Carbohydrates

Complete Table 1 by adding a tick (✓) if the feature is present in the biological molecules.

Lactase catalyses the hydrolysis of lactose.

A student investigated the effect of galactose on lactase activity.

The student used the following method.

1. Mix together (in tube X) solutions of lactose, galactose and lactase in a 15: 3: 1 volume ratio so that the total volume is 90 cm³

2. Leave the mixture at 30 °C for 10 minutes.

3. Measure the glucose concentration in the mixture.

4. Repeat steps 1 to 3 with two different mixtures.

Calculate the volume of galactose solution used in tube X.

Table 2 shows the student’s results.

Galactose is a competitive inhibitor of lactase.

Use this information and your knowledge of enzymes to explain the results obtained in Table 2.

Give:

• one change the student could make in tube X to confirm that galactose is a competitive inhibitor

• the result you would expect.

Molasses is a solution obtained from sugar beet plants. The sugars present in molasses are sucrose, glucose and fructose.

Give the number of different types of monosaccharides present in molasses.

A student used the biochemical test for reducing sugars on a clear sample of molasses.

Describe the biochemical test for a reducing sugar.

Explain the result expected from the test on the sample of molasses.

Description of biochemical test

Explanation of expected result

‘Free sugar’ is the sugar in food and drinks released when food is crushed or when sugar is added to food at home or by manufacturers.

Scientists recommend that no more than 5% of the energy consumed per day should come from ‘free sugar’.

The mean daily energy requirement for a 10-year-old child is 8100 kJ

The ‘free sugar ’ in one tablespoon of molasses contains 250 kJ of energy.

Calculate the number of tablespoons of molasses required for a 10-year-old child to reach the recommendation for energy consumed in ‘free sugar’ per day.

A scientist used the apparatus in Figure 3 to investigate osmosis.

Use your understanding of osmosis to explain why the air pressure in the tube increased.

The scientist repeated the investigation, but made one change to the molasses. The scientist did not change the volume of molasses at the start of the investigation.

The scientist observed that the air pressure inside the tube increased by 160 kPa compared with 800 kPa in the first investigation.

Suggest the change the scientist made to the molasses to cause this smaller increase in air pressure.

Use the air pressure figures in a calculation to support your answer.

Chitin is a polysaccharide. The chitin monomer is a β-glucose molecule with one OH group replaced by an NHCOCH₃ group. NHCOCH₃ can be represented by N(Ac).

Figure 1 shows the monomer that forms chitin and the chitin polymer.

Chitin has a similar structure to cellulose.

Use Figure 1 to describe three ways the structure of chitin is similar to the structure of cellulose.

Chitin keeps the tracheae open in the tracheal system of gas exchange in an insect. Gas exchange does not occur in the tracheae.

Explain the importance of one adaptation of the gas exchange surface in the tracheal system of an insect.

Lignin is a polymer found in the walls of xylem vessels in plants. Lignin keeps the xylem vessel open as a continuous tube.

Explain the importance of the xylem being kept open as a continuous tube.

A student investigated the effect of different sugars on the rate of respiration in yeast. Yeast normally respires glucose.

Figure 2 shows the method she used for her first experiment.

Other than those stated, suggest two variables the student needed to keep constant in her investigation.

Figure 3 shows the result she obtained for yeast in glucose solution.

Use Figure 2 and Figure 3 to calculate the rate of carbon dioxide production in mm s^–1 for yeast in glucose solution.

Give your answer in standard form and to 2 significant figures.

Show your working.

The student repeated the experiment using yeast in maltose solution. She found the rate of carbon dioxide production was slower than with yeast in glucose solution. Suggest why.

Describe the structure and function of the nucleus.

Name the main polymer that forms the following cell walls.

Scientists investigated the effect of the number of fungal species in soil on the diversity of plant species.

Table 1 shows their raw data for soil containing 14 fungal species.

Suggest one reason the scientists used biomass instead of the number of individuals of each plant species when collecting data to measure diversity.

