Questions & Answers

Q1(a).
Short Answer 1 mark

Fig. 1.1 is a diagram of part of a Bowman's capsule and a glomerular capillary .

Q1(i).
MCQ 2 marks

Identify structures A and B. A B

A ........................................................................................................................................
B ........................................................................................................................................
Q1(ii).
Short Answer 1 mark

The glomerular filtrate is produced in the Bowman' s capsule by the process of ultrafiltration. State the conditions required for ultrafiltration.

Q1(b).
Short Answer 3 marks

As the newly formed glomerular filtrate passes along the proximal convoluted tubule, selective reabsorption takes place. The fluid remaining in the proximal convoluted tubule eventually forms urine. T able 1.1 lists three of the components of blood plasma that enter the glomerulus in a healthy person. Complete Table 1.1 by writing: plasma plasma Y ou may use each response once, more than once or not at all. Table 1.1 component of blood plasma entering glomerulus component in newly formed glomerular filtrate component in urine glucose large plasma proteins urea

Q1(c).
Short Answer 1 mark

Sweating during exercise can lead to a response in the posterior pituitary gland and in the kidney. State the response to sweating that occurs in the posterior pituitary gland and the response that occurs in the kidney. [Total: 9] * 0000800000003 *

Q2.
Short Answer 1 mark

Several dif ferent processes can affect allele frequencies in populations.

Q2(a).
Short Answer 1 mark

Outline the processes that may affect allele frequencies in wildlife populations.

Q2(b).
Short Answer 1 mark

The process of selective breeding changes allele frequencies in a population. Plant breeders use selective breeding to improve crop plants such as maize. Explain how selective breeding can be used to obtain a variety of maize where the plants are vigorous and of uniform height. [Total: 9] * 0000800000004 *

Q3.
Short Answer 1 mark

The treatment of many diseases has been improved by the use of genetic technology in medicine.

Q3(a).
Short Answer 1 mark

Explain what is meant by genetic engineering.

Q3(b).
Short Answer 1 mark

(i) Leber Congenital Amaurosis (LCA) is an inherited eye disease. In LCA, the photoreceptor cells in the retina die at an early age. This causes impaired vision (reduced eyesight) in children, which can progress to blindness. Mutations in dif ferent genes cause dif ferent forms of LCA. One form of this disease, LCA2, is caused by a mutation in the RPE65 gene. Gene therapy has been used to treat LCA2. Outline how an inherited eye disease, such as LCA2, is treated with gene therapy.

Q3(ii).
Short Answer 1 mark

Suggest why the eye is a suitable organ for gene therapy.

Q3(c).
Short Answer 1 mark

LCA10 is a dif ferent form of LCA caused by a recessive mutation in the CEP290 gene. This gene codes for the protein CEP290, which is involved in the correct functioning of photoreceptor cells in the retina. The mutation in CEP290 causes an error to be made when the primary transcript is spliced to form messenger RNA (mRNA). The abnormal mRNA that is formed has an extra sequence of RNA nucleotides, known as exon X , between exon 26 and ex on 27. Exon X contains a ST OP codon. Fig. 3.1 compares the ef fect of the mutation in CEP290 with the normal gene expression. normal LCA10 exon intron exon section of DNA of CEP290 exon mutationexon exon intron exon primary transcript exon mutationexon mRNA exon exon exon CEP290 polypeptide product shortened CEP290 polypeptide In 2022, research was carried out into possible treatment of LCA10 using genetic technology .

Q3(i).
Short Answer 1 mark

A human clinical trial investigated a treatment of LCA10 using a short RNA nucleotide sequence known as Sepofarsen. Sepofarsen binds to the section of the primary transcript containing the CEP290 mutation so that normal splicing occurs and functional CEP290 protein is synthesised. the greatest loss of vision (treated eye) for a period of 12 months. vision chart. Fig. 3.2 shows the results of the clinical trial over 12 months. exon exon 27 * 0000800000006 * -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0.1 0.2 1 2 3 4 5 6 time / months 7 8 9101112 change in visual acuity / arbitrary units impaired acuity improved acuity Key:eye not treated eye treated Describe the results of the clinical trial data shown in Fig. 3.2.

