Mitosis

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.

Strains of bacteria with antibiotic resistance are common in hospitals.

Use your knowledge of selection to explain why.

Antibiotic resistance alleles are located in plasmids.

There are 11 different types of plasmid inside cells of the bacterium Bacillus thuringiensis.

A scientist determined the:

• length of each type of plasmid in kilobase pairs (kbp)

• mean plasmid copy number (mean number of each type of plasmid per cell).

Table 5 shows the scientist’s results for 5 of these plasmids.

The Spearman rank correlation coefficient (r_s) between plasmid length and mean plasmid copy number is determined using this equation.

where Σd² = the sum of all d² figures n = 5 (the highest rank number)

Use information in Table 5 and the formula to calculate the r_s value for this set of data.

Show your working.

The scientist determined that the r_s value between plasmid length and mean plasmid copy number for all 11 types of plasmid is –0.94

What can you conclude from this r_s value?

The length of the single circular (non-plasmid) DNA molecule in a B. thuringiensis cell is 5400 kbp

The total length of all plasmid DNA molecules in a B. thuringiensis cell is 65% greater than the length of the single circular DNA molecule.

Calculate the total number of kbp in all the plasmid molecules in a B. thuringiensis cell .

Show your working.

A growing population of B. thuringiensis cells includes a:

• germination phase

• growth phase

• dormant phase.

A scientist incubated a culture of B. thuringiensis cells at 20 °C

At intervals, the scientist:

• measured the cloudiness of the culture

• determined the mean plasmid B copy number per cell.

Cloudiness is a measure of the number of cells in a culture.

Figure 11 shows the scientist’s results.

Using information in Figure 11, what can you conclude about binary fission and the

replication of plasmid B in the different phases of a growing population of B. thuringiensis ?

Describe binary fission in bacteria.

Microbiologists investigated the effect of two temperatures on the population growth of a species of bacterium.

They grew separate bacterial cultures in identical nutrient media at 20 °C and 24 °C for 8 hours.

Figure 1 shows the estimated final population size of each culture.

The bacterial population at 24 °C is larger than that at 20 °C

Calculate this difference as a percentage increase.

The actual number of bacteria can be calculated from the log₁₀ value by using the 10^x button on your calculator.

Give your answer to the nearest whole number.

Show your working.

Percentage increase =

Bedaquiline is a drug that inhibits ATP synthase in mycobacteria by stopping hydrogen ion (proton) movement.

Explain how bedaquiline can kill these bacteria.

Describe the appearance and behaviour of chromosomes during prophase and during anaphase of mitosis.

A scientist used an optical microscope to determine the mitotic index in cells at different distances from the tip of onion roots.

Figure 4 shows the results.

Figure 4

Figure 4 not reproduced here due to third-party copyright restrictions

The figure is taken from the following website:

https://legacy.nimbios.org//~gross/bioed/webmodules/mitoticindex.htm#:~:text= Interpretation%3A%20In%20general%2C%20the%20mitotic,of%20the%20root's %20dividing%20cells

Complete the word equation used to determine each mitotic index in Figure 4.

Mitotic Index =

The scientist used data from Figure 4 to calculate a correlation coefficient (r). The scientist then used a statistical test to determine the probability (P) associated with the value of r.

r = –0.98 (P < 0.05)

What can you conclude from this result?

What can you conclude about the effect that distance from the root tip has on the proportion of cells in different stages of the cell cycle?

Use information in Figure 4.

Scientists investigated a drug called MiTMAB as a treatment for cancer. MiTMAB inhibits cytokinesis.

Figure 8 shows drawings of cancer cells seen with an optical microscope from a:

• sample treated with MiTMAB

• control sample.

Figure 8

The cells in drawing A can be identified as those treated with MiTMAB.

Explain why.

MiTMAB acts as a non-competitive inhibitor of an enzyme called dynamin.

Suggest how MiTMAB can cause dynamin to become inactive.

