The Roles of Genes in Determining the Phenotype (CIE A Level Biology)

Shepherd's purse (Capsella bursa-pastoris) is a flowering plant that belongs to the mustard family. Fruit shape in this plant is determined by two alleles, namely allele T for a triangular fruit shape, which is dominant over allele t for top-shaped fruit. Two plants, both heterozygous for fruit shape, were crossed.

Complete the genetic diagram below (Fig.1) to show the results of this genetic cross.

Fig. 1

Explain why the parental phenotype was triangular shaped fruit with reference to their genotype.

A plant with triangular shaped fruit was crossed with a plant that has top-shaped fruit. All 35 offspring of this cross had triangular shaped fruit.

State whether there can be a certainty that the original parent plant with triangular shaped fruit had a genotype of TT and provide a reason for your answer.

The alleles for fruit shape in C. bursa-pastoris is an example of complete dominance.

Suggest the possible phenotype of heterozygous individuals if these alleles were codominant.

Salvia is a plant that forms part of the sage family and contains almost 1 000 different species. Flower colour in Salvia is controlled by two genes, A/a and B/b, located on different chromosomes. Allele B gives rise to purple flowers and is dominant over the allele for pink flowers b. In order for these colours to be expressed in the phenotype allele A must be present as well, otherwise, the result is white flowers.

Table 1 lists some of the genotypes that are possible in Salvia plants.

Table 1

Identify the phenotypes of the plants listed in Table 1.

Explain the interaction between gene loci that is responsible for the phenotypes identified in part a).

A homozygous pink-flowered Salvia is crossed with a homozygous white-flowered Salvia and all the offspring had purple flowers. Construct a Punnett square to show this cross.

Two purple Salvia plants from the genetic cross from part c) were crossed.

State the phenotypic ratios of the F₂ generation.

Aubergine (Solanum melongena) belongs to the same family as potatoes and tomatoes. They are a widely grown vegetable crop across Southeast Asia, Africa and the Mediterranean. Due to their importance as a food crop, scientists have been studying the inheritance of two genes (stem prickliness and fruit shape) that can have an influence on the profitability of the crop.

The allele P, for a non-prickly stem, is dominant over the allele p, for a prickly stem. The allele R, for round fruit, is dominant over the allele r, for linear fruit. The scientists examined 4000 offspring produced from the crosses between parent plants heterozygous for both genes.

Identify the missing phenotypes and genotypes of the offspring in Table 1

Table 1

State the expected phenotypic ratio for the genetic cross from part a) by completing Table 2, assuming that the genes were located on different chromosomes.

Table 2

The scientists found that the phenotypic ratios observed in the offspring did not correspond with the expected ratios listed in part b).

Suggest two possible reasons why the observed ratios do not reflect the scientists expected ratios.

In some plant species, height is partially controlled by the Le gene. The allele Le produces tall plants and is dominant over allele le, which produces short plants.

Explain the reason why the recessive allele le will result in short plants.

Factor VIII is a coagulating agent that plays an important role in blood clotting and is coded for by the F8 gene. There are abnormal alleles of this gene that lead to blood clotting abnormalities. A person that inherits these abnormal alleles will suffer from haemophilia. The F8 gene is an example of a sex-linked gene.

Define the term 'sex-linked gene'.

A gene, such as F8, can affect the phenotype of an organism.

Describe the mechanism by which the F8 gene can affect the phenotype of a person.

A woman that is a carrier of haemophilia and a haemophiliac male decide to have a child together.

Use the following symbols and complete the genetic diagram (Fig.1) showing the cross between them.

X^H = normal allele

X^h = allele for haemophilia

Fig. 1

Haemophilia in male offspring will always be inherited from their mother.

Explain why it is impossible for male offspring to inherit haemophilia from their father.

The black pigment melanin contributes to hair, skin and eye colour. Melanin is produced by cells known as melanocytes.

Tyrosinase is an enzyme involved in the production of melanin.

A study was carried out to investigate the effect of an extract of the starfish, Patiria pectinifera, on the activity of tyrosinase.

Table 1 shows the results of this study.

Table 1

Suggest how the starfish extract affects the activity of tyrosinase.

The dominant allele of the TYR gene codes for the enzyme tyrosinase.

In people with albinism, the melanocytes do not produce melanin. This is caused by a recessive allele of the TYR gene.

Explain what is meant by the terms recessive and allele.

[2]

(ii)

Construct a genetic diagram to show how a man and a woman, who both produce melanin, could have a child with albinism.

         Use appropriate symbols in your answer and state what they represent.

[3]

Mammals such as sheep, Ovis aries, and goats, Capra hircus, are important agricultural animals that are sometimes kept together in mixed flocks. Very occasionally, live offspring are born from a mating between a male sheep and a female goat.

In sheep 2n = 54 and in goats 2n = 60.

Normal (wild-type) goats have a gold and black coat colour pattern, known as bezoar, and are also horned (have horns). Domestic goats may have a white coat and may be hornless (do not have horns).

These variations are coded for by two unlinked genes:

• white coat colour, coded for by the dominant allele of the gene A/a.
• hornless, coded by the dominant allele of the gene H/h.

A cross between a white hornless goat and a bezoar horned goat produced offspring of four different phenotypes.

Draw a genetic diagram to show the genotypes of the two parents, their gametes and the offspring, and the phenotypes of the offspring.

