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Cambridge IGCSE BIOLOGY 0610/41 Paper 4 Theory (Extended) May/June 2020 1 hour 15 minutes You must answer on the question paper. No additional materials are needed. INSTRUCTIONS Answer all questions. Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. Write your name, centre number and candidate number in the boxes at the top of the page. Write your answer to each question in the space provided. Do not use an erasable pen or correction fluid. Do not write on any bar codes. You may use a calculator. You should show all your working and use appropriate units. INFORMATION The total mark for this paper is 80. The number of marks for each question or part question is shown in brackets [ ]. This document has 20 pages. Blank pages are indicated. DC (ST/CT) 180481/3 © UCLES 2020 [Turn over *0883915607*

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Page 1: IGCSE Biology (0610) Past Papers Solutions - 0610/41 Biology - … · 2020. 10. 27. · Cambridge IGCSE BIOLOGY 0610/41 Paper 4 Theory (Extended) May/June 2020 1 hour 15 minutes You

Cambridge IGCSE™

BIOLOGY 0610/41

Paper 4 Theory (Extended) May/June 2020

1 hour 15 minutes

You must answer on the question paper.

No additional materials are needed.

INSTRUCTIONS ● Answer all questions. ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, centre number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● You may use a calculator. ● You should show all your working and use appropriate units.

INFORMATION ● The total mark for this paper is 80. ● The number of marks for each question or part question is shown in brackets [ ].

This document has 20 pages. Blank pages are indicated.

DC (ST/CT) 180481/3© UCLES 2020 [Turn over

*0883915607*

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1 The gas exchange system is one of the organ systems of the human body.

Fig. 1.1 shows parts of the gas exchange system during breathing in and breathing out.

breathing in breathing out

intercostalmuscles

sternum

diaphragm

vertebrae

Fig. 1.1

(a) Complete Table 1.1 to show:

• the functions of the diaphragm and the intercostal muscles during breathing in and breathing out

• the pressure changes in the thorax.

Use these words:

contract relax increases decreases.

Table 1.1

diaphragmintercostal muscles pressure change

in the thoraxinternal external

breathing in

breathing out

[4]

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Contract
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Relax Relax Decreases
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Relax
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Contract/ relax
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Relax
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Increases
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Fig. 1.2 shows part of the gas exchange surface of a human.

movement of air

magnification ×350

X

Fig. 1.2

(b) State two features of the gas exchange surface that are visible in Fig. 1.2.

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

2 ................................................................................................................................................[2]

(c) The cells labelled X on Fig. 1.2 form a tissue.

(i) Define the term tissue.

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Well supplied by blood
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Good venation in air
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A tissue is made up of a group of cells that usually look similar to one another and come from
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he same region in a developing embryo. The group of cells that make up a tissue have
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physiological functions that work together in a coordinated way to support the special;
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functions. Different kinds of tissues have different physical properties.
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(ii) Cartilage is another tissue found in the gas exchange system.

State the functions of cartilage in the gas exchange system.

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[Total: 10]

2 Biological washing powders contain enzymes that break down food stains.

(a) Complete Table 2.1 by naming the enzymes that break down three substances in food stains and by stating the product or products.

Table 2.1

substance enzyme product(s)

starch

fat

protein

[3]

Some students compared how effective biological and non-biological washing powders are at removing stains at temperatures between 10 °C and 60 °C.

• Pieces of stained cloth were washed using two different washing powders.• The degree of stain removal was measured by using a light meter to record the

percentage of light reflected from the cloth.• A light meter gave a value of 100% when the cloth was completely clean.• Any stain left on the cloth reduced the percentage of light reflected.

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amylase
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lipase
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pepsin
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maltose
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fatty acid
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amino acid
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Cartilage forms incomplete rings around trachea. It reduces resistant to movement of air and
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also helps in sound production in larynx .
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The results of the students’ investigation are shown in Fig. 2.1.

percentageof lightreflectedafterwashing

temperature of washing / °C

Key:non-biological washing powderbiological washing powder

4010 20 30 40 50 60 70

50

60

70

80

90

100

110

Fig. 2.1

(b) Compare the effectiveness of the two washing powders at removing stains.

Use the information in Fig. 2.1 in your answer.

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From the above graph , we observe that biological washing powder is more effective at lower
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temperatures i.e between 10 degree 40 degree celsius. Biological washing powder removes all stains
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between 30 degree and 40 degree celsius whereas non-biological washing powder removes stain only
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60 degree celsius . Effectiveness of washing powder is similar at higher temperatures however for
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biological washing powder it increases 40 to 44 degree celsius and there is no such decrease for non
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biological washing powder .
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(c) The students suggested that the enzymes in the biological washing powder were denatured at high temperatures.

Explain why enzyme molecules do not function when they are denatured.

