Dihybrid crosses, linkage and epistasis

Plants are often used to investigate genetic inheritance.

Genetic crosses are used to investigate the inheritance of two genes at the same time. This is known as dihybrid inheritance.

Gregor Mendel investigated dihybrid inheritance in pea plants. When parent plants that were both heterozygous for two different genes were crossed, the ratio of phenotypes in the offspring was 9: 3: 3: 1.

Sweet pea is another species of plant.

• Sweet pea flower colour is coded for by a gene with two alleles: blue (B) and red (b).

• Sweet pea pollen shape is coded for by a gene with two alleles: long (L) and round (l).

When parents that were both heterozygous for the two genes were crossed, the ratio of phenotypes in the offspring was 3 (blue and long): 1 (red and round).

The inheritance of flower colour and pollen shape in sweet pea is not affected by epistasis.

Explain what can be concluded from the phenotypic ratio not being the expected 9: 3: 3: 1 for Mendelian dihybrid inheritance.

Explain why epistasis affects the offspring ratio in a genetic cross.

In pea plants, flower colour is controlled by two alleles: purple (dominant) and white (recessive).

• 600 pea plants are growing in a large greenhouse.

• 564 of the plants have purple flowers.

Use the Hardy–Weinberg principle to calculate the number of heterozygous individuals in the greenhouse.

Use the equations:

p + q = 1

p² + 2pq + q² = 1

Estimates made by using the Hardy–Weinberg principle can be inaccurate.

Use the information above to suggest one reason why the answer calculated in part (b)(i) might be accurate.

On a biology field trip, some students are studying an area of mixed woodland.

Three tree species dominate the woodland: oak, ash and sycamore.

There are many small herbaceous plants in the woodland, but the students choose to collect data on only one: wood sorrel.

The students choose an oak tree, an ash tree and a sycamore tree and count every individual wood sorrel plant within a 3 m radius of each tree.

The results are shown in the table.

The null hypothesis of the students’ investigation is that the species of tree would not affect the number of wood sorrel growing near it.

The students conclude that the conditions under oak trees are the most favourable of the three trees studied and that conditions under sycamore trees are the least favourable.

Calculate the |² value for the students’ data. (i)

You may use the empty columns in the table to help with your calculation.

The critical value for |² in this investigation at p = 0.05 is 5.991. (ii)

Discuss what can be concluded from the |² calculation in part. (a)(i)

The teacher suggested that, in order to have confidence in their results, the students should include data from more than one tree for each species.

The students collect all their results on the same day.

Identify two other variables the students should attempt to control when carrying out an investigation such as this.

The teacher described the mixed woodland as a climax community that had resulted from succession.

Succession involves a series of changes in community over time, known as seres.

* The first sere is the pioneer community. The final sere is the climax community.

The species that make up the pioneer and climax communities are different.

Compare the plant and animal species that make up the climax and the pioneer communities.

Extra answer space if required.

Succession involves changes in the species that live in an area over time.

State one other change that takes place in the same area over the same time.

Suggest what was present in the area that is now occupied by mixed woodland before the pioneer community arrived.

Scientists crossed two plants that were heterozygous for two different genes. Both plants had the same genotype, AaBb.

The genotypes and ratios of the offspring resulting from this cross were:

AABB AaBb aabb

1: 2: 1

Which option could explain these observations?

A Autosomal linkage

B Crossing over

C Epistasis

D Mutation

Students investigated the effect of light on the growth of garden cress seedlings.

• A total of 120 seedlings were divided into 2 groups of 60.

• Group A was grown in darkness for 2 days.

• Group B was grown for 1 day in darkness and then for 1 day in white light using the set‑up shown in Fig. 3.1.

The results of the students’ experiment are shown in Tables 3.1 and 3.2.

* Describe and explain the results shown in Tables 3.1 and 3.2.

Additional answer space if required.

The students wanted to test whether there was a significant difference between the stem lengths of the seedlings in Group A and the seedlings in Group B.

