Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

AQA A Level Biology Paper 1 June 2018, Exams of Advanced Education

An examination paper or question set for the aqa a level biology paper 1 exam from june 2018. It covers a range of topics in biology, including cell biology, genetics, biochemistry, and physiology. Several short-answer questions that require the student to demonstrate their understanding of key biological concepts and their ability to apply this knowledge to analyze and explain various scenarios. The questions cover topics such as chromosome structure, the genetic code, enzyme-substrate interactions, immune responses, dna replication, and the human gas exchange system. Overall, this document could be a valuable resource for a level biology students preparing for their exams, as it provides insight into the types of questions they may encounter and the level of understanding expected of them.

Typology: Exams

2024/2025

Available from 10/25/2024

cate-mentor
cate-mentor 🇺🇸

1.9K documents

1 / 5

Toggle sidebar

Related documents


Partial preview of the text

Download AQA A Level Biology Paper 1 June 2018 and more Exams Advanced Education in PDF only on Docsity!

AQA A Level Biology Paper 1 June 2018

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. [2 marks] - water moves into the cells/ cytoplasm by osmosis cell/ cytoplasm gets bigger 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.[1 mark] - differences in base sequence What is a homologous pair of chromosomes? [1 mark] - chromosomes that carry the same genes Give two ways in which the arrangement of prokaryotic DNA is different from the arrangement of the human DNA. [2 marks] - (Prokaryotic DNA) is

  1. Circular (as opposed to linear)
  2. Not associated with proteins/ histones
  3. Only one molecule/ piece of DNA Describe the method the student would have used to obtain the results in Figure 3. Start after all the cubes of potatoes have been cut. Also consider variables he should have controlled. [3 marks] - Method to ensure all the cut surfaces of the eight cubes are exposed to the sucrose solution method of controlling temperature

method of drying cubes before measuring measure of mass of tubes at stated time intervals What is meant by 'species richness'? [1 mark] - (a measure of)the number of (different) species in a community Formation of an enzyme-substrate complex increases the rate of reaction. Explain how. [2 marks] - reduces activation energy due to bending bonds Lyxose binds to the enzyme. Suggest a reason for the difference in the results shown in Figure 5 with and without lyxose. [3 marks] - (binding) alters the treaty structure of the enzyme; (this causes) active site to change (shape) (so) more (successful) E-S complexes form (per minute) OR E-S complexes form more quickly OR Further lowers the activation energy The genetic code is described as degenerate. What is meant by this? [1 mark] - more than one codon codes for a single amino acid 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.

What does this result suggest about the difference between the observed and expected results and what can the scientists therefore conclude? [2 marks] - there is a less than 0.05/5% probability that the difference(s) (between observed and expected) occurred by chance calculated value is greater than critical value so the null hypothesis can be rejected the scientists conclude that the proportion of plants that produce 2n gametes does change from one breeding cycle to the next When a person is bitten by a venomous snake, the snake injects a toxin into the person. Antivenom is injected as treatment. Antivenom contains antibodies against the snake toxin. This treatment is an example of passive immunity. Explain how the treatment with antivenom works and why it is essential to use passive immunity, rather than active immunity. [2 marks] - 1. (Antivenom/Passive immunity) antibodies bind to the toxin/venom/antigen and (causes) its destruction;

  1. Active immunity would be too slow/slower; Use your knowledge of directional selection to explain the results shown in Table 3. [3 marks] - 1. The scientists selected/used for breeding plants that produced 2n gametes;
  2. (So these plants) passed on their alleles (for production of 2n gametes to the next generation);
  3. The frequency of alleles for production of 2n gametes increased (in the population). A mixture of venoms from several snakes of the same species is used. Suggest why. [2 marks] - may be different form of antigen (within one species) different antibodies (needed in the antivenom) During vaccination, each animal is initially injected with a small volume of venom. Two weeks later, it is injected with a larger volume of venom.

Use your knowledge of the humoral immune response to explain this vaccination programme. [3 marks]

    1. B cells specific to the venom reproduce by mitosis;
  1. (B cells produce) plasma cells and memory cells;
  2. The second dose produces antibodies (in secondary immune response) in higher concentration and quickly Describe the role of two names enzymes in the process of semi-conservative replication of DNA. [ marks] - (DNA) helciase causes breaking of hydrogen; DNA polymerase joins the (DNA) nucleotides; forming phosphodiester bonds Describe the gross structure of the human gas exchange system and how we breathe in and out. [ marks] - trachea, bronchi, bronchioles, alveoli; breathing in - diaphragm contracts and external intercostal muscles contract; (causes) volume increase and pressure decrease in thoracic cavity (to below atmospheric, resulting in air moving in); breathing out- diaphragm relaxes and internal intercostal muscles contract; (causes) volume decrease and pressure increase in thoracic cavity (to above atmospheric, resulting in air moving 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.[5 marks] - both contain ester bonds (between glycerol and fatty acid)

both contain glycerol; fatty acids on both may be saturated or unsaturated; both are insoluble in water; both contain C, H and O but phospholipids also contain P; Triglyceride has three fatty acids and phospholipid has two fatty acids plus phosphate group; triglycerides are hydrophobic and phospholipids have hydrophilic and hydrophobic region; phospholipids form bilayer in water but triglycerides don't 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. [4 marks] - lactose is formed from glucose and galactose joined by condensation (reaction) joined by glycosidic bond; added to polypeptide in Golgi (apparatus)