Horns on agricultural animals such as goats and cattle can be dangerous to the farmer and to other animals. Horns are often prevented from growing in 5-day-old animals by a stressful procedure called disbudding.

Genetic modification can cause a deletion in the allele h coding for horns in cattle embryos, so that the allele no longer codes for a functional protein and the embryos grow into cattle that are hornless.

(i)

State an ethical advantage of this example of genetic modification.

(ii)

Suggest why genetic modification that causes a deletion in the horned allele, in established breeds of dairy cattle, is preferable to selective breeding for hornless animals.

Gibberellin is a plant growth hormone that has a role in germination and in stem elongation.

Outline how gibberellin is involved in activating genes for stem elongation.

Outline the role played by gibberellin in the germination of wheat seeds.

The length of stem in pea plants is controlled by a single gene. Pea plants can be either tall or short.

A study was carried out to investigate the effect of applying gibberellin to short pea plants.
Two groups of short pea seedlings were used, group P and group Q.

• Group P consisted of 20 seedlings to which a paste containing gibberellin had been applied two days after germination.
• Group Q consisted of 20 seedlings to which a paste without gibberellin had been applied two days after germination.
• The length of stem of the pea plants was recorded at intervals over 20 days.

The results are shown in Fig. 1.

With reference to Fig. 1, describe the results of the investigation and compare the growth rate of plants in group P and group Q.

Explain the role of the gene controlling stem length in pea plants.

In fruit flies (Drosophila melanogaster) wing length and body colour are each controlled by a single gene with two alleles. Allele L for long wings is dominant over allele l for vestigial wings, while allele G for grey body colour is dominant over allele g coding for ebony body colour.

Two homozygous fruit flies were crossed, one had a grey body colour and long wings while the other had an ebony body colour and vestigial wings.

State the number of offspring that would display a grey body colour and vestigial wings if 400 offspring were produced from this cross and explain your answer.

Two fruit flies from the cross in part a) were crossed and 4 800 offspring were produced.

Calculate the expected number of offspring that would display the phenotypes listed in Table 1. Assume that the genes for the body colour and wing length are not linked and show your working.

Table 1

The results for the cross from part b) were different from what was expected. Scientists decided to perform a chi-squared (χ²) test to determine if the difference is significant.

The formulae to calculate χ² are as follows:

Key:

Σ = sum of
O = observed value
E = expected value

Complete Table 2 and use this to calculate the value of χ² .

Table 2

Phenotype of offspring	Observed (O)	Expected (E)	(O - E)	(O - E)2
Grey body, long wings	2 580
Grey body, vestigial wings	900
Ebony body, long wings	1 010
Ebony body, vestigial wings	310

χ² = .....................................

The scientists calculated the degrees of freedom (v) by using the following formula:

v = n - 1

The symbol "n" refers to the number of classes in the data set.

Table 3 shows the values for χ² at different levels of probability and for different degrees of freedom.

Table 3

State the conclusion the scientists can make about the significance of their results and explain your answer.

In the fruit fly, Drosophila melanogaster, two different genes control body colour and eye colour.

• G/g are alleles of the body colour gene.
• G results in grey body, g results in black body.
  
  
• R/r are alleles of the eye colour gene.
• R results in red eyes, r results in brown eyes.

Each gene is autosomal.

A dihybrid cross was carried out using a fly with a grey body and red eyes crossed with a fly with a black body and brown eyes. Both parents were homozygous for both genes. The offspring from the F1 generation were crossed to obtain the F2 offspring.

A statistical test showed that the results of the cross were significantly different from those expected.

State the name of the statistical test used and state the expected phenotypic ratio for the F2 generation

A test cross can be carried out in order to identify flies from an F2 generation that are heterozygous for both genes.

Draw a genetic diagram to show how a test cross between a heterozygous grey‐bodied, red‐eyed F2 fly and a fly with a black body and brown eyes can produce four different offspring phenotypes.

Use the symbols G/g and R/r.

The results of the test cross in (b) are shown in Table 1. These results are significantly different from the expected results.

Table 1

Describe how these results are different from the expected results and explain why they are different.

Some neurones in the brain produce a neurotransmitter known as dopamine. Parkinson’s disease occurs when the neurones that produce dopamine die. A person with the disease may experience difficulty in coordinating movement, often seen as tremors (shaking) in different parts of the body.

Parkinson’s disease typically occurs in people older than 55 years. Younger people with these symptoms are said to have early onset Parkinson’s disease (EOPD).

Recessive mutations in a gene known as PINK1, located on chromosome 1, an autosome, are believed to be one cause of EOPD. A person with this form of EOPD has a homozygous recessive genotype.

Draw a genetic diagram of a cross between two individuals who are heterozygous at the PINK1 gene locus.

Include the following:

• key to symbols used for alleles
• parental genotypes
• gametes
• offspring genotypes
• ratio of offspring phenotypes

PINK1 codes for a protein kinase enzyme that is important in the functioning of mitochondria in neurones.

Most recessive PINK1 mutations are base substitutions which lead to the production of a non-functioning protein kinase enzyme.

Explain how a base substitution mutation can lead to the production of a non-functioning protein kinase.

One rare, dominant mutation of the PINK1 gene codes for a product that inhibits the normal protein kinase.

Explain how this mutation causes EOPD in a heterozygote.