Biological Molecules (OCR A Level Biology): Exam Questions

In plants the glucose produced by photosynthesis is changed into starch for storage. Glucose and starch are both carbohydrates.

(i) Complete the table below to show the main differences in structure between glucose and starch.

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(ii) Carbohydrates contain only the elements carbon, hydrogen and oxygen. Polypeptides contain carbon, hydrogen, oxygen and nitrogen. Name one other element present in polypeptides.

 [1]

Lipids are an important group of biological molecules.

Lipoproteins are roughly-spherical structures that transport lipids in the blood. Fig. 21 shows a simplified drawing of a section from the widest part of a lipoprotein.

Fig. 21

(i) Calculate the number of phospholipid molecules in the outer surface of the lipoprotein shown in Fig. 21.

Assume that the pattern of proteins and phospholipids shown in Fig. 21 is continued across the whole surface of the lipoprotein. Use the formula: Surface area of sphere =

[2]

number of phospholipid molecules = ...........................................

(ii) Lipoproteins with fewer phospholipids and more protein in the outer layer are known as high density lipoproteins.

Lipoproteins with a larger number of phospholipids but less protein are known as low density lipoproteins.

Use this information to explain why lipids can increase the buoyancy of aquatic animals.

[1]

Complete the passage by choosing the most appropriate word from the list.

bile       carbon         hydrogen         insoluble

nitrogen      oxygen       permeability        production       solid      soluble

stability          storage           vitamins

Lipids have many roles in living organisms. Some are used for energy ................................ in adipose cells. Unsaturated fatty acids contain at least one double bond between two ................................ atoms and so contain fewer ................................ atoms. All lipids are ................................ in water so need to be transported in the blood by lipoproteins. Cholesterol molecules increase the ............................ of membranes, and cholesterol is also used to synthesise steroid hormones and ............................ .

Triglycerides are a type of lipid molecule that can be broken down during hydrolysis reactions.

Using the structure of triglyceride molecules as an example, explain what is meant by hydrolysis.

Rubredoxin is a protein found in bacteria. It contains around 50 amino acids. One iron ion is bound by the sulphur atoms of four cysteine amino acids.

The structure of rubredoxin is shown in Fig. 20.1.

Fig. 20.1

(i)   Rubredoxin is known as a conjugated protein.

Use Fig. 20.1 to explain what is meant by the term conjugated protein.

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(ii) Using the information provided about rubredoxin, state two similarities between the structures of rubredoxin and haemoglobin.

[2]

(iii) Rubredoxin and haemoglobin have different secondary and tertiary structures. Using the information provided about rubredoxin, state two other differences between the structures of rubredoxin and haemoglobin.

[2]

Ferritin is a protein that is used to regulate iron levels within plant tissues.

It is a large spherical structure which can hold many iron (Fe³⁺) ions at its centre.

Iron can be toxic to plant tissues. Ferritin prevents the build-up of iron.

Fig. 20.2 shows the internal structure of ferritin.

Fig. 20.2

(i) Ferritin molecules can hold 4500 Fe³⁺ ions in the inner sphere.

It is thought that the Fe³⁺ ions are unable to occupy the total available volume of the inner sphere because other molecules are present in the inner sphere.

The volume of the inner sphere of the ferritin molecule is 268nm³.

The volume of an Fe³⁺ ion is 9.04 × 10^–4 nm³.

Calculate the volume of the inner sphere not occupied by Fe³⁺ ions.

volume = .................................................. nm³ [3]

(ii) Explain how hydrophilic and hydrophobic interactions contribute to the spherical shape of ferritin.

 [1]

Fig. 20 shows the disaccharide lactose, which is found in milk.

Fig. 20

Another disaccharide is maltose. Maltose and lactose both contain the same number of atoms of each element, C, H and O.

(i) State two other structural similarities between lactose and maltose.

