CBSE Class 12 Biology: sample paper 2023-24

Section A

MCQ 1-12, assertion-reason 13-16

1. Q1 · 1 mark · MCQSexual Reproduction in Flowering Plants

   Remnants of nucellus are persistent during seed development in:

   1. (A)pea
   2. (B)groundnut
   3. (C)wheat
   4. (D)black pepper

   Try it first, then show our solutionSolution

   1. Persistent remnants of the nucellus in a seed form the perisperm, which is found in black pepper and beet. Pea and groundnut have no persistent nucellus, and wheat stores its food in the endosperm.

      1 mark

   Answer: (D) black pepper

   Where the marks are

   • Identifies black pepper as a perispermic seed (1 mark)

2. Q2 · 1 mark · MCQSexual Reproduction in Flowering Plants

   The wall layer of microsporangium which nourishes the pollen grain is:

   1. (A)epidermis
   2. (B)endothecium
   3. (C)middle layers
   4. (D)tapetum

   Try it first, then show our solutionSolution

   1. The tapetum is the innermost wall layer of the microsporangium. Its cells are dense and often multinucleate, and they nourish the developing pollen grains.

      1 mark

   Answer: (D) tapetum

   Where the marks are

   • Tapetum is the nourishing layer (1 mark)

3. Q3 · 1 mark · MCQCompetency-basedMolecular Basis of Inheritance

   A short piece of DNA, having 20 base pairs, was analyzed to find the number of nucleotide bases in each of the polynucleotide strands. Some of the results are shown in the table.

   	Adenine	Cytosine	Guanine	Thymine
   Strand 1	4	4
   Strand 2		5

   (Column heading: Number of nucleotide bases)

   How many nucleotides containing Adenine were present in strand 2?

   1. (A)2
   2. (B)4
   3. (C)5
   4. (D)7

   Try it first, then show our solutionSolution

   1. Each strand has 20 nucleotides. In strand 1, C = 4, and C in strand 2 is 5, so G in strand 1 must be 5 (G pairs with C of strand 2). Then T in strand 1 = 20 − (4 + 4 + 5) = 7.

      ½ mark

   2. T in strand 1 pairs with A in strand 2, so A in strand 2 = 7.

      ½ mark

   Answer: 7, option (D)

   Where the marks are

   • Uses complementary base pairing to find T in strand 1 = 7 (0.5 mark)
   • Concludes A in strand 2 = 7 (0.5 mark)

   Open Q3 on its own page

4. Q4 · 1 mark · MCQPrinciples of Inheritance and Variation

   In a certain species of insects, some have 13 chromosomes, and the others have 14 chromosomes. The 13 and 14 chromosome bearing organisms are

   1. (A)males and females, respectively
   2. (B)females and males, respectively
   3. (C)all males
   4. (D)all females

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   1. This species shows XO-type sex determination, as in grasshoppers. Males are XO and so carry one chromosome fewer (13), while females are XX and carry 14.

      1 mark

   Answer: (A) males and females, respectively

   Where the marks are

   • Recognises XO-type sex determination: male 13, female 14 (1 mark)

5. Q5 · 1 mark · MCQEvolution

   At a particular locus, the frequency of allele A is 0.8 and that of allele a is 0.2. What would be the frequency of heterozygotes in a random mating population at equilibrium?

   1. (A)0.32
   2. (B)0.16
   3. (C)0.24
   4. (D)0.48

   Try it first, then show our solutionSolution

   1. Under Hardy–Weinberg equilibrium, the heterozygote frequency is $2pq=2(0.8)(0.2)=0.32$.

      1 mark

   Answer: 0.32, option (A)

   Where the marks are

   • Applies $2pq$ correctly to get 0.32 (1 mark)

6. Q6 · 1 mark · MCQEvolution

   Variations caused due to mutations are

   1. (A)random and directionless
   2. (B)random and directional
   3. (C)random and small
   4. (D)random, small and directional

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   1. According to de Vries, mutations arise suddenly and are random and directionless. Darwinian variations, by contrast, are small and directional.

      1 mark

   Answer: (A) random and directionless

   Where the marks are

   • States that mutations are random and directionless (1 mark)

7. Q7 · 1 mark · MCQMolecular Basis of Inheritance

   What is the smallest part of a DNA molecule that can be changed by a point mutation?

   1. (A)Oligonucleotide
   2. (B)Codon
   3. (C)Gene
   4. (D)Nucleotide

   Try it first, then show our solutionSolution

   1. A point mutation changes a single base pair. The smallest unit of DNA it can alter is therefore one nucleotide.

      1 mark

   Answer: (D) Nucleotide

   Where the marks are

   • Identifies the nucleotide as the smallest unit changed (1 mark)

8. Q8 · 1 mark · MCQCompetency-basedPrinciples of Inheritance and Variation

   What should be the genotype of the indicated member?

   1. (A)AA
   2. (B)Aa
   3. (C)XY
   4. (D)aa

   Figure in the paper: Pedigree chart: a normal male and female at top produce four children in the second generation: an unaffected male, an affected (shaded) male, an unaffected female (married to an unaffected male) and an affected (shaded) female marked (?), plus an unaffected male. The unaffected female and her husband have three children: an affected (shaded) male and two unaffected females. See it in the official paper.

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   1. Two unaffected parents have affected children, so the trait is autosomal recessive. The affected female (?) must therefore be homozygous recessive, aa.

      1 mark

   Answer: (D) aa

   Where the marks are

   • Infers a recessive trait and gives the genotype aa (1 mark)

9. Q9 · 1 mark · MCQCompetency-basedBiotechnology and its Applications

   A patient was advised to have a kidney transplant. To suppress the immune reaction, the doctor would administer him:

   1. (A)statins produced from Monascus purpureus
   2. (B)statins produced from Streptococcus thermophilus
   3. (C)cyclosporin A produced from Trichoderma polysporum
   4. (D)cyclosporin A produced from Clostridium butylicum

   Try it first, then show our solutionSolution

   1. Cyclosporin A, produced by the fungus Trichoderma polysporum, is the immunosuppressant given to organ-transplant patients. Statins from Monascus purpureus lower blood cholesterol instead.

      1 mark

   Answer: (C) cyclosporin A from Trichoderma polysporum

   Where the marks are

   • Correct drug and correct source organism (1 mark)

10. Q10 · 1 mark · MCQCompetency-basedBiotechnology: Principles and Processes

    Identify the activity of endonuclease and exonuclease in the given image.

    1. (A)Option a) of the image
    2. (B)Option b) of the image
    3. (C)Option c) of the image
    4. (D)Option d) of the image

    Figure in the paper: Table with columns Endonuclease and Exonuclease and four rows a) to d). Each cell shows a double-stranded DNA with 5' and 3' ends labelled, with an arrow to the product: endonuclease cuts within the strand, exonuclease removes nucleotides from the ends; the rows differ in where cuts are shown and in the strand polarity labels. See it in the official paper.

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    1. An endonuclease makes cuts at specific positions within the DNA, whereas an exonuclease removes nucleotides from the ends. According to the scheme, row (d) shows both activities correctly.

