Meiosis

Figure 7 shows how human sperm cells are produced in the testes from diploid A₁ epithelial cells.

Use Figure 7 to calculate the minimum number of A₁ cells needed to produce

Give your answer in standard form.

Show your working.

A₁ cells

A PS cell goes through interphase before a meiotic division.

A scientist determined that the mass of DNA in one PS cell at the start of interphase is 8 arbitrary units.

Draw a bar chart in Figure 8 showing the mass of DNA in one cell in each of these phases:

• at the start of interphase

• at the end of interphase

• at the end of the first meiotic division

• at the end of the second meiotic division.

A PS cell contains 46 chromosomes.

Figure 9 shows the appearance of six of those chromosomes just as a PS cell enters meiosis.

Using the information in Figure 9, draw chromosomes in Figure 10 as they would appear in one cell at the end of the first meiotic division.

A mutation can produce a gamete with one extra chromosome.

Name the type of mutation that produces a gamete with one extra chromosome.

Explain how meiosis can produce a gamete with one extra chromosome.

Give three structural differences between an mRNA molecule and a tRNA molecule.

Table 3 shows mRNA codons and the amino acid coded by each codon.

Figure 5 shows the mRNA base sequence produced when part of a gene coding for an enzyme is transcribed.

Use Table 3 to give the amino acids coded by the mRNA base sequence in Figure 5

A mutation occurred in the part of the gene transcribed in Question 06.2.

Figure 6 shows the mRNA base sequence produced when the identical part in the mutated gene is transcribed.

The amino acids coded by this mRNA base sequence form part of the enzyme’s active site.

Use all the information in this question to:

• name the type of mutation that occurred to produce the mutated gene

• give the change in DNA caused by this mutation

• explain the effect this mutation will have on the function of the enzyme.

Explanation of effect on function of enzyme

Describe how a quaternary protein is formed from its monomers.

Do not include the process of translation in your answer.

Describe the structure of DNA and the structure of a chromosome.

Mutation can result in an increase in genetic variation within a species.

Describe and explain the other processes that result in increases in genetic variation within a species.

Describe viral replication.

Complete Table 2 by putting a tick () where the feature is part of a cell cycle involving mitosis or a cell cycle involving binary fission.

Figure 6 represents a cell undergoing meiosis. It shows the chromosomes in the parent cell and in the two cells formed after the first meiotic division.

The second division of meiosis proceeds normally except that non-disjunction occurs in the chromosome labelled N.

Complete Figure 6 to show the chromosomes inside the daughter cells formed after the second meiotic division.

Doctors studied babies born with a mutation caused by chromosome non-disjunction during gamete formation in their mother.

They determined each mother’s age at the time of childbirth and whether the non-disjunction happened in the first meiotic division (MM1 error) or in the second meiotic division (MM2 error).

Figure 7 shows the doctors’ results.

Figure 7

Figure 7 not reproduced here due to third-party copyright restrictions

A student concluded that there were more mothers of age ˃37 with MM2 errors than with MM1 errors.

Using Figure 7 and suitable calculations show why this conclusion is not valid.

Describe how mRNA is formed by transcription in eukaryotes.

Describe how a polypeptide is formed by translation of mRNA.

Define ‘gene mutation’ and explain how a gene mutation can have:

• no effect on an individual

• a positive effect on an individual.

Ulva lactuca is an alga that lives on rocks on the seashore. It is regularly covered by seawater.

Figure 6 shows a diagram of one Ulva lactuca alga.

Unlike plants, Ulva lactuca does not have xylem tissue.

Suggest how Ulva lactuca is able to survive without xylem tissue.

Ulva lactuca has a haploid and a diploid form.

Figure 7 shows the life cycle of Ulva lactuca.

On Figure 7 complete each box with an appropriate letter to show the type of cell division happening between each stage in the life cycle. Use ‘ T ’ to represent mitosis and ‘ E ’ to represent meiosis.