When active, dynamin has two functions:

it stimulates cytokinesis

it inhibits cell death.

The scientists treated actively growing cultures of cancer cells with MiTMAB.

They incubated:

• one sample of 2500 cells without MiTMAB as a control

• eight samples, each with 2500 cells and a different concentration of MiTMAB.

After 72 hours, the scientists measured the number of cells in each sample.

Figure 9 shows the scientists’ results.

A negative value for proportion of control growth means that fewer than 2500 cells were counted after 72 hours.

Use all the information given to explain the results shown in Figure 9.

0.01 dm³ of MiTMAB solution was added to the treated cells.

Calculate the increase in mass of MiTMAB (in μg) added to the cells to reduce the cell growth from equal to the control to 0.0 of the control.

Show your working.

Dengue fever is a human disease caused by the dengue virus.

Scientists designed an ELISA test to detect antibodies to the dengue virus in a patient’s blood sample.

Figure 10 shows a diagram of this test and some information about how it works.

Figure 11 shows the negative and positive results that were produced 20 minutes after the use of the test shown in Figure 10.

Suggest what is on the test at line T and explain what causes the line to appear in a positive test.

A line at C shows that the test has worked.

Suggest one reason why a line at C shows the test has worked.

Figure 12 shows a flowchart of how the anti-human antibodies with enzyme attached are produced.

Suggest why the fused cells allow continuous production of monoclonal antibodies.

Evaluate the ethics of the production process shown in Figure 12.

Early identification of dengue fever can be difficult as many other diseases produce

the same symptoms. Early identification is important because people suffering with dengue fever can become ill very quickly and may need hospital treatment.

Scientists compared the effectiveness of three diagnostic tests for dengue fever.

• Laboratory-based test – a patient’s blood sample is sent from the doctor’s clinic to a laboratory for testing.

• Current test used in the doctor’s clinic.

• New test to be used in the doctor’s clinic – the ELISA test shown in Figures 10 and 11 (on page 24).

The scientists’ results are shown in Table 3.

A blood sample from each patient with confirmed dengue fever at each time after onset of symptoms was tested with all three diagnostic tests.

The scientists recommend that the new test is used for the identification of dengue fever in all countries around the world.

Discuss this recommendation. Use all the information given.

The dengue virus causes damage to capillaries so that blood proteins move out of the capillaries into the tissue fluid.

Explain how this would affect the return of tissue fluid into the capillaries.

Below are four statements about the structure of prokaryotic cells.

1. No prokaryotic cell has DNA that is associated with proteins.

2. No prokaryotic cell has membrane-bound organelles.

3. All prokaryotic cells have one or more flagella.

4. All prokaryotic cells have smaller ribosomes than eukaryotic cells.

Which statements about the structure of prokaryotic cells are correct?

Tick () one box.

A student investigated the effect of two antibiotics on the growth of the bacterium Micrococcus luteus.

During the investigation, the student:

• transferred 9 cm³ of a liquid culture of M. luteus into each of three bottles

• added the antibiotic chloramphenicol to the first bottle

• added the antibiotic novobiocin to the second bottle

• added no antibiotic to the third bottle.

After 24 hours, he diluted the contents of each bottle by 1 in 100 000 (10^–5). He then transferred 0.25 cm³ samples from the first bottle onto each of 3 separate agar plates. He repeated this with 0.25 cm³ samples from the second bottle and the third bottle, resulting in 9 agar plates in total.

He incubated the plates for 48 hours.

Table 1 shows the number of colonies of bacteria he counted on each plate after 48 hours’ incubation.

Calculate the mean number of bacteria in the undiluted bottle of liquid culture containing novobiocin.

Give your answer in standard form. Show your working.

Starting with a single bacterium, calculate how many generations it would take to produce at least the number of bacteria you have calculated for your answer in Question 03.2

You can assume no bacteria die.

You could use the ln or log button on your calculator to calculate your answer.