       1 mark

    Answer: Option (d)

    Where the marks are

    • Selects the row showing an internal cut and an end removal (1 mark)

11. Q11 · 1 mark · MCQBiotechnology and its Applications

    The main objective of production of pest resistant GM crops is to

    1. (A)encourage eco-friendly pesticides
    2. (B)reduce pesticide accumulation in food chain
    3. (C)eliminate pests from the field without the use of manual labour
    4. (D)retain maximum nutritional content in the crop that would be otherwise consumed by pest

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    1. Pest-resistant GM crops reduce the need to spray chemical pesticides. This in turn lowers the accumulation of pesticides in the food chain (biomagnification).

       1 mark

    Answer: (B)

    Where the marks are

    • Links GM crops to reduced pesticide accumulation in the food chain (1 mark)

12. Q12 · 1 mark · MCQCompetency-basedEcosystem

    Observe the contents 1,2,3 and 4 of soil samples A,B and C shown in the graph. If the temperature and soil moisture of all soil samples are identical, which soil sample (s) will show faster decomposition?

    1. (A)Soil Sample A
    2. (B)Soil Sample B
    3. (C)Soil Samples A and B both
    4. (D)Soil Sample C

    Figure in the paper: Bar graph of Content (%) (0-20) for Soil A, B, C, each with bars 1-4. 1 indicates lignin content, 2 indicates chitin, 3 indicates Nitrogen content and 4 indicates sugar content. Approx. values: Soil A: 8, 15, 5, 20; Soil B: 15 (10), 20, 7, 11.5; Soil C: 8, 10, 17, 20. See it in the official paper.

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    1. Decomposition is faster when the detritus is rich in nitrogen and water-soluble sugars, and slower when it contains more lignin and chitin.

       ½ mark

    2. Soil C has the lowest lignin and chitin together with the highest nitrogen and sugar, so it will decompose fastest.

       ½ mark

    Answer: (D) Soil Sample C

    Where the marks are

    • States the factors that speed up or slow down decomposition (0.5 mark)
    • Applies them to choose Soil C (0.5 mark)

    Open Q12 on its own page

13. Q13 · 1 mark · Assertion-reasonCompetency-basedSexual Reproduction in Flowering Plants

    Assertion: Primary endosperm nucleus is diploid.
    Reason: It is the product of double fertilisation.

    1. (A)Both A and R are true and R is the correct explanation of A.
    2. (B)Both A and R are true and R is not the correct explanation of A.
    3. (C)A is true but R is false.
    4. (D)A is false but R is true.

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    1. The primary endosperm nucleus forms when a male gamete fuses with the two polar nuclei (triple fusion), so it is triploid. The assertion is therefore false.

       ½ mark

    2. Triple fusion is one part of double fertilisation, so the reason is true.

       ½ mark

    Answer: (D) A is false but R is true

    Where the marks are

    • States that the PEN is triploid (0.5 mark)
    • Accepts that it results from double fertilisation (0.5 mark)

    Open Q13 on its own page

14. Q14 · 1 mark · Assertion-reasonCompetency-basedMolecular Basis of Inheritance

    Assertion: Ribosomal RNA is synthesized in the nucleus of the cell.
    Reason: It is translated with the enzyme RNA polymerase III.

    1. (A)Both A and R are true and R is the correct explanation of A.
    2. (B)Both A and R are true and R is not the correct explanation of A.
    3. (C)A is true but R is false.
    4. (D)A is false but R is true.

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    1. rRNA is synthesised in the nucleus, within the nucleolus, so the assertion is true.

       ½ mark

    2. rRNA is transcribed by RNA polymerase I and is never translated. RNA polymerase III transcribes tRNA and 5S rRNA. The reason is therefore false.

       ½ mark

    Answer: (C) A is true but R is false

    Where the marks are

    • Confirms that rRNA is made in the nucleus (0.5 mark)
    • Identifies the error in R: rRNA is transcribed by RNA polymerase I, not translated (0.5 mark)

    Open Q14 on its own page

15. Q15 · 1 mark · Assertion-reasonCompetency-basedHuman Health and Disease

    Assertion: Smoking can raise blood pressure and increase heart rate.
    Reason: Nicotine stimulates adrenal glands to release adrenaline and nor-adrenaline into the blood circulation, both of which raise blood pressure and increase heart rate.

    1. (A)Both A and R are true and R is the correct explanation of A.
    2. (B)Both A and R are true and R is not the correct explanation of A.
    3. (C)A is true but R is false.
    4. (D)A is false but R is true.

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    1. Nicotine stimulates the adrenal medulla to release adrenaline and nor-adrenaline, and these hormones raise blood pressure and heart rate. Both statements are true, and R correctly explains A.

       1 mark

    Answer: (A)

    Where the marks are

    • Recognises that both statements are true and that R explains A (1 mark)

16. Q16 · 1 mark · Assertion-reasonCompetency-basedBiotechnology and its Applications

    Assertion: PCR is a powerful technique to identify genetic disorders.
    Reason: PCR can detect mutations in low amounts of DNA.

    1. (A)Both A and R are true and R is the correct explanation of A.
    2. (B)Both A and R are true and R is not the correct explanation of A.
    3. (C)A is true but R is false.
    4. (D)A is false but R is true.

    Try it first, then show our solutionSolution

    1. PCR amplifies even tiny amounts of DNA, which lets mutations be detected at an early stage. This is exactly why PCR is useful for identifying genetic disorders, so both statements are true and R explains A.

       1 mark

    Answer: (A)

    Where the marks are

    • Recognises that both statements are true and that R explains A (1 mark)

Section B

1. Q17 · 2 marks · Very short answerHuman Reproduction

   Explain the process of hormonal regulation of spermatogenesis.

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   1. At puberty, the hypothalamus sharply increases its secretion of gonadotropin releasing hormone (GnRH), and this starts spermatogenesis.

      ½ mark

   2. GnRH acts on the anterior pituitary. The pituitary then secretes two gonadotropins: luteinising hormone (LH) and follicle stimulating hormone (FSH).

      ½ mark

   3. LH acts on the Leydig (interstitial) cells, which make and release androgens (testosterone). Androgens stimulate spermatogenesis.

      ½ mark

   4. FSH acts on the Sertoli cells, which release factors that help in spermiogenesis (conversion of spermatids into sperms).

      ½ mark

   Answer: Hypothalamic GnRH → anterior pituitary releases LH and FSH; LH → Leydig cells → androgens → spermatogenesis; FSH → Sertoli cells → factors aiding spermiogenesis.

   Where the marks are

   • GnRH from hypothalamus rises at puberty (0.5 mark)
   • GnRH stimulates anterior pituitary to release LH and FSH (0.5 mark)
   • LH acts on Leydig cells to secrete androgens that stimulate spermatogenesis (0.5 mark)
   • FSH acts on Sertoli cells to secrete factors that help spermiogenesis (0.5 mark)

   Open Q17 on its own page

2. Q18 · 2 marks · Very short answerCompetency-basedMolecular Basis of Inheritance

   The diagram below shows the sequence of amino acids in part of a haemoglobin molecule.