Ulva prolifera also produces haploid, mobile single cells that can fuse to form a zygote.

Suggest and explain one reason why successful reproduction between Ulva prolifera and Ulva lactuca does not happen.

Draw the general structure of an amino acid.

Table 1 shows mRNA codons and the amino acids coded for by each codon. It also shows some properties of the R group of each amino acid.

The genetic code is described as degenerate.

What is meant by this? Use an example from Table 1 to illustrate your answer.

A scientist investigated changes in the amino acid sequence of a human enzyme resulting from mutations. All these amino acid changes result from single base substitution mutations. This enzyme is a polypeptide 465 amino acids long.

Table 2 shows the result of three of the base substitutions.

What is the minimum number of bases in the gene coding for this polypeptide?

A change from Glu to Lys at amino acid 300 had no effect on the rate of reaction catalysed by the enzyme. The same change at amino acid 279 significantly reduced the rate of reaction catalysed by the enzyme.

Use all the information and your knowledge of protein structure to suggest reasons for the differences between the effects of these two changes.

Figure 6 shows a faulty form of meiosis that can occur in some plants.

Complete Figure 7 to show the chromosome content of the cells that would result from a normal meiotic division of the diploid parent cell shown in Figure 6.

The scientists tested their null hypothesis using the chi-squared statistical test. After 1 cycle their calculated chi-squared value was 350 The critical value at P=0.05 is 3.841

What does this result suggest about the difference between the observed and expected results and what can the scientists therefore conclude?

Use your knowledge of directional selection to explain the results shown in Table 3.

In women, the first division of meiosis produces one daughter cell that has almost all of the cytoplasm. The other daughter cell consists of a nucleus surrounded by a very small amount of cytoplasm and a cell-surface membrane. This very small daughter cell is called a polar body. Polar bodies do not usually divide. The same process occurs in the second division of meiosis, resulting in one egg cell and two polar bodies.

The diagram in Figure 3 shows the formation of an egg cell and two polar bodies during meiosis. It also shows what happens to one pair of homologous chromosomes. This pair carries two alleles of gene A.

Complete Figure 3 by putting A or a in the boxes. One box has been completed for you with A.

combination of alleles on the chromosome in the first polar body in Figure 3.

A scientist measured the diameter of a polar body and the diameter of the nucleus inside it. The diameter of the polar body was 10.4 µm and the diameter of the nucleus was 7.0 µm. The density of mitochondria in the cytoplasm of the polar body (outside of the nucleus) was 0.08 mitochondria per µm³.

Calculate the number of mitochondria in the polar body. You should assume polar bodies and nuclei are spherical.

The formula for the volume of a sphere is πr³ where π = 3.14

Show your working.

Mitochondrial diseases are caused by faulty mitochondria. All of a person’s mitochondria are inherited from their mother via the egg cell. An egg cell contains

One proposed treatment to prevent passing on faulty mitochondria involves

• removing the nucleus from an egg cell donated by a woman with healthy mitochondria

• replacing this nucleus with the contents of the polar body from a woman whose egg cells are affected by mitochondrial disease.

Suggest how this treatment prevents inheritance of mitochondrial diseases.

If most of the mitochondria in a cell are faulty, this prevents many important enzyme-catalysed reactions taking place or slows them down.

Suggest and explain one reason why.

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Contrast how an optical microscope and a transmission electron microscope work and contrast the limitations of their use when studying cells.

Figure 6 shows an image from an optical microscope of meiosis occurring in a flower bud of a flowering plant. W and Z are undergoing meiosis.

Explain the appearance of W and Z.

An environmental scientist investigated a possible relationship between air pollution and the size of seeds produced by one species of tree.

He was provided with a very large number of seeds collected from a population of trees in the centre of a city and also a very large number of seeds collected from a population of trees in the countryside.

Describe how he should collect and process data from these seeds to investigate whether there is a difference in seed size between these two populations of trees.