (b) State the stage of meiosis in which (i) crossing over occurs and (ii) homologous chromosomes are separated. [2]
3. (c) State two ways in which the cells produced by meiosis differ from those produced by mitosis. [2]

1. (a) Explain how independent assortment in meiosis produces genetic variation among gametes. [2]
2. (b) A plant species has a diploid number of 14. Calculate the number of genetically different gametes one plant could produce by independent assortment alone. [1]
3. (c) Calculate the number of different chromosome combinations possible in a zygote produced from two plants of this species, by independent assortment and random fertilisation (ignore crossing over). [1]
4. (d) Explain why the real number of genetically different gametes is far larger than your answer to (b). [1]

A student examined squashed anther cells of an onion (2n = 16) undergoing meiosis. Cell P: pairs of homologous chromosomes (bivalents) lined up across the middle of the cell. Cell Q: two groups of chromosomes, each made of two chromatids, moving to opposite poles. Cell R: four groups of single chromatids, in two cells, being pulled to the poles.

1. (a) Identify the stage of meiosis shown in each cell. [3]
2. (b) State the number of chromosomes that will be present in each nucleus at the end of meiosis. [1]
3. (c) Explain why the separation of chromatids in meiosis II also contributes to genetic variation. [2]

The mass of DNA in a nucleus of a cell from the testis of a mammal was measured during one meiotic division. Mass in arbitrary units: before the S phase (G1) – 20; end of interphase – 40; after telophase I – 20; after telophase II – 10.

1. (a) Explain the changes in mass of DNA shown by these data. [4]
2. (b) A sperm cell fertilises an egg. State the DNA mass, in arbitrary units, of the zygote nucleus before it replicates, and explain the significance of this for the species. [2]

1. (a) Describe how meiosis and fertilisation together produce genetic variation in the offspring of sexually reproducing organisms. [6]

A dihybrid cross between two plants heterozygous for flower colour and leaf shape was expected to give a 9 : 3 : 3 : 1 ratio. Observed offspring: 93, 29, 31 and 7 (total 160).χ² = Σ (O − E)² ÷ E. Critical values at p = 0.05: 1 df – 3.84; 2 df – 5.99; 3 df – 7.82.

1. (a) Calculate the expected numbers for each class. [1]
2. (b) Calculate the value of χ². Give your answer to 2 decimal places. [2]
3. (c) State the number of degrees of freedom and explain what the result shows. [2]

A breeder in Karachi mated BbEe Labrador dogs (B black, b chocolate; ee dogs are yellow whatever their B/b alleles) and recorded 160 puppies over several years: 98 black, 22 chocolate, 40 yellow. The expected ratio is 9 black : 3 chocolate : 4 yellow.χ² = Σ (O − E)² ÷ E. Critical values at p = 0.05: 1 df – 3.84; 2 df – 5.99; 3 df – 7.82.

1. (a) State the null hypothesis. [1]
2. (b) Calculate χ². Give your answer to 2 decimal places. [3]
3. (c) State a conclusion, using the critical values. [2]

A geneticist has two true-breeding stocks of Drosophila: one with normal (red) eyes and one with a new mutant eye colour, 'sepia-like'. She wants to find out whether the mutant allele is recessive, and whether the gene is on the X chromosome or on an autosome.

1. (a) Describe how she could carry out breeding experiments to answer these questions. Include the crosses, how the flies are handled, and the results she would expect for each possibility. [7]

A student counted the kernels on one ear of maize produced by a test cross between a plant heterozygous for purple kernels (Pp) and a plant with yellow kernels (pp). There were 212 purple and 188 yellow kernels.χ² = Σ (O − E)² ÷ E. Critical value at p = 0.05 for 1 df is 3.84.

1. (a) Explain why a χ² test, not a t-test, is the appropriate statistical test for these data. [2]
2. (b) Calculate χ² and state a conclusion. [3]

A plant breeder thinks that seed coat colour in a variety of chickpea is controlled by one gene, with black seed coat dominant to cream. She crosses two plants that are both heterozygous.

1. (a) State the prediction she would make about the offspring if her hypothesis is correct. [1]
2. (b) State a null hypothesis for a χ² test on her results. [1]
3. (c) Suggest why she should grow a large number of offspring. [1]

Students will grow a non-pathogenic laboratory strain of E. coli on agar plates with and without lactose, and test for β-galactosidase activity using a colourless substrate that turns yellow when hydrolysed.

1. (a) Identify two hazards in this investigation and describe a precaution for each. [4]
2. (b) Explain why plates are incubated at about 25 °C rather than 37 °C in school laboratories. [1]