   1. a)

      If the base T* was substituted with A, how would it affect the haemoglobin chain?

      0 marks
   2. b)

      Name the condition and the effects associated with the above substitution.

      0 marks

   Figure in the paper: Haemoglobin chain: Val, His, Leu, Thr, Pro, Glu, Glu; below it the mRNA, below that DNA triplets CAT, GTA, AAT, TGA, GGA, CTT (T in the middle marked with * ), CTC. Key: Val=valine, His=histidine, Leu=leucine, Thr=threonine, Pro=proline, Glu=glutamic acid. See it in the official paper.

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   1. a) The DNA triplet for glutamic acid is CTT, which gives the mRNA codon GAA. If the middle T* is replaced by A, the triplet becomes CAT. The mRNA codon then becomes GUA, which codes for valine. So in this chain, valine replaces glutamic acid (Glu → Val) at that position.

      ½ mark

   2. b) The condition is sickle-cell anaemia, caused by this point mutation (single base substitution) in the β-globin gene.

      ½ mark

   3. Effect: the mutant haemoglobin (HbS) polymerises under low oxygen tension.

      ½ mark

   4. This changes the RBC from a biconcave disc into an elongated, sickle-shaped cell.

      ½ mark

   Answer: a) CTT → CAT (mRNA GAA → GUA), so glutamic acid is replaced by valine. b) Sickle-cell anaemia: mutant Hb polymerises under low O₂ and the RBCs become sickle-shaped.

   Where the marks are

   • States the triplet becomes CAT/GUA and valine replaces glutamic acid (0.5 mark)
   • Names sickle-cell anaemia (0.5 mark)
   • Mutant haemoglobin polymerises (0.5 mark)
   • RBC changes from biconcave disc to sickle shape (0.5 mark)

   Open Q18 on its own page

3. Q19 · 2 marks · Very short answerCompetency-basedHuman Health and Disease

   The graph given below indicates the administration of the first (L) and second dose (M) of a vaccine. The corresponding response of the body is indicated by X and Y. Interpret the graph and explain the reason for such a response shown by the body.

   Figure in the paper: Graph of concentration of antibody (y) vs time (x). First dose L is followed by a small peak X; second dose M is followed by a much higher, steeper peak Y. See it in the official paper.

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   1. After the first dose (L), the body meets the antigen of the weakened or inactivated pathogen for the first time. It gives a slow, low-intensity primary immune response (X) with little antibody.

      1 mark

   2. After the second dose (M), the memory B and T cells formed during the first contact recognise the same antigen. They produce a rapid, highly intensified secondary (anamnestic) response (Y). This response is faster and stronger, with far more antibody, so the pathogen is removed more effectively. This is why booster doses are given.

      1 mark

   Answer: X = weak primary immune response to the first exposure; Y = fast, strong secondary response due to immunological memory formed after the first dose.

   Where the marks are

   • Interprets X as low-intensity primary response on first contact with antigen (1 mark)
   • Interprets Y as rapid, intensified secondary response due to memory cells (1 mark)

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4. Q20 · 2 marks · Very short answerCompetency-basedBiotechnology: Principles and Processes

   The image below shows the result of plating bacteria in chromogenic medium after incorporating the gene of interest in plasmid. Some plates had blue colonies; some plates had white colonies. A single bacterium extracted from Plate I,II,III is shown below:

   On the basis of your observations

   1. a)

      Identify the plate(s) which is/are white. Give a reason.

      0 marks
   2. b)

      Identify the plate(s) which is/are blue. Give a reason.

      0 marks

   Figure in the paper: Bacterium + plasmid + gene of interest. Bacteria from plate I contains a plasmid with the gene of interest inserted (green bar on plasmid); bacteria from plate II contains a plasmid without the insert; bacteria from plate III contains no plasmid. See it in the official paper.

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   1. a) Plate I has white colonies. The gene of interest is inserted inside the lacZ (β-galactosidase) gene of the plasmid. This causes insertional inactivation, so no β-galactosidase is made. The chromogenic substrate is not acted on and the colonies stay white (these are the recombinants).

      1 mark

   2. b) Plate II has blue colonies. The plasmid has no insert, so its lacZ gene is intact. β-galactosidase is made and converts the chromogenic substrate into a blue product (non-recombinants).

      ½ mark

   3. Plate III also has blue colonies. The bacterium has no plasmid (no foreign DNA), so its own β-galactosidase activity is unaffected and the colonies appear blue.

      ½ mark

   Answer: a) Plate I is white: insertional inactivation of β-galactosidase. b) Plates II and III are blue: plasmid without insert / no plasmid, so β-galactosidase is active.

   Where the marks are

   • Plate I white because the insert inactivates the β-galactosidase gene (1 mark)
   • Plate II blue because the plasmid lacks the insert (0.5 mark)
   • Plate III blue because there is no plasmid (0.5 mark)

   Open Q20 on its own page

5. Answer one of these 2 alternatives.

   Q21(a) · 2 marks · Very short answerCompetency-basedEcosystem

   Biomass of a standing crop of phytoplankton is 4 kg/m${}^2$ which supports a large standing crop of zooplankton having a biomass 11 kg/m${}^2$. This is consumed by small fishes having biomass 25 kg/m${}^2$ which are then consumed by large fishes with the biomass 37 kg/m${}^2$.

   Draw an ecological pyramid indicating the biomass at each stage and also name the trophic levels. Mention whether it is an upright or inverted pyramid.

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   1. Pyramid of biomass, drawn from bottom to top (each bar's width proportional to biomass):

      Level (top to bottom)	Organisms	Biomass
      T4 – Tertiary consumers (top carnivores)	Large fishes	37 kg/m² (widest bar, at top)
      T3 – Secondary consumers	Small fishes	25 kg/m²
      T2 – Primary consumers (herbivores)	Zooplankton	11 kg/m²
      T1 – Producers	Phytoplankton	4 kg/m² (narrowest bar, at base)

      1½ marks

   2. The base (producers) is the smallest bar and the bars widen upwards. So it is an inverted pyramid of biomass, which is typical of aquatic ecosystems because the small standing crop of phytoplankton turns over rapidly.

      ½ mark

   Answer: Pyramid with phytoplankton (producers) 4 → zooplankton (primary consumers) 11 → small fish (secondary consumers) 25 → large fish (tertiary consumers) 37 kg/m²; it is an inverted pyramid of biomass.

   Where the marks are

   • Correct pyramid drawn with biomass values at each level (1 mark)
   • Trophic levels correctly named (0.5 mark)
   • States it is inverted (0.5 mark)

   Open Q21(a) on its own page

   OR

   Q21(b) · 2 marks · Very short answerCompetency-basedEcosystem

   Use the information provided in the table given below to answer the following questions:

   Tropic level	Net Production (KJm${}^{-2}$y${}^{-1}$)	Respiration (KJm${}^{-2}$y${}^{-1}$)
   Top Carnivore	50	35
   Carnivores	420	378
   Herbivores	4490	4041
   Producers	45000	40,367

   a) Calculate the gross primary productivity.
   b) Analyze the trend in the Net Production from Producers to Top Carnivore. Give a reason for your observation.