[2]

(ii) Complete the table below to identify differences between the structures of lactose and maltose. The first one has been completed for you. Lactose Maltose one glucose monomer and one galactose monomer both monomers are glucose            

[3]

One of the monomers of lactose is galactose. The bacterium E. coli usually uses glucose as a respiratory substrate.

Under certain circumstances, E.coli is able to use galactose as a respiratory substrate by breaking down lactose into glucose and galactose and then using both glucose and galactose as respiratory substrates.

(i) Explain how the structure of galactose allows it to be used as a respiratory substrate.

 [3]

(ii) E. coli usually grows in conditions where the extracellular concentration of lactose is low.

In such conditions lactose does not easily cross the bacterial cell surface membrane.

Suggest and explain why lactose is unable to cross membranes.

 [2]

(iii) In order for lactose to enter the cytoplasm of E.coli a protein is required. The E. coli living in the digestive system of young mammals are more likely to contain this protein than E.coli living in the digestive system of old mammals.

Suggest an explanation for this observation.

 [2]

Lactose is a reducing sugar. Benedict’s reagent can be used to detect the presence of lactose in a solution. A colorimeter can be used to measure the concentration of lactose. The colorimeter first needs to be calibrated. Describe how a method that uses Benedict’s reagent and a colorimeter could be calibrated to measure the concentration of lactose in an unknown sample.

Plant cell walls are made of cellulose. Cellulose is a polymer of β-glucose.

Give three properties of cellulose that make it suitable as the basis of plant cell walls.

Cellulose cannot be digested by animals. Some mammals have bacteria in their stomachs that produce enzymes that can digest cellulose.

Explain whether the action of these enzymes is intracellular or extracellular.

Bread contains a mixture of polypeptides known as gluten.

Gluten consists of two types of polypeptide: gliadins and glutenins.

(i) The table below contains statements about the structures of gluten polypeptides.

In the boxes next to each statement, write the level of protein structure (primary, secondary, tertiary, or quaternary) to which the statement refers.

[2]

(ii) Coeliac disease is caused by an immune reaction to gliadins in a person’s digestive system. The immune system produces antibodies that bind to part of the gliadin polypeptides, which causes inflammation.

Some people who stop eating foods that contain gluten still occasionally experience the symptoms of coeliac disease.

What can you conclude about:

• the structure of the antibody that causes coeliac disease; and

• what the antibody binds to when producing the symptoms of coeliac disease?

[2]

Crude oil contains hydrocarbons. Three hydrocarbons commonly present in crude oil are shown in Fig. 22.

Fig. 22

Compound W shows some structural similarities with fatty acids.

State one structural difference between compound W and a saturated fatty acid.

Crude oil is often spilled from ships into the sea causing great damage to wildlife. The chemicals in crude oil are harmful to many species and do not break down quickly in the environment.

Some bacteria can break down the hydrocarbons in crude oil. These bacteria have been used by conservationists at sites where oil has been spilled.

(i) The rate of hydrocarbon breakdown by bacteria can be increased by spraying the oil with detergent. Detergents break up oil into droplets, thereby increasing their surface area.

• Student A concluded that the detergent speeded up the rate of hydrocarbon breakdown only because it increased the surface area of hydrocarbon upon which the bacteria could grow.

• Student B concluded that the detergents also increased the growth of the bacterial population by an alternative mechanism.

Use the information in Fig. 22 and your knowledge of bacterial growth requirements to provide support for student B’s conclusion.

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(ii) Suggest one piece of evidence that would further support student B’s conclusion.

[1]

Milk contains lactose. Lactose cannot be absorbed in the small intestine. The intestinal cells of mammalian infants produce lactase, an enzyme that splits lactose into glucose and galactose. These monosaccharides can pass into the blood.

Fig. 18.1 shows a molecule of the disaccharide lactose and the products of its breakdown in digestion.

Fig. 18.1

(i) Suggest why galactose and glucose cannot pass through the plasma membrane into intestinal cells by simple diffusion through the phospholipid bi-layer.