M. luteus is not resistant to chloramphenicol.

Suggest two reasons why the bacteria were able to grow in the culture containing chloramphenicol.

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.

A student prepared a stained squash of cells from the root tips of garlic to calculate a mitotic index. He:

1. cut the end 5 mm from 10 garlic roots

2. placed the root tips into a Petri dish containing 5 cm³ of hydrochloric acid for 12 minutes

3. rinsed the root tips in distilled water

4. placed one of the root tips on a microscope slide and added toluidine blue stain

5. placed a coverslip onto the microscope slide, and gently pressed the coverslip downwards on the root tip

6. observed the root tip using an optical microscope.

Suggest why the student soaked the root tips in hydrochloric acid in step 2.

Pressing the coverslip downwards enabled the student to observe the stages of mitosis clearly.

Explain why.

Figure 6 shows the student’s drawing of one field of view.

Name the stage of mitosis shown in cell G. Explain the appearance of this cell.

Use Figure 6 to calculate a mitotic index for the cells in this field of view.

Other students in the class followed the same method, but calculated different mitotic indices.

Apart from student errors, suggest two explanations why.

Figure 7 shows the mean distance between centromeres and the poles (ends) of the spindle during mitosis.

Calculate the rate of movement of the centromeres during phase E.

Give your answer in μm minute^–1 and to 3 decimal places.

Name the three phases of mitosis shown by C, D and E on Figure 7.

Describe the role of the spindle fibres and the behaviour of the chromosomes during each of these phases.

Describe the role of DNA polymerase in the semi-conservative replication of DNA.

Figure 5 shows the percentage of rat cells undergoing DNA replication. Some cells contained a protein called cyclin D and some cells did not contain cyclin D. All cells were in early interphase at time 0

It took less time for 25% of cells with cyclin D to be undergoing DNA replication than for 25% of cells without cyclin D.

Use Figure 5 to calculate this time difference as a percentage decrease.

Show your working.

Cyclin D stimulates the phosphorylation of DNA polymerase, which activates the DNA polymerase.

Describe how an enzyme can be phosphorylated.

Some tumour cells contain higher than normal concentrations of cyclin D.

Use Figure 5 to suggest why higher than normal concentrations of cyclin D could result in a tumour.

Scientists investigated the cell cycle in heart cells taken from mice 6 days before their birth and then at 4, 14 and 21 days after their birth.

Their results are shown in Table 4. Age 0 days = day of birth.

Describe and explain the data in Table 4.

The scientists determined the percentage of heart cells undergoing DNA replication by using a chemical called BrdU. Cells use BrdU instead of nucleotides containing thymine during DNA replication.

Describe how BrdU would be incorporated into new DNA during semi-conservative replication.

Cells with BrdU in their DNA are detected using an anti-BrdU antibody with an enzyme attached.

Use your knowledge of the ELISA test to suggest and explain how the scientists identified the cells that have BrdU in their DNA.

Uncontrolled cell division can cause tumours to form.

Figure 7 shows the growth pattern followed by a type of tumour.

Use Figure 7 to calculate the percentage of maximum growth this type of tumour reaches before it can be detected.

You will need to use the 10^x button on your calculator.

Figure 7 can also be used to calculate the age of this type of tumour.

At diagnosis, a patient had a tumour of 3.98 10¹¹ cells. Calculate the age of the tumour.

You will need to use the log₁₀ button on your calculator.

Trexall is a drug that can be used to slow the development of various forms of cancer.

Trexall slows cell division by interacting with an enzyme called dihydrofolate reductase (DR).

DR is involved in making nucleotides; the substrate for DR is folic acid.

Figure 8 shows the chemical structure of Trexall.

Figure 9 shows the chemical structure of folic acid.

Use the information provided to suggest how Trexall slows cell division.

Doctors investigated how the concentration of Trexall given to patients affected the growth of lung tumours. The doctors measured the volume of tumours at the beginning of the study and after 8 months.