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(d) Forensic scientists often try to find DNA on items of stained clothing. The DNA can be used to identify individual people.

Suggest why DNA can be used to identify individual people.

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[Total: 11]

3 (a) Dialysis tubing is an artificial membrane, which is similar to the lining of the intestine.

A student investigated the diffusion of glucose through dialysis tubing by using the apparatus shown in Fig. 3.1.

rubber bandto secure thedialysis tubing

water outsidethe dialysis tubing

knot at the end ofthe dialysis tubing

glucose solution

dialysis tubing

Fig. 3.1

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The active side of enzymes molecules changes shape which prevents binding of substrate ultimately
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preventing the formation of enzyme-substrate complex .
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Individual people have unique DNA which contains genes made up of sequences of bases .
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The student took samples of the water outside the dialysis tubing at 5 minute intervals and tested the samples with Benedict’s solution.

The results are shown in Table 3.1.

Table 3.1

time / minutes results of the Benedict’s tests on the water outside the dialysis tubing

0 blue

5 green

10 yellow

15 red

(i) Describe and explain the results shown in Table 3.1.

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..................................................................................................................................... [3]

(ii) The student repeated the investigation with a higher concentration of glucose in the dialysis tubing.

Predict the results that the student would observe.

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...........................................................................................................................................

..................................................................................................................................... [1]

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Appearance of blue Colour at the time 0 indicates absence of glucose this ensures that no glucose
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on outer surface of dialysis tubing as a result of an error .
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Green,yellow,red appearance indicate presence of glucose. Glucose diffuses out of the dialysis
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cubing as the dialysis permeable to glucose.
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Benedict's solution changes colour with the concentration of glucose .
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(b) Fig. 3.2 shows a drawing of a cell from the lining of the small intestine. The lumen is the space inside the intestine where food is digested.

A

C

B

not to scale

lumen of the intestine

Fig. 3.2

State the names of the three labelled structures in Fig. 3.2 and describe the role of each structure in the intestinal cell.

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Microvilli
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Rough endoplasmic reticulum
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Mitochondrion
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A) Allow movement substances into the cell, increase the surface area for absorption for diffusion
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B) Site of protein synthesis
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C) Aerobic respiration
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(c) The cholera bacterium can survive in the small intestine and the large intestine. The bacterium releases a toxin that interacts with receptors on the surface of cells.

Fig. 3.3 shows the effect of the toxin. The arrows indicate the direction of movement.

Key:toxinion X

bacterium

not to scale

lumen of the intestine

Fig. 3.3

The toxin stimulates the secretion of ion X out of the intestinal cell.

(i) State the name of ion X.

..................................................................................................................................... [1]

(ii) Describe the effects on the body of the secretion of ion X into the lumen of the intestine.

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[Total: 15]

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Chloride
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A) Diarrhea
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B) Dehydration
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C) Loss of water of osmosis
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D) Loss of other ions such as Na + , K + etc
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4 Johnson grass, Sorghum halepense, is wind-pollinated.

(a) Fig. 4.1 shows some Johnson grass flowers.

Fig. 4.1

(i) State the genus of Johnson grass.

..................................................................................................................................... [1]

(ii) Describe two features visible in Fig. 4.1 that show that Johnson grass flowers are adapted for wind-pollination.

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

...........................................................................................................................................

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

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Sorghum
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Feathery stigma
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Anthers hanging outside the flowers .
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(b) Fig. 4.2 shows a section through a carpel shortly after pollination.

D

A

C

B

E

Fig. 4.2

(i) State the names of the parts of the carpel labelled C, D and E.

C ........................................................................................................................................

D ........................................................................................................................................

E ........................................................................................................................................[3]

(ii) Complete the sentences:

Pollen grains are formed in anthers. During their formation the number of chromosomes

in the nuclei is halved by the process of .............................................. . This means the

male nucleus A in the pollen tube is described as a .............................................. nucleus.

When nucleus A .............................................. with nucleus B, the chromosome number

doubles to form a .............................................. nucleus. The name of this process is

.............................................. . Then the .............................................. divides by the

process of .............................................. to form an embryo.[7]

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ovary
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ovule
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style
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Reduction division
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Haploid
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Combines
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Diploid
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Fertilisation
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Zygote
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Mitosis
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(c) Discuss the advantages of sexual reproduction to a wild population of flowering plants such as Johnson grass.

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(d) Sexual reproduction requires energy.

State three uses of energy in organisms other than in reproduction.