[1]

(ii) What two substances are required to break the glycosidic bond in lactose?

1 .............................................

2 .............................................

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A gene codes for the production of lactase. This gene is normally switched off after an infant moves to adult food. Almost all adult mammals are unable to digest lactose. They are said to be lactose intolerant. Humans are an exception.

Most humans have a genetic mutation that prevents the shutdown of lactase production.

State what structural detail of a polypeptide is altered by gene mutations.

Describe how two water molecules bond together.

Water makes up a large component of blood.

Explain why the properties of water make it an important component of blood.

The water component (plasma) is separated and frozen for storage when blood is donated. 

Discuss how the freezing process changes the properties of the blood plasma. 

A colorimeter is an instrument which measures how much light is absorbed or transmitted through a solution. A more concentrated solution will absorb more and transmit less light, so the result can be used to calculate the concentration of the solution.

A student used Benedict’s test and a colorimeter to investigate the concentration of glucose in a series of known and unknown solutions. The results of the colorimeter are shown in Table 1 below. The unknown solutions are solution A and solution B.

Table 1: Light transmittance (%) in a series of solutions.

Explain why using a colorimeter alongside Benedict’s test will help the student in determining glucose concentration. 

Plot the results of Table 1 above and draw a calibration curve. 

Estimate the concentration of glucose in solution A. 

Solution B was made by mixing a sample of solution A with an unknown substance C. Substance C has a glucose concentration of 0 mmol dm^-3

Solution B was made by mixing a sample of solution A with an unknown substance C. Substance C has a glucose concentration of 0 mmol dm^-3

Suggest what this information might tell the student about what substance C might be. 

A student investigated the composition of collagen and elastin using thin-layer chromatography.

The proteins were digested (broken down) in a solution before the investigation. 

The student chose four abundant amino acids from their results. These are shown in Figure 3.1 below. Using a known standard, they can identify three amino acids, but one is unidentified.

Figure 3.1

Figure 1: Diagram of thin layer chromatography results of collagen and elastin. Glycine, proline and hydroxyproline are shown alongside one unidentified amino acid.

Explain why glycine appears at the top of the thin layer chromatography plate.

Calculate the R_f value for the unidentified amino acid. Give your answer to 2 decimal places. 

Compare and contrast the structure of collagen and elastin, using the information provided and your own knowledge.

The oxyhaemoglobin dissociation curve shows the relationship between oxygen available (PaO₂) and the amount of oxygen carried by haemoglobin (percentage saturation). 

Figure 4.1 shows the oxyhemoglobin dissociation curve for a healthy person (A) and for a person who has sickle cell disease (B).

Figure 4.1

Figure 1: Oxyhaemoglobin dissociation curve in a healthy person (A) and a person with sickle cell disease (B).

The partial pressure of oxygen in the blood at which the haemoglobin is 50% saturated (P₅₀).

Calculate the difference between the P₅₀ of a healthy person, and P₅₀ of a person with sickle cell disease. Give your answer to 1 decimal point. 

Interpret the difference in oxygen transport between a healthy person and a person with sickle cell disease. 

Sickle cell disease is a genetic disease which causes haemoglobin to have a sickle or crescent shape.

Explain how a genetic mutation might alter the protein structure of haemoglobin. 

Suggest how this change might lead to the result we see in Figure 4.1. 

Figure 5.1 shows an adipocyte, a cell type which stores fat in mammals. Adipocytes are roughly spherical.

Figure 5.1

Calculate the volume of the cell in Figure 5.1. Give your answer in millimetres cubed standard form to 3 significant figures.

The fat reservoir in adipocytes consists mainly of triglycerides and is surrounded by a phospholipid monolayer.

Explain how the structure of triglycerides makes them suitable for energy storage. 

Suggest why phospholipids form a monolayer around the fat droplet.