   Try it first, then show our solutionSolution

   1. a) GPP = NPP + R = $45000+40367=85367{\text{kJ m}}^{-2}{\text{y}}^{-1}$

      1 mark

   2. b) Net production falls steadily from producers (45000) to herbivores (4490), carnivores (420) and top carnivores (50). At each level only about 10% of the energy is passed on (10% law), because most energy is lost as heat in respiration.

      1 mark

   Answer: a) GPP = 85367 kJ m⁻² y⁻¹. b) Net production decreases at each higher trophic level (about 10% transfer) due to respiratory heat loss.

   Where the marks are

   • GPP = NPP + respiration = 85367 kJ m⁻² y⁻¹ (1 mark)
   • States the decreasing trend with reason: respiration/heat loss/10% law (1 mark)

   Open Q21(b) on its own page

Section C

1. Q22 · 3 marks · Short answerCompetency-basedSexual Reproduction in Flowering Plants

   The figure given below shows 3 sperms A, B and C.

   1. a)

      Which one of the three sperms will gain entry into the ovum?

      0 marks
   2. b)

      Describe the associated changes induced by it on P and Q.

      0 marks

   Figure in the paper: Figure: Ovum surrounded by few sperms. Ovum surrounded by granulosa cells; sperms A, B, C approach, A is in contact with the layer P (zona pellucida region), Q is the ovum cytoplasm centre, R is another labelled part, perivitelline space labelled. See it in the official paper.

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   1. a) Sperm A, the one in contact with the zona pellucida, will enter the ovum.

      ½ mark

   2. b) Change in P (zona pellucida): Contact of sperm A with the zona pellucida (P) causes changes in the membrane that block the entry of additional sperms (B and C). This ensures monospermy.

      ½ mark

   3. The acrosomal secretions of sperm A help it pass through P and the plasma membrane into the cytoplasm of Q. This entry triggers completion of meiosis II of the secondary oocyte (Q).

      1 mark

   4. Change in Q (secondary oocyte): Meiosis II is unequal and forms a second polar body and a haploid ovum (ootid). The haploid nucleus of sperm A then fuses with the haploid nucleus of the ovum, forming a diploid zygote.

      1 mark

   Answer: a) Sperm A. b) The zona pellucida changes to block entry of other sperms; sperm entry completes meiosis II in Q, giving a second polar body and a haploid ovum, which fuses with the sperm nucleus to form a diploid zygote.

   Where the marks are

   • Identifies sperm A (0.5 mark)
   • Zona pellucida changes to block polyspermy (0.5 mark)
   • Acrosomal secretions help entry; entry induces completion of meiosis II (1 mark)
   • Second polar body and haploid ovum form; fusion gives diploid zygote (1 mark)

   Open Q22 on its own page

2. Q23 · 3 marks · Short answerHuman Reproduction

   Explain the phases in embryonic development from the morula stage till the establishment of pregnancy in a human female.

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   1. The zygote divides by cleavage, and the embryo with 8–16 blastomeres is called a morula.

      ½ mark

   2. The morula keeps dividing and becomes a blastocyst as it moves further towards the uterus.

      ½ mark

   3. The blastomeres of the blastocyst form an outer layer called the trophoblast.

      ½ mark

   4. An inner group of cells attached to the trophoblast forms the inner cell mass, which differentiates into the embryo.

      ½ mark

   5. The trophoblast attaches to the endometrium. The uterine cells then divide rapidly and cover the blastocyst.

      ½ mark

   6. The blastocyst becomes embedded in the endometrium. This is called implantation, and it leads to pregnancy.

      ½ mark

   Answer: Morula (8–16 cells) → blastocyst (trophoblast + inner cell mass) → trophoblast attaches to endometrium → uterine cells cover it → implantation → pregnancy.

   Where the marks are

   • Morula defined (0.5 mark)
   • Morula becomes blastocyst while moving to uterus (0.5 mark)
   • Outer trophoblast layer (0.5 mark)
   • Inner cell mass, which forms the embryo (0.5 mark)
   • Trophoblast attaches to endometrium; uterine cells cover blastocyst (0.5 mark)
   • Embedding = implantation, leading to pregnancy (0.5 mark)

   Open Q23 on its own page

3. Q24 · 3 marks · Short answerCompetency-basedHuman Reproduction

   A pregnant human female was advised to undergo MTP. It was diagnosed that the fetus she was carrying had developed from a zygote having 45 chromosomes with only one X chromosome.

   1. a)

      What is this condition called and how does it arise?

      0 marks
   2. b)

      Why was she advised to undergo MTP?

      0 marks

   Try it first, then show our solutionSolution

   1. a) The condition is Turner's syndrome (karyotype 45, XO).

      ½ mark

   2. It is an aneuploidy of the sex chromosomes in which one X chromosome is missing. It arises from non-disjunction of sex chromosomes during gametogenesis, so a gamete lacking a sex chromosome fuses with a normal X-bearing gamete.

      ½ mark

   3. b) MTP was advised because the child would be a sterile female with serious defects:

      • Rudimentary ovaries, so she would be infertile
      • Lack of secondary sexual characters (poorly developed breasts) and delayed or absent menstruation
      2 marks

   Answer: a) Turner's syndrome (45, XO), an aneuploidy caused by non-disjunction of the sex chromosomes. b) The child would have rudimentary ovaries (sterility), lack secondary sexual characters and have absent/delayed menstruation.

   Where the marks are

   • Names Turner's syndrome (0.5 mark)
   • Explains absence of one X / aneuploidy due to non-disjunction (0.5 mark)
   • Any two symptoms such as rudimentary ovaries, lack of secondary sexual characters, infertility (1 + 1 marks)

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4. Q25 · 3 marks · Short answerCompetency-basedEvolution

   The graphs below show three types of natural selection. The shaded areas marked with arrows show the individuals in the population which are not selected. The dotted vertical lines show the statistical means.

   1. a)

      What names are given to the types of selection shown in graphs A, B and C.

      0 marks
   2. b)

      After the selection has operated for several generations in the above populations indicated as Graph A, B and C, graphically illustrate the probable results.

      0 marks

   Figure in the paper: Three bell curves of number of individuals vs character. Graph A: both extreme tails shaded. Graph B: only the left tail shaded. Graph C: the central region around the mean shaded. See it in the official paper.

   Try it first, then show our solutionSolution

   1. a) Graph A: stabilising selection (both extremes removed). Graph B: directional selection (one extreme removed). Graph C: disruptive selection (the mean/intermediates removed).

      1½ marks

   2. b) Probable results after several generations, described as graphs:

      • Graph A (stabilising): the curve becomes narrower and taller, with the same mean. More individuals now have the mean character value.
      • Graph B (directional): the whole curve, with its peak and mean, shifts to the right (away from the eliminated left tail).
      • Graph C (disruptive): the curve splits into two peaks at the two extremes, with a dip at the original mean.
      1½ marks

   Answer: a) A: stabilising, B: directional, C: disruptive. b) A: narrower, taller peak at the same mean; B: peak shifted to the right; C: two peaks at the extremes.