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

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

3 ................................................................................................................................................[3]

[Total: 21]

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Protein synthesis
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Transport in phloem
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Cell division
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Sexual reproduction gives genetic diversity and variation . It allows mutation to be expressed,
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develop adaptations to new conditions , allows natural selection to occur. In sexual reproduction
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there is cross pollination
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. There is diversity in the genetic makeup of the individuals produced by sexual
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reproduction . Since both the parents are involved, the newly formed individuals have the attributes of both.
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Variations are more successful in sexual mode than in asexual one.
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The species produced by
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sexual
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reproduction survive more than those produced by asexual reproduction. This is because genetic
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variations help them to adapt to different environments.
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Hence, a diseases
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is less likely to affect all the
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individuals in the population.
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5 Ciliates are classified in the kingdom Protoctist. Bacteria are classified in the kingdom Prokaryote.

(a) State two structural features that distinguish the cells of a protoctist from a prokaryote.

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

...................................................................................................................................................

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

...................................................................................................................................................[2]

(b) Fig. 5.1 shows five species of ciliate that are found in sewage treatment works.

A Chilodonella

C Euplotes D Paramecium E Vorticella

not to scale

B Didinium

rows ofcilia

cilia

ciliafused

together

Fig. 5.1

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Presence of nuclear membrane
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Cell walls if present have different composition .
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Fig. 5.2 is a dichotomous key to identify the ciliates shown in Fig. 5.1.

1 has a ring of cilia at one end of the organism

START

2

3 has star-like structures inside the organism

organism C

organism A

organism D

organism B

organism E

4

yesyes

no

no

no

no

yes

yes

Fig. 5.2

Complete the key in Fig. 5.2 by writing suitable statements:

• for box 2 to distinguish species B and E• for box 4 to distinguish species A and C.

text for box 2 .............................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

text for box 4 .............................................................................................................................

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Organism has two rings of cilia
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organism has a covering of cilia
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has two rings of cilia .
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has a covering of cilia
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.
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(c) Didinium is a predatory ciliate. A video recording was made of one Didinium feeding on a Paramecium. Fig. 5.3 shows a sequence of still photographs taken from the video.

Paramecium

Didinium

Fig. 5.3

Complete the table by putting a tick (✓) by each characteristic of life that can be seen in the still photographs from the video in Fig. 5.3.

excretion nutrition

growth reproduction

movement respiration

[1]

(d) Fig. 5.4 is a food web for some of the microorganisms in a sewage treatment works.

Didinium

Paramecium

Vorticella

rotifers

photosynthetic bacteria decomposerbacteria

nematodes

Fig. 5.4

(i) Construct one food chain with three trophic levels that use energy derived from the breakdown of sewage. Do not draw the organisms.

..................................................................................................................................... [1]

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Bacteria Paramecium Didinium
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Line
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Line
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Stamp
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Stamp
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(ii) The water that passed out of the sewage works was often cloudy with suspended matter.

Scientists discovered that ciliates reduce the cloudiness of water during sewage treatment.

Suggest how the ciliates reduce the cloudiness of the water using the information in Fig. 5.4.

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(iii) Explain how sewage treatment reduces the spread of disease.

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(iv) Nitrifying bacteria are found in sewage works.

Explain the importance of nitrifying bacteria in the nitrogen cycle.

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[Total: 14]

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Nitrifying bacteria perform nitrogen cycle . They are involved in conversion of ammonium ions
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to nitrate ions which are further converted to nitrite ions . The nitrate ions are biologically
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available to plants for absorption . Plants absorb it and use it to make proteins .
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A) By removal of harmful bacteria and pathogens of sewage
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.
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e.g. Cholera bacteria or water borne disease or parasites
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B) Stopping the spread of pathogens via water
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C) Use of chlorination .
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Ciliates eat many bacteria. They may also eat dead and decomposing material .
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6 Colour blindness is a characteristic that is inherited. Colour blindness is more common in males than in females.

Fig. 6.1 is a pedigree diagram showing the inheritance of colour blindness in a family.

1 2

3

5 6 7 8

4

male with normalcolour visionmale with colourblindnessfemale with normalcolour vision

Key:

Fig. 6.1

(a) Define the term inheritance.

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(b) (i) Using the symbols B and b, state the genotypes of individual 5 and individual 8 in the pedigree diagram.

5 ........................................................................................................................................

8 ........................................................................................................................................[3]

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X Y
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X Y
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b
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B
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Transmission of genetic information from generation to generation is called as inheritance .
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(ii) Individual 3 is a carrier of colour blindness because she has one copy of the allele for colour blindness but has normal colour vision.

Describe the evidence from Fig. 6.1 that shows that individual 3 is a carrier.

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(iii) There was no history of colour blindness in the parents and grandparents of individuals 1 and 2.

Suggest how colour blindness first occurred in the family in Fig. 6.1.

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[Total: 9]

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Occurrence of Colour blindness in family could be due to mutation or due to appearance
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of recessive allele.
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Colour blindness is a sex-linked characteristic . She is heterozygous for the gene (Bb) . She
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has normal allele for normal colour vision but has passed on the relative allele to her sons
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.
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She has two X chromosomes which have the gene to colour vision ; father pases on his Y
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chromosome .