Q3(ii).
Short Answer 1 mark

Another method being investigated to treat LCA10 is to use a gene editing tool known as the CRISPR/Cas9 system. The CRISPR/Cas9 system uses a short length of RNA called guide RNA. Guide RNA is complementary to the target DNA and is linked to a nuclease enzyme called Cas9. Cas9 breaks phosphodiester bonds in DNA. The cell repair mechanisms repair the cut in DNA after the modification has taken place. Explain how this method used to treat LCA10 is an example of gene editing. [Total: 13] * 0000800000008 * Question 4 starts on page 10 * 0000800000009 *

Q4.
Short Answer 1 mark

Genetic crosses can be used to investigate patterns of inheritance.

Q4(a).
MCQ 3 marks

A mutation in a gene involved in fruit colour in tomato plants, Solanum lycopersicum, results in the production of yellow fruits instead of red fruits. A genetic cross was carried out between a pure breeding plant with a red fruit and a pure breeding plant with a yellow fruit to produce the F1 generation. All offspring plants produced red fruits. The F1 plants were then crossed with each other and the seeds produced were planted to obtain the F2 generation. Construct a genetic diagram to show the cross of the F1 generation that produced this F2 generation. Use the symbols R and r for the alleles.

A mutation in a gene involved in fruit colour in tomato plants, Solanum lycopersicum, results
A genetic cross was carried out between a pure breeding plant with a red fruit and a pure
Q4(b).
MCQ 1 mark

A theoretical dihybrid cross involves two genes located on dif ferent autosomes. Each gene has two alleles, one dominant and one recessive. A parent, homozygous dominant for both genes, is crossed with a parent that is homozygous recessive for both genes. This produces F1 individuals that are then crossed to produce the F2 generation. State the phenotypic ratio of this dihybrid F2 generation and explain why some of these offspring phenotypes are different from the original parental phenotypes. ratio explanation [Total: 6] * 0000800000010 *

A theoretical dihybrid cross involves two genes located on dif ferent autosomes. Each gene
A parent, homozygous dominant for both genes, is crossed with a parent that is homozygous
Q5.
Short Answer 1 mark

Phenotypic variation exists in natural populations. There are many causes of variation. Natural selection determines which phenotypes are advantageous.

Q5(a).
Short Answer 1 mark

Variation in a particular characteristic can be described as either discontinuous or continuous. T able 5.1 contains a list of statements that apply to discontinuous variation, continuous variation or both. leave the box empty if the statement does not apply. Table 5.1 statement discontinuous variation continuous variation often involves one gene only environmental factors may affect gene expression there is an additive effect of genes that contributes to the phenotype there are distinct differences between the various forms of a characteristic [4] * 0000800000011 *

Q5(b).
Short Answer 1 mark

There is variation in the quantity of vitamin D stored in the body . V itamin D has an important role in keeping bones healthy . The main storage form of vitamin D A study was carried out on 262 healthy women to investigate if the concentration of supplements. The age range of the women in the sample was 40 to 72 years old. T able 5.2 shows the results of the study. Table 5.2 group mean concentration of / ng cm standard deviation sampled during summer n = 138 32.7 7.6 sampled during winter n = 124 28.5 8.3 whole sample n = 262 30.7 8.2 Additional analysis showed that there was no significant correlation between age and

Q5(i).
Short Answer 1 mark

Explain what is meant by standard deviation. the winter, as shown in Table 5.2. Calculate the value of t using the formula provided. key to symbols: ¯ x = mean s = sample standard deviation n = sample size (number of observations) Give your answer to four significant figures. There is space for your working. fp * 0000800000012 *

Q5(iii).
Short Answer 1 mark

The critical value at the 0.0001 probability level is 3.773. State the conclusion that can be made about the results of the study shown in Table 5.2 and explain how the result of your calculation in (b)(ii) can be used to support this conclusion.

Q5(iv).
Short Answer 1 mark

Suggest the likely causes of variation in quantity of vitamin D stored in the body in this sample of women. [Total: 12] * 0000800000013 *

Q6(a).
Short Answer 1 mark

DCPIP is an indicator and can be used to determine the rate of respiration of organisms such

Q6(i).
Short Answer 1 mark

State the category of indicators to which DCPIP belongs.

Q6(ii).
Short Answer 1 mark

Describe the colour change that occurs in DCPIP during experiments to determine the rate of respiration.