   Where the marks are

   • Correctly names all three types of selection (1.5 marks)
   • Correct resultant curves for A, B and C (0.5 × 3 = 1.5 marks)

   Open Q25 on its own page

5. Q26 · 3 marks · Short answerCompetency-basedBiodiversity and Conservation

   The aeration tank of a sewage treatment plant is not functioning properly. Explain in detail the impact of this on the treatment of sewage and BOD of the effluent.

   Try it first, then show our solutionSolution

   1. Secondary (biological) treatment of sewage, which takes place in the aeration tank, will be badly affected.

      ½ mark

   2. If the aeration tank does not work, air will not be pumped into the effluent.

      ½ mark

   3. Without oxygen, useful aerobic microbes cannot grow vigorously into flocs (masses of bacteria joined with fungal filaments into mesh-like structures).

      ½ mark

   4. So the organic matter in the effluent will not be consumed by these microbes.

      ½ mark

   5. The BOD will not fall. BOD is the amount of oxygen that would be used if all the organic matter in one litre of water were oxidised by bacteria.

      ½ mark

   6. Since higher BOD means greater polluting potential, the effluent will remain highly polluted with organic matter and high BOD. It will also not form activated sludge properly and cannot be discharged safely.

      ½ mark

   Answer: No aeration means no aerobic flocs, so organic matter is not degraded and BOD stays high; the effluent remains polluted.

   Where the marks are

   • Secondary treatment affected (0.5 mark)
   • No air pumped in (0.5 mark)
   • Aerobic microbes fail to form flocs (0.5 mark)
   • Organic matter not consumed (0.5 mark)
   • BOD not reduced, with BOD explained (0.5 mark)
   • Effluent stays polluted/high BOD (0.5 mark)

   Open Q26 on its own page

6. Answer one of these 2 alternatives.

   Q27(a) · 3 marks · Short answerCompetency-basedBiotechnology and its Applications

   A farmer grew 2 varieties of corn crop in field A and B. He grew normal corn crops in field A and GM corn crops in field B. He observed corn borers attacked only in field A. To control it, spores of Bt were sprayed in field A.
   a) Name the gene in the spores responsible for the control of this pest.
   b) What effect will the spores of Bt have on the insect pest?
   c) How has field B developed resistance against this pest?

   Try it first, then show our solutionSolution

   1. a) The gene is *cry*IAb, whose toxin controls the corn borer.

      ½ mark

   2. b) Bt spores contain the toxin as an inactive crystalline protoxin.

      ½ mark

   3. When the insect eats it, the alkaline pH of the insect gut solubilises the crystals and activates the toxin.

      ½ mark

   4. The active toxin binds to the midgut epithelial cells, creating pores. The cells swell and lyse, and the insect dies.

      ½ mark

   5. c) Field B has Bt corn (GM crop). The Bt toxin gene was isolated, cloned and inserted into the corn genome by recombinant DNA technology. The plant makes the toxin itself, so it is resistant to corn borers.

      1 mark

   Answer: a) cryIAb. b) The inactive crystal toxin is activated by alkaline gut pH and lyses midgut epithelial cells, killing the insect. c) The Bt cry gene was inserted by rDNA technology, so the GM corn expresses the toxin and resists the pest.

   Where the marks are

   • Names cryIAb (0.5 mark)
   • Toxin present as inactive crystals (0.5 mark)
   • Activated by alkaline pH of insect gut (0.5 mark)
   • Midgut epithelial cells become porous, swell and lyse (0.5 mark)
   • Bt gene inserted by rDNA technology, plant expresses toxin (1 mark)

   Open Q27(a) on its own page

   OR

   Q27(b) · 3 marks · Short answerCompetency-basedBiotechnology and its Applications

   Lipoprotein lipase deficiency (LPLD) is a genetic disorder in which a person has a defective gene for lipase. This leads to high triglycerides, stomach pain, fat deposits under the skin. It may eventually affect the liver, pancreas and may also cause diabetes. The disorder occurs if a child acquires defective genes from both parents (autosomal recessive). ERT (enzyme replacement treatment) is one of the treatments offered to patients with LPLD.
   a) (i) What procedure is followed in ERT?
   (ii) What could be one possible drawback of ERT?
   b) How can LPLD be treated using Biotechnology? Elaborate.

   Try it first, then show our solutionSolution

   1. a)(i) In ERT, the functional enzyme (lipoprotein lipase) is given to the patient by injection.

      ½ mark

   2. a)(ii) ERT is not a permanent cure. The injected enzyme is used up, so the patient needs repeated, lifelong doses.

      ½ mark

   3. b) LPLD can be treated by gene therapy.

      ½ mark

   4. Gene therapy is a collection of methods that correct a gene defect diagnosed in a child or embryo.

      ½ mark

   5. A normal (functional) lipase gene is delivered into the person's cells or tissues, for example using a disarmed viral vector. The gene takes over the function of the defective gene and compensates for it, so the patient makes functional enzyme. If introduced at an early embryonic stage, it could be a permanent cure.

      1 mark

   Answer: a)(i) Functional lipase is injected; (ii) not curative, so repeated doses are needed. b) Gene therapy: a normal lipase gene is delivered into the patient's cells to compensate for the defective gene.

   Where the marks are

   • ERT: functional enzyme given by injection (0.5 mark)
   • Drawback: not completely curative / repeated doses (0.5 mark)
   • Names gene therapy (0.5 mark)
   • Defines gene therapy (0.5 mark)
   • Normal gene delivered to compensate for the non-functional gene (1 mark)

   Open Q27(b) on its own page

7. Q28 · 3 marks · Short answerBiodiversity and Conservation

   Give three reasons as to why the prokaryotes are not given any figures for their diversity by the ecologists.

   Try it first, then show our solutionSolution

   1. Classifying and identifying the huge diversity of microbes is very difficult. Conventional taxonomic methods are not suitable for this.

      1 mark

   2. Many microorganisms cannot be cultured under laboratory conditions, so they cannot be studied and counted.

      1 mark

   3. Biochemical and molecular techniques suggest that prokaryotic species could run into millions or billions, living in very diverse habitats (air, water, soil, extreme environments). More advanced techniques are still needed to estimate this diversity reliably.

      1 mark

   Answer: (1) Classification/identification of microbes is very difficult; (2) many cannot be cultured in the lab; (3) molecular estimates suggest an enormous (millions to billions) diversity that current methods cannot count.

   Where the marks are

   • Difficulty in identification/classification by conventional methods (1 mark)
   • Many microbes cannot be cultured in the laboratory (1 mark)
   • Enormous diversity in varied habitats, estimated only by molecular techniques (1 mark)

   Open Q28 on its own page

Section D

case-based

1. Q29 · 4 marks · Case-basedCompetency-basedBiotechnology: Principles and Processes

   The structure below shows pUC18 which is similar to pBR322 in its function. However, they differ in some of their restriction sites and number of ori. The ori number for pBR322 is approximately 20.