Figure 10 shows the results of this investigation. The bars represent ± 2 standard deviations. A value of ± 2 standard deviations from the mean includes over 95% of the data.

The scientists measured the percentage change in tumour volume.

Suggest why they recorded both percentage change and tumour volume.

Percentage change

A lung cancer patient received 15 mg of Trexall per week. After treatment, the diameter of his lung tumour was 35.8 mm

Assuming the tumour was spherical, use the mean percentage change in tumour volume shown in Figure 10 to calculate the volume of the patient’s tumour before treatment with Trexall.

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

To reduce the size of tumours, would it be better to use 30 mg of Trexall per week, or 20 mg of Trexall per week?

Explain your answer.

Trexall can also be used to slow the development of rheumatoid arthritis (a pain-causing joint disease).

Scientists investigated the effectiveness of Trexall as a pain relief treatment in 12 rheumatoid arthritis patients. All of the patients were female. They randomly divided the patients into two groups:

• Group R received Trexall tablets for 35 days

• Group S was a control group.

They asked both groups to rate their pain on a scale of 0–10 (0 being no pain and 10 being the worst pain possible) at the start and then every 7 days for 35 days. They calculated mean scores for each group.

Their results can be seen in Table 3.

Apart from age and general health, give two important factors when choosing patients for this investigation.

A student analysed Table 3 and concluded that Trexall was effective in reducing pain in arthritis patients.

Evaluate the student’s conclusion.

Turn over for Section B

Section B

one question.

You are advised to spend no more than 45 minutes on this section.

Give two pieces of evidence from Figure 1 that this cell was undergoing mitosis. Explain your answers.

When preparing the cells for observation the scientist placed them in a solution that had a slightly higher (less negative) water potential than the cytoplasm. This did not cause the cells to burst but moved the chromosomes further apart in order to reduce the overlapping of the chromosomes when observed with an optical microscope.

Suggest how this procedure moved the chromosomes apart.

The dark stain used on the chromosomes binds more to some areas of the chromosomes than others, giving the chromosomes a striped appearance.

Suggest one way the structure of the chromosome could differ along its length to result in the stain binding more in some areas.

In Figure 2 the chromosomes are arranged in homologous pairs. What is a homologous pair of chromosomes?

Give two ways in which the arrangement of prokaryotic DNA is different from the arrangement of the human DNA in Figure 1.

In mammals, in the early stages of pregnancy, a developing embryo exchanges substances with its mother via cells in the lining of the uterus. At this stage, there is a high concentration of glycogen in cells lining the uterus.

Describe the structure of glycogen.

During early pregnancy, the glycogen in the cells lining the uterus is an important energy source for the embryo.

Suggest how glycogen acts as a source of energy.

Do not include transport across membranes in your answer.

Suggest and explain two ways the cell-surface membranes of the cells lining the uterus may be adapted to allow rapid transport of nutrients.

In humans, after the gametes join at fertilisation, every cell of the developing embryo undergoes mitotic divisions before the embryo attaches to the uterus lining.

The first cell division takes 24 hours.

The subsequent divisions each take 8 hours.

After 3 days, the embryo has a total volume of 4.2 × 10^–3 mm³.

What is the mean volume of each cell after 3 days? Express your answer in standard form.

Show your working.

Bacteria are often used in industry as a source of enzymes. One reason is because bacteria divide rapidly, producing a large number of them in a short time.

Describe how bacteria divide.

Washing powders often contain enzymes from bacteria. These enzymes include proteases that hydrolyse proteins in clothing stains.

Figure 1 shows the effect of temperature on a protease that could be used in washing powder.

Explain the shape of the curves at 50 C and 60 C.

Some proteases are secreted as extracellular enzymes by bacteria.

Suggest one advantage to a bacterium of secreting an extracellular protease in its natural environment.

Explain your answer.

Mammals have some cells that produce extracellular proteases. They also have cells with membrane-bound dipeptidases.

Describe the action of these membrane-bound dipeptidases and explain their importance.