Q6(b).
Short Answer 1 mark

An investigation was carried out to determine the ef fect of temperature on the rate of respiration of yeast. was measured. The results are shown in Fig. 6.1. 1020304050 temperature / °C time taken for DCPIP to change colour / minutes

Q6(i).
Short Answer 1 mark

State two variables that need to be kept constant in this experiment.

Q6(ii).
Short Answer 1 mark

Explain the results shown in Fig. 6.1. [Total: 7] * 0000800000015 *

Q7(a).
Short Answer 1 mark

Paper chromatography is a technique that can be used to separate and identify dif ferent chloroplast pigments. Describe how the results of paper chromatography can be used to identify chloroplast pigments.

Q7(b).
Short Answer 1 mark

Carotene and xanthophyll are chloroplast pigments. Describe the role played by these pigments in photosynthesis. [Total: 13] * 0000800000017 *

Q8(a).
Short Answer 1 mark

Fig. 8.1 is a diagram of a motor neurone.

Q8(i).
Short Answer 2 marks

Name cell X and part Y. X Y

Q8(ii).
Short Answer 1 mark

Name a type of cell that forms a synapse with structure Z.

Q8(b).
MCQ 1 mark

Table 8.1 shows some of the events that occur during muscle contraction. They are not listed in the correct order . Table 8.1 event description of event A calcium ions dif fuse out of sarcoplasmic reticulum B myosin heads bind to actin C sarcolemma depolarised D sarcomere shortens E calcium ions bind to troponin G binding sites on actin exposed H myosin heads tilt I tropomyosin moves J troponin changes shape * 0000800000018 * Complete Table 8.2 to show the correct order of the events that occur during muscle contraction. Two of the events have been completed for you. Table 8.2 correct order letter of event 5 J 10 D [4] channels on the presynaptic knob. One symptom of LEMS is weaker muscle contraction. Suggest and explain why LEMS leads to weaker muscle contraction. [Total: 11] * 0000800000019 *

A calcium ions dif fuse out of sarcoplasmic reticulum
B myosin heads bind to actin
C sarcolemma depolarised
D sarcomere shortens
Q9.
Short Answer 1 mark

The red ruffed lemur, Varecia rubra, is a mammal found only in the rainforests of the Masoala region in north east Madagascar. Fig. 9.1 shows a red ruffed lemur.

Q9(a).
Short Answer 1 mark

The International Union for Conservation of Nature (IUCN) Red List of Threatened Species states that the red ruf fed lemur is critically endangered. Suggest why the red ruf fed lemur has become critically endangered.

Q9(b).
Short Answer 1 mark

Many zoos around the world operate captive breeding programmes for the red ruf fed lemur. 1970 to 2020. 1970 1980 1990 2000 2010 2020 year number of red ruffed lemurs Describe the results shown in Fig. 9.2.

Q9(c).
Short Answer 1 mark

Captive breeding programmes for endangered mammals such as the red ruf fed lemur can vary in their success rate.

Q9(i).
Short Answer 1 mark

Suggest problems that may af fect the success of captive breeding programmes of mammals like the red ruffed lemur. programmes. Suggest why this is done. [Total: 10] * 0000800000022 *

Q10(a).
Short Answer 1 mark

Blood glucose concentration is maintained around a set point by homeostasis. Explain the principles of homeostasis.

Q10(b).
Short Answer 1 mark

Glycogen phosphorylase catalyses the conversion of glycogen to glucose in liver cells. The production of glycogen phosphorylase is coded for by the gene PYGL. A mutation in PYGL leads to a condition called glycogen storage disease type VI (GSDVI), in which glycogen is not broken down efficiently. Suggest and explain why cell signalling by glucagon is likely to be af fected in the liver cells of a person with GSDVI. The boundaries and names shown, the designations used and the presentation of material on any maps contained in this question paper/insert do not imply official endorsement or acceptance by Cambridge Assessment International Education concerning the legal status of any country , territory, or area or any of its authorities, or of the delimitation of its frontiers or boundaries. reasonable ef fort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge.

Q10(c).
Short Answer 1 mark

Glycogen synthase catalyses the conversion of glucose to glycogen in liver cells. The production of glycogen synthase is coded for by the gene GYS2. A mutation in GYS2 leads to a condition called glycogen storage disease type 0 (GSD0) in which glycogen is not formed efficiently. Suggest what the consequences would be if a person with GSD0 has a meal rich in glucose. [Total: 10] * 0000800000024 *