   1. a)

      How are puc18 and pBR322 used in biotechnological studies?

      0 marks

   2. OR

      a)

      What will be the impact if ori in the above structure gets damaged?

      0 marks
   3. b)

      The lac z gene has many recognition sites. Study the segment of DNA given below and answer the questions
      5’... ATC GTA AAG CTT CAT…3’
      3’... TAG CAT TTC GAA GTA…5’
      i) Applying your knowledge of palindrome sequences identify and mark the possible region where the restriction enzyme X will act.
      ii) Restriction enzyme Y was used to extract gene of interest from a plant. This gene needs to be inserted in the given DNA segment which has been treated with restriction enzyme X. Will there be a successful recombination? Explain with a reason.

      0 marks
   4. c)

      Which one of the two (pUC18 and pBR322) would you prefer for biotechnological studies? Justify.

      0 marks

   Figure in the paper: Circular map of plasmid pUC18 showing AmpR gene, lacZ gene and ori (label: ori more than or equal to 100). See it in the official paper.

   Try it first, then show our solutionSolution

   1. a) pUC18 and pBR322 are used as cloning vectors. The gene of interest from a desired organism is inserted into them and carried into a host cell, where it is replicated or expressed.

      a) (OR) If ori is damaged, the plasmid cannot replicate in the host, so no copies of the linked DNA (gene of interest) will be made.

      1 mark

   2. b)(i) The palindrome, which reads the same 5'→3' on both strands, is the HindIII site:
      5'… ATC GTA /AAG CTT/ CAT …3'
      3'… TAG CAT /TTC GAA/ GTA …5'
      Enzyme X acts at 5'-AAGCTT-3' / 3'-TTCGAA-5'.

      1 mark

   3. b)(ii) No, recombination will not be successful. Enzyme Y produces sticky ends different from those made by enzyme X, so the ends are not complementary and cannot be joined by ligase. The same restriction enzyme must cut both the vector and the source DNA.

      1 mark

   4. c) pUC18 is preferred because its copy number is higher (ori ≥ 100 copies, against about 20 for pBR322). It therefore gives many more copies of the recombinant DNA.

      1 mark

   Answer: a) Used as cloning vectors (OR: damaged ori means no replication, so no copies). b)(i) AAGCTT/TTCGAA; (ii) No, different enzymes give non-complementary sticky ends. c) pUC18, due to its higher copy number.

   Where the marks are

   • Used as vectors to carry gene of interest / ori damage means no replication (1 mark)
   • Correct palindrome marked on both strands (1 mark)
   • No, the same restriction enzyme is needed for complementary sticky ends (0.5 + 0.5 marks)
   • pUC18 with justification of higher copy number (0.5 + 0.5 marks)

   Open Q29 on its own page

2. Q30 · 4 marks · Case-basedCompetency-basedOrganisms and Populations

   Observe the graph given below.
   The graph represents inter-specific interaction between two species of Paramecia competing for the same resource in a culture medium. Paramecium caudatum and Paramecium aurelia were grown in separate cultures as well as in mixed cultures. It was found that each species grew in numbers according to the logistic equation.

   1. a)

      Which species is competitively superior? Support it with the data provided in the graph.

      0 marks
   2. b)

      State the underlying principle for the above result and name the scientist associated with this principle.

      0 marks
   3. c)

      Explain the mechanism in which two or more species competing with each other can co - exist.

      0 marks

   4. OR

      c)

      Graphs A and B shown below depict interaction of two species. Which graph indicates Mutualism? Give reason.

      0 marks

   Figure in the paper: Graph 1: number of paramecia (0-100) vs time (days, 0-18). P. aurelia alone reaches about 100 (dashed), in mixed culture about 90; P. caudatum alone reaches about 62 (dashed), in mixed culture peaks near 27 at day 5 then declines to near 0. Graphs A and B (OR part): population size vs time with species introduced at a dotted line; in A both species A and B increase after introduction; in B species B increases steeply while A stays flat then declines slightly. See it in the official paper.

   Try it first, then show our solutionSolution

   1. a) P. aurelia is competitively superior.

      1 mark

   2. Evidence from the graph:

      • Grown alone, P. aurelia reaches about 100 individuals, while P. caudatum reaches only about 60.
      • In mixed culture, P. aurelia still rises to about 90, whereas P. caudatum peaks at about 27 around day 5 and then declines to almost zero.
      1 mark

   3. b) This illustrates Gause's Competitive Exclusion Principle: two closely related species competing for the same limited resources cannot coexist indefinitely, and the competitively inferior one is eliminated. The scientist is G.F. Gause.

      1 mark

   4. c) Competing species can coexist by resource partitioning. They avoid competition by feeding at different times or using different foraging patterns. For example, MacArthur's five warbler species coexist on the same tree by foraging in different parts of it.

      c) (OR) Graph A shows mutualism. After the two species are brought together, both populations increase, so both benefit (+/+). In Graph B only one species gains.

      1 mark

   Answer: a) P. aurelia (about 90–100 individuals; P. caudatum falls to near 0 in mixed culture). b) Competitive exclusion principle, G.F. Gause. c) Resource partitioning (OR: Graph A, since both species increase).

   Where the marks are

   • Names P. aurelia (1 mark)
   • Supports with graph data (1 mark)
   • States the competitive exclusion principle and names Gause (1 mark)
   • Explains resource partitioning / identifies Graph A as mutualism with reason (1 mark)

   Open Q30 on its own page

Section E

1. Answer one of these 2 alternatives.

   Q31(a) · 5 marks · Long answerCompetency-basedReproductive Health

   Placed below are case studies of some couples who were not able to have kids. These couples are not ready for adoption or taking gametes from donors. After thoroughly examining the cases, which Assisted Reproductive Technology will you suggest to these couples as a medical expert? Explain briefly with justification of each case.

   Couple	Test reports of Female partner	Test reports of male partner
   Couple 1	Normal reports	Normal sperms in testes, Missing connection in epididymis and Vas deferens
   Couple 2	Blockage in the fallopian tube	Normal reports
   Couple 3	Normal reports	Poor semen parameters in terms of count, motility and morphology
   Couple 4	low ovarian reserve	Normal reports
   Couple 5	Sterilization in male	Morphologically abnormal sperms

   Try it first, then show our solutionSolution

   1. Couple 1: Sperms are normal in the testes, but there is no connection through the epididymis or vas deferens, so the semen has no sperms.

