Data analysis: biological molecules and cells

Question 1

Newborn calves have very few antibodies in their blood. They obtain antibodies from colostrum, the first milk produced by their mother. For a short time after birth, cells lining the calf's small intestine can absorb whole antibody molecules, but this ability decreases with time.

Farm scientists fed four groups of 12 newborn calves the same volume of colostrum at different times after birth. They measured the concentration of antibodies in each calf's blood 48 hours after birth. They also recorded how many calves in each group developed a gut infection during their first month. The table shows their results.

Calculate the percentage decrease in the mean concentration of antibodies when colostrum was fed at 24 hours compared with 1 hour. Give your answer to 2 significant figures.

Explain why the antibodies from colostrum give calves only short-term protection against infection.

A farmer concluded that feeding colostrum within 1 hour of birth prevents gut infections in calves. Evaluate this conclusion.

Accepted Answer

Percentage decrease = (32 - 6) ÷ 32 × 100 = 26 ÷ 32 × 100 = 81.25 = 81%

This is passive immunity: the antibodies were made by the mother, not by the calf. The calf's own B cells are not activated, so no plasma cells or memory cells are produced, and the absorbed antibodies are broken down over time and are not replaced.

The data support a link: only 1 of the 12 calves fed at 1 hour developed an infection, compared with 7 of 12 fed at 24 hours, and the early-fed calves had the most antibodies. However, one early-fed calf still became infected, so early feeding reduces the risk rather than preventing infection, and the groups were small.

Question 2

Scientists investigated whether preparing cells for a transmission electron microscope (TEM) changes their size. Tetrahymena is a single-celled organism with no cell wall. The scientists measured the lengths of 50 living Tetrahymena cells with an optical microscope. They prepared other cells from the same culture for the TEM: they fixed the cells with a chemical, removed the water using ethanol, embedded the cells in resin and cut very thin sections. They then measured the lengths of 50 cells in electron micrographs of these sections.

The table shows their results. A statistical test showed that the difference between the mean lengths was significant (P < 0.001).

In one electron micrograph, the image of a Tetrahymena cell had a length of 86 mm. The magnification of the micrograph was × 2000. Calculate the actual length of this cell in μm.

Explain why specimens for a TEM must have the water removed and must be cut into very thin sections.

The scientists concluded that preparing Tetrahymena cells for the TEM makes the cells shrink. Evaluate this conclusion.

Accepted Answer

Image length = 86 mm = 86 000 μm
Actual length = image size ÷ magnification = 86 000 ÷ 2000 = 43 μm

The inside of a TEM is a vacuum, so that electrons are not scattered by air molecules; any water in the specimen would evaporate in the vacuum, so it must be removed. The image is formed by electrons passing through the specimen, so it must be very thin.

The difference in mean length (48 μm compared with 39 μm) is significant, so it is unlikely to be due to chance, and removing water could make the cells shrink. However, a thin section often does not pass through the middle of a cell, so the length in a section can be smaller than the real length even if the cell has not shrunk. Also, different cells were measured with each microscope, so the conclusion is not fully supported.

Question 3

Lactase tablets can be swallowed with milk by people who do not produce lactase. Scientists compared two lactase preparations used in these tablets: lactase F, from a fungus, and lactase Y, from a yeast. They measured the initial rate of hydrolysis of lactose by each enzyme at different pH values, at 37 °C. In every test they used the same concentration of lactose and the same mass of enzyme protein. The graph shows their results.

Calculate the percentage decrease in the rate of reaction of lactase Y when the pH is lowered from 7 to 4. Give your answer to 3 significant figures.

Explain why lactase Y has a very low rate of reaction at pH 3.

After a meal containing milk, the stomach contents stay at about pH 4 for an hour or more. The contents of the small intestine are at about pH 6 to 7. A company claims that tablets containing lactase F are better than tablets containing lactase Y for people who do not produce lactase. Use the graph and your own knowledge to evaluate this claim.

Accepted Answer

Percentage decrease = (76 - 10) ÷ 76 × 100 = 66 ÷ 76 × 100 = 86.8%

At pH 3 there is a high concentration of H⁺ ions. These change the charges on the R groups of amino acids in lactase Y, so ionic and hydrogen bonds holding the tertiary structure break. The shape of the active site changes, so lactose is no longer complementary to it and very few enzyme-substrate complexes form.

For the claim: at pH 4, the pH in the stomach after a milky meal, lactase F hydrolyses lactose about 9 times faster than lactase Y (86 compared with 10), so much more lactose would be hydrolysed in the stomach. Lactase F is still quite active at pH 6 (60).

Against the claim: in the small intestine (pH 6 to 7) lactase Y is faster (70 and 76 compared with 60 and 22). Both enzymes are proteins and may be hydrolysed by proteases in the stomach. The data come from test tubes at constant pH, not from people, so the tablets need to be tested in a trial.

Question 4

Glucose produced by hydrolysing the cellulose in wheat straw can be fermented to make ethanol for fuel. In plant cell walls, cellulose molecules form microfibrils, which are surrounded by other polymers, including lignin.

Scientists investigated four ways of treating 10 g samples of straw before adding the enzyme cellulase: chopped only (untreated), ground into a fine powder, heated with dilute acid, and heated with dilute alkali, which removes most of the lignin. They incubated each sample with the same concentration of cellulase at 50 °C and measured the mass of glucose released over 48 hours. The graph shows their results.

Each 10 g sample of straw contained 4.0 g of cellulose. Complete hydrolysis of this cellulose would produce 4.4 g of glucose. Calculate the percentage of the cellulose that was hydrolysed in 48 hours in the alkali-treated straw. Give your answer to 3 significant figures.

Describe a control that the scientists should have included for each treatment, and explain why it is needed.

Suggest explanations for the effects of grinding and of alkali treatment on the release of glucose.

A biofuel company wants to choose one of these treatments. Evaluate the use of alkali treatment.

Accepted Answer

Mass of glucose at 48 hours (alkali) = 3.6 g
Percentage hydrolysed = 3.6 ÷ 4.4 × 100 = 81.8%

For each treatment, the scientists should have incubated treated straw in the same way but with boiled (denatured) cellulase, or water, instead of cellulase. This would show that the glucose was released by cellulase and not by the treatment itself: heating with acid, for example, can hydrolyse some glycosidic bonds.

Grinding increases the surface area of the straw, so more cellulose is exposed and more enzyme-substrate complexes form per minute. Lignin surrounds the cellulose microfibrils, so removing it with alkali lets cellulase reach and bind to much more of the cellulose.

Alkali treatment released the most glucose (3.6 g, 82% of the maximum) and released it fastest (1.6 g in the first 8 hours, compared with 0.2 g for untreated straw). However, heating with alkali costs energy and chemicals, and the alkali would have to be neutralised before the enzyme and yeast were added. Only one temperature and one cellulase concentration were tested, so the company would need further trials.

Sample questions with model answers

Question 1

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Question 2

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Question 3

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Question 4

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