State the name of the most appropriate statistical test for the students to use.

Justify your choice of statistical test given in part (ii).

Table 3.2 records the direction of growth as:

• away from light

• neither away from nor towards light

• towards light.

The students used the chi‑squared test to determine whether the direction of root growth was significantly different from their expectations.

Their null hypothesis was:

There is no difference between the expected direction of root growth and the observed direction of root growth.

The calculated chi‑squared value was 8.10.

The students compared their chi‑squared value of 8.10 to the values in Table 3.3.

What can the students conclude about their results based on a chi‑squared value of 8.10?

The solution in which the seedlings were grown contained various dissolved ions.

The table below lists two of the functions of these ions in the seedlings.

Complete the table below with the formula of each ion.

The students also investigated the effect of plant hormone concentration on root growth.

State the name of a plant hormone that would be expected to affect root growth.

In the investigation, the students controlled light, temperature and mineral concentration.

State one other factor that the students should have controlled in this investigation.

The onion plant, Allium cepa, is grown as a food crop around the world.

The table below contains statements about the root cells of an onion.

Place ticks (✓) in the boxes in the table to indicate whether the statements are true or false.

Fig. 1 shows a cross section of the root of an onion plant.

Identify the tissues shown at M and N.

The colour of onion bulbs is determined by two genes, A/a and B/b.

• A is a dominant allele and codes for the production of a red pigment.

• Onion bulbs that are homozygous for the recessive allele, a, produce no pigment and are white.

• B is a dominant allele that inhibits the expression of allele A.

• The recessive allele, b, allows the production of the red pigment.

A white onion plant was cross-pollinated with a red onion plant. All 15 offspring had the genotype AaBb.

Identify the following:

The genotype of the white onion plant........

The genotype of the red onion plant........

The phenotype of the offspring........

State the type of gene interaction shown by the genes A/a and B/b.

Suggest how allele B inhibits the expression of allele A.

The sweet pea plant has been used to study inheritance since the nineteenth century. The seeds of the sweet pea can vary in colour and shape.

The gene that controls colour has two alleles:

• Y is dominant and produces yellow seeds.

• y is recessive and produces green seeds.

The gene that controls shape has two alleles:

• R is dominant and produces round seeds.

• r is recessive and produces wrinkled seeds.

In the nineteenth century, Gregor Mendel crossed a pea plant that was heterozygous for both seed colour and shape with a pea plant that had green and wrinkled seeds.

List the gametes that would be produced by a sweet pea plant that was heterozygous for both seed colour and shape.

List the genotypes of the offspring that were produced from Mendel’s cross and state the corresponding phenotypes.

When Mendel crossed two pea plants that were heterozygous for both seed colour and shape, the ratio of phenotypes in the offspring was:

• 9 yellow round

• 3 green round

• 3 yellow wrinkled

• 1 green wrinkled.

Some students tried to recreate this investigation using a modern variety of plant that showed the same phenotypic variation in seed colour and shape.

The students crossed two of the modern plants that were heterozygous for both seed colour and shape. The results of this cross were:

• 58 yellow and round

• 31 green and round

• 21 yellow and wrinkled

• 2 green and wrinkled

The students used the chi-squared test to compare their data to the expected 9: 3: 3: 1 ratio.

Table 17 shows a |² probability table.

After analysing the results, the students stated that the inheritance of the seed colour and shape in their investigation was different from that in Mendel’s investigation.

Using Table 17, discuss whether the results of the investigation and the chi-squared test support the students’ statement.

A ratio that is different from the expected 9: 3: 3: 1, in a cross such as this, can be the result of epistasis.

Suggest and explain one reason, other than epistasis, why the phenotype ratio might not be 9: 3: 3: 1.

The yellow colour in peas is the result of an enzyme that breaks down chlorophyll, which is green.

• The Y allele codes for the production of an enzyme that breaks down chlorophyll.

• The y allele is the result of a mutation in the Y allele.

• The y allele codes for an inactive form of this enzyme.