      • ART: Collect sperms surgically from the testis/epididymis and fertilise the ovum by IVF (or ICSI).
      • Then transfer the embryo by ZIFT (zygote or embryo up to 8 blastomeres into the fallopian tube) or IUT (embryo with more than 8 blastomeres into the uterus).
      1 mark

   2. Couple 2: The female's blocked fallopian tube prevents sperm and ovum from meeting at the site of fertilisation.

      • ART: IVF followed by IUT (test-tube baby). The embryo with more than 8 blastomeres goes directly into the uterus, bypassing the blocked tube.
      1 mark

   3. Couple 3: The male's sperms have poor count, motility and morphology.

      • ART: ICSI (intracytoplasmic sperm injection). A selected sperm is injected directly into the ovum in the lab, and the embryo is then transferred.
      • If only the count were low, artificial insemination (IUI) could be tried.
      1 mark

   4. Couple 4: The female has low ovarian reserve and the male is normal.

      • ART: Stimulate the ovaries to recover the few available ova, then perform IVF with the partner's sperm.
      • Transfer the embryo by ZIFT (up to 8 blastomeres) or IUT (more than 8 blastomeres).
      1 mark

   5. Couple 5: The female has poor ovarian reserve and the male has morphologically abnormal sperms.

      • ART: Select the best ova and normal-looking sperms, and perform ICSI by injecting a sperm directly into each ovum.
      • Then transfer the embryo (ZIFT/IUT).
      1 mark

   Answer: Couple 1: sperm retrieval + IVF/ICSI, then ZIFT/IUT; Couple 2: IVF + IUT; Couple 3: ICSI; Couple 4: IVF + ZIFT/IUT; Couple 5: ICSI.

   Where the marks are

   • Couple 1: IVF with sperms from testis/epididymis, ZIFT/IUT with reason (1 mark)
   • Couple 2: IVF followed by IUT, bypassing the blocked tube (1 mark)
   • Couple 3: ICSI for poor semen parameters (1 mark)
   • Couple 4: IVF followed by ZIFT/IUT (1 mark)
   • Couple 5: ICSI with selected sperms/ova (1 mark)

   Open Q31(a) on its own page

   OR

   Q31(b) · 5 marks · Long answerCompetency-basedReproductive Health

   Given below are certain situations. Analyse the situation and suggest the name of suitable contraceptive device along with mode of action.

   Situation	Requirement of contraceptive for -	Name of contraceptive device	Mode of action
   1	blocking the entry of sperms through cervix
   2	spacing between children
   3	effective emergency contraceptive
   4	terminal method to prevent any more pregnancy in female
   5	sterilization in male

   Try it first, then show our solutionSolution

   1. Situation 1 (block sperm entry through cervix):

      • Device: Diaphragm / cervical cap / vault (barrier methods)
      • Mode of action: covers the cervix during coitus and stops sperms entering the uterus
      1 mark

   2. Situation 2 (spacing between children):

      • Device: IUD such as Cu-T, Cu-7, Multiload 375 (copper-releasing) or Progestasert, LNG-20 (hormone-releasing)
      • Mode of action: Cu ions suppress sperm motility and fertilising capacity and increase phagocytosis of sperms in the uterus; hormone-releasing IUDs make the uterus unsuitable for implantation and the cervix hostile to sperms
      1 mark

   3. Situation 3 (effective emergency contraceptive):

      • Device: Progestogen or progestogen–estrogen pills, or an IUD, used within 72 hours of coitus
      • Mode of action: pills inhibit ovulation and implantation and change cervical mucus to block sperms; IUDs act through Cu ions or hormones as above
      1 mark

   4. Situation 4 (terminal method in female):

      • Device: Tubectomy
      • Mode of action: a small part of each fallopian tube is cut and tied, which blocks transport of the ovum and prevents conception
      1 mark

   5. Situation 5 (sterilisation in male):

      • Device: Vasectomy
      • Mode of action: a small part of each vas deferens is cut and tied, which blocks sperm transport
      1 mark

   Answer: 1: Diaphragm/cervical cap/vault (covers cervix); 2: Cu-T/Cu-7/Multiload/LNG-20 IUDs (Cu ions kill or immobilise sperms, hormones prevent implantation); 3: emergency pills or IUD within 72 h (inhibit ovulation/implantation); 4: Tubectomy (blocks ovum transport); 5: Vasectomy (blocks sperm transport).

   Where the marks are

   • Correct device for each of 5 situations (0.5 × 5 = 2.5 marks)
   • Correct mode of action for each of 5 situations (0.5 × 5 = 2.5 marks)

   Open Q31(b) on its own page

2. Answer one of these 2 alternatives.

   Q32(a) · 5 marks · Long answerMolecular Basis of Inheritance

   Given below is a stretch of DNA showing the coding strand of a structural gene of a transcription unit?
   5’--ATG ACC GTA TTT TCT GTA GTG CCC GTA CTT CAG GCA TAA—3’
   a) Write the corresponding template strand and the mRNA strand that will be transcribed, along with its polarity.
   b) If GUA of the transcribed mRNA is an intron, depict the sequence involved in the formation of mRNA /the mature processed hnRNA strand.
   i. In a bacterium
   ii. In humans
   c) Upon translation, how many amino acids will the resulting polypeptide have? Justify.

   Try it first, then show our solutionSolution

   1. a) Template strand (complementary and antiparallel to the coding strand):
      3'– TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC CGT ATT –5'

      1 mark

   2. mRNA (same as the coding strand, with U in place of T):
      5'– AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UAA –3'

      1 mark

   3. b)(i) In a bacterium: prokaryotic genes are not split (no introns), so no splicing occurs. The mRNA is used as transcribed:
      5'– AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UAA –3'

      1 mark

   4. b)(ii) In humans: the hnRNA is processed in three ways:

      • Splicing: all three GUA introns are removed and the exons are joined.
      • Capping: a methyl guanosine cap is added at the 5' end.
      • Tailing: a poly-A tail is added at the 3' end.

      Mature mRNA: 5'– m⁷GpppAUG ACC UUU UCU GUG CCC CUU CAG GCA UAA – poly(A) tail –3'

      1 mark

   5. c) The mature human mRNA has 10 codons. UAA is a stop codon and codes for no amino acid, so the polypeptide has 9 amino acids. (For the unspliced bacterial mRNA, 12 codons come before UAA, giving 12 amino acids.)

      1 mark

   Answer: a) Template 3'-TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC CGT ATT-5'; mRNA 5'-AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UAA-3'. b) Bacterium: unchanged; humans: 5'-m⁷GpppAUG ACC UUU UCU GUG CCC CUU CAG GCA UAA-polyA-3'. c) 9 amino acids (UAA is a stop codon).

   Where the marks are

   • Correct template strand with polarity (1 mark)
   • Correct mRNA with polarity (1 mark)
   • Bacterial mRNA unchanged, as no splicing occurs (1 mark)
   • Human mature mRNA with GUA removed, 5' cap and poly-A tail (1 mark)
   • 9 amino acids because UAA is a stop codon (1 mark)

   Open Q32(a) on its own page

   OR

   Q32(b) · 5 marks · Long answerCompetency-basedPrinciples of Inheritance and Variation

   In shorthorn cattle, the coat colours red or white are controlled by a single pair of alleles. A calf which receives the allele for red coat from its mother and the allele for white coat from its father is called a 'roan'. It has an equal number of red and white hairs in its coat.
   a) Is this an example of codominance or of incomplete dominance?
   b) Give a reason for your answer.
   c) With the help of genetic cross explain what will be the consequent phenotype of the calf when
   i. red is dominant over white
   ii. red is incompletely dominant.

   Try it first, then show our solutionSolution

   1. (a) The roan calf shows codominance.