The scientists used this equation to calculate the plant species index of diversity.

where n = shoot biomass of each plant species and N = total shoot biomass of all plant species

Use this equation to calculate the index of diversity for the data in Table 1.

Index of diversity

Figure 1 shows the plant species index of diversity the scientists calculated when the soil contained 0, 1, 2, 4 and 8 fungal species.

Figure 1

Figure 1 not reproduced here due to third-party copyright restrictions

Sometimes farmers stop growing crops on an area of land to allow the natural ecosystem to recover. The plant species index of diversity of these areas previously used to grow crops is different from nearby land that has never been used to grow crops.

Suggest and explain how the plant species index of diversity would be different in these areas previously used to grow crops.

Use Figure 1 and your knowledge of the effect of farming on biodiversity in your answer.

Clostridium difficile is a bacterial species that causes disease in humans.

Antibiotic-resistant strains of C. difficile have become a common cause of infection acquired when in hospital.

Explain how the use of antibiotics has led to antibiotic-resistant strains of bacteria becoming a common cause of infection acquired when in hospital.

Scientists suggested that factors, other than antibiotic use, led to the increase in antibiotic-resistant C. difficile infections. One suggested factor is people eating more trehalose in their diet.

Trehalose is a disaccharide formed from two glucose molecules.

Name another disaccharide formed from two glucose molecules.

Scientists investigated the effect of trehalose on the growth rate of C. difficile.

They grew populations of non-resistant and antibiotic-resistant C. difficile on separate agar plates with:

• no carbohydrate added

• trehalose added.

They measured the growth rate of the C. difficile.

Figure 2 shows the scientists’ results.

Describe how the scientists could use aseptic techniques to transfer 0.3 cm³ of C. difficile in liquid culture from a bottle onto an agar plate.

Use Figure 2 to evaluate whether more trehalose in the diet could be a factor in the increased number of antibiotic-resistant C. difficile infections.

Describe the transport of carbohydrate in plants.

Compare and contrast the structure of starch and the structure of cellulose.

Describe the complete digestion of starch by a mammal.

Alpha-gal is a disaccharide found in red meat.

Alpha-gal is made of two galactose molecules. Galactose has the chemical formula

Give the chemical formula for the disaccharide, alpha-gal, and describe how it is formed from two galactose molecules.

Some people eat red meat for many years without having any reaction, then have an allergic reaction to the alpha-gal in red meat.

An allergic reaction is caused by an immune response.

Draw a labelled diagram of an antibody and identify the specific alpha-gal binding site.

A tick is a small animal that bites humans and feeds on their blood. This results in proteins from the tick saliva entering the human body.

Scientists have suggested one hypothesis for the allergic reaction to alpha-gal in red meat. They think that an earlier immune response to a tick bite can cause a person to have an allergic reaction to alpha-gal in red meat.

Suggest how one antibody can be specific to tick protein and to alpha-gal.

Scientists took blood samples from one man over several weeks and measured the concentration of antibody in the man’s blood. During this time, the man had two tick bites and had an allergic reaction to alpha-gal in red meat.

The scientists’ results are shown in Figure 7.

Figure 7

This figure has been removed due to third-party copyright restrictions.

The scientists’ hypothesis was that an earlier immune response to tick protein causes the allergic reaction.

Consider whether Figure 7 supports this hypothesis.

Explain five properties that make water important for organisms.

Describe the biochemical tests you would use to confirm the presence of lipid, non-reducing sugar and amylase in a sample.

Describe the chemical reactions involved in the conversion of polymers to monomers and monomers to polymers.

Give two named examples of polymers and their associated monomers to illustrate your answer.

Describe the gross structure of the human gas exchange system and how we breathe in and out.

Mucus produced by epithelial cells in the human gas exchange system contains triglycerides and phospholipids.

Compare and contrast the structure and properties of triglycerides and phospholipids.

Mucus also contains glycoproteins. One of these glycoproteins is a polypeptide with the sugar, lactose, attached.

Describe how lactose is formed and where in the cell it would be attached to a polypeptide to form a glycoprotein.

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.