      ½ mark

   2. (b) Reason: The roan coat has red hairs and white hairs side by side, in equal numbers. No hair is an intermediate (pink) colour. Both alleles are therefore fully and separately expressed in the heterozygote (Rr), and neither hides the other. This is the definition of codominance.

      ½ mark

   3. **(c)(i) Red ($R$) completely dominant over white ($r$):**

      Parents: $RR$ (red, mother) $\times$ $rr$ (white, father)

      Gametes: $R$ and $r$

      F${}_1$ (the calf): $Rr$. All F${}_1$ calves are red, because the dominant allele $R$ masks $r$ and the phenotype is the same as the red parent.

      1 mark

   4. Selfing F${}_1$ ($Rr\times Rr$) to show the full pattern:

      	$R$	$r$
      $R$	$RR$ red	$Rr$ red
      $r$	$Rr$ red	$rr$ white

      F${}_2$ phenotypic ratio is red : white = 3 : 1. The genotypic ratio is $RR:Rr:rr$ = 1 : 2 : 1.

      1 mark

   5. (c)(ii) Red incompletely dominant over white:

      Parents: $RR$ (red) $\times$ $rr$ (white)

      Gametes: $R$ and $r$

      F${}_1$ (the calf): $Rr$. All F${}_1$ calves are pink, an intermediate shade, because one $R$ allele cannot produce enough pigment for a full red coat.

      1 mark

   6. Crossing F${}_1$ ($Rr\times Rr$):

      	$R$	$r$
      $R$	$RR$ red	$Rr$ pink
      $r$	$Rr$ pink	$rr$ white

      F${}_2$ ratio is red : pink : white = 1 : 2 : 1. The phenotypic ratio equals the genotypic ratio.

      1 mark

   Answer: (a) Codominance. (b) Both red and white hairs appear separately in the heterozygote, with no blending. (c)(i) Calf $Rr$ is red; F${}_2$ red : white = 3 : 1. (ii) Calf $Rr$ is pink (intermediate); F${}_2$ red : pink : white = 1 : 2 : 1.

   Where the marks are

   • Identifies codominance (0.5 mark)
   • Reason: both alleles expressed independently as separate red and white hairs, no intermediate colour (0.5 mark)
   • Complete dominance cross RR × rr giving Rr red calf (1 mark)
   • F2 Punnett square with 3 : 1 ratio (1 mark)
   • Incomplete dominance cross giving Rr pink calf (1 mark)
   • F2 Punnett square with 1 : 2 : 1 ratio (1 mark)

   Open Q32(b) on its own page

3. Answer one of these 2 alternatives.

   Q33(a) · 5 marks · Long answerHuman Health and Disease

   Explain the role of Primary and Secondary Lymphoid organs with the help of suitable examples.

   Try it first, then show our solutionSolution

   1. Lymphoid organs are organs in which lymphocytes originate and/or mature and proliferate.

      ½ mark

   2. Primary lymphoid organs are the bone marrow and the thymus. In these organs, immature lymphocytes differentiate into antigen-sensitive lymphocytes.

      ½ mark

   3. The bone marrow is the main lymphoid organ. All blood cells, including lymphocytes, are produced here, and B-lymphocytes mature here.

      ½ mark

   4. The thymus is a lobed organ lying near the heart, beneath the breastbone. It provides the micro-environment for maturation of T-lymphocytes. It is large at birth and shrinks with age.

      ½ mark

   5. After maturation, lymphocytes migrate to the secondary lymphoid organs. Examples are the spleen, lymph nodes, tonsils, Peyer's patches of the small intestine and the appendix.

      ½ mark

   6. Secondary lymphoid organs are the sites where lymphocytes interact with antigens. The lymphocytes then proliferate to become effector cells.

      ½ mark

   7. The spleen is a large bean-shaped organ containing mainly lymphocytes and phagocytes. It filters the blood by trapping blood-borne microorganisms, and it also stores a large reservoir of erythrocytes.

      ½ mark

   8. Lymph nodes are small solid structures found at intervals along the lymphatic system. They trap microorganisms and other antigens that enter the lymph and tissue fluid.

      ½ mark

   9. Antigens trapped in lymph nodes activate the lymphocytes present there, which triggers the immune response.

      ½ mark

   10. MALT (mucosa-associated lymphoid tissue) lines the respiratory, digestive and urogenital tracts. It makes up about 50% of the body's lymphoid tissue.

       ½ mark

   Answer: Primary lymphoid organs (bone marrow, thymus) are where lymphocytes are formed and mature into antigen-sensitive cells. Secondary lymphoid organs (spleen, lymph nodes, tonsils, Peyer's patches, appendix, MALT) are where lymphocytes meet antigens and proliferate into effector cells.

   Where the marks are

   • Definition of lymphoid organs (0.5 mark)
   • Primary organs named with their role in differentiation (0.5 mark)
   • Bone marrow role (0.5 mark)
   • Thymus location and role (0.5 mark)
   • Secondary organs named (0.5 mark)
   • Role of secondary organs: antigen interaction and proliferation (0.5 mark)
   • Spleen structure and function (0.5 mark)
   • Lymph nodes trap antigens (0.5 mark)
   • Activation of lymphocytes in lymph nodes (0.5 mark)
   • MALT, about 50% of lymphoid tissue (0.5 mark)

   Open Q33(a) on its own page

   OR

   Q33(b) · 5 marks · Long answerHuman Health and Disease

   With the help of a flow chart illustrate how an infected animal cell can survive while viruses are being replicated or released.

   Try it first, then show our solutionSolution

   1. The example is a retrovirus (HIV) infecting an animal cell such as a macrophage.

      Step 1: The virus enters the animal cell, and its RNA genome is released into the cytoplasm.

      1 mark

   2. ↓

      Step 2: The viral enzyme reverse transcriptase uses the viral RNA as a template to make viral DNA.

      1 mark

   3. ↓

      Step 3: The viral DNA is incorporated into the host cell's genome.

      1 mark

   4. ↓

      Step 4: The host cell's machinery now transcribes the integrated DNA, producing new viral RNA and viral proteins. These assemble into new virus particles.

      1 mark

   5. ↓

      Step 5: New viruses are released and can infect other cells, such as helper T-lymphocytes. The infected cell itself survives while viruses are being replicated and released. A macrophage therefore acts as an 'HIV factory' that keeps producing virus.

      1 mark

   Answer: Flow chart: Virus enters cell → viral RNA released → reverse transcriptase makes viral DNA → viral DNA integrates into host genome → new viral RNA and proteins form new viruses → new viruses released to infect other cells, while the infected cell (e.g. macrophage, an 'HIV factory') survives.

   Where the marks are

   • Entry of retrovirus and release of viral RNA (1 mark)
   • Reverse transcriptase forms viral DNA (1 mark)
   • Viral DNA integrates into host genome (1 mark)
   • New viral RNA and particles produced (1 mark)
   • New viruses released to infect other cells while the infected cell survives, e.g. macrophage as HIV factory (1 mark)

   Open Q33(b) on its own page