CBSE Class 12 Biology Question Paper 2026 Solved Code 57-1-1

SECTION A

1. Identify the plant in which emasculation is not required for artificial hybridization. [1 Mark]
(A) Rice
(B) Wheat
(C) Pea
(D) Papaya

Answer: (D) Papaya

Teacher's Note:
a) Papaya is dioecious, so male and female flowers are on separate plants and there is no chance of self-pollination.
b) Emasculation (removal of anthers) is needed only in bisexual flowers such as rice, wheat and pea.

2. Which of the following is not a component of immune system ? [1 Mark]
(A) Lymphocytes
(B) Plasma Proteins
(C) Red Blood Cells
(D) Bone Marrow

Answer: (C) Red Blood Cells

Teacher's Note:
a) RBCs carry oxygen; they play no direct role in fighting pathogens.
b) Bone marrow is a primary lymphoid organ where lymphocytes are formed.

3. Select the option that shows correct statements : [1 Mark]
(i) Corpus luteum secretes progesterone.
(ii) Only FSH attains a peak level in middle of cycle.
(iii) LH surge induces rupture of graafian follicle.
(iv) Luteal phase of menstrual cycle is also called proliferative phase.
Options :
(A) Only (ii) is correct.
(B) (i) and (iii) are correct.
(C) Only (iv) is correct.
(D) (ii) and (iv) are correct.

Answer: (B) (i) and (iii) are correct.

Teacher's Note:
a) Both LH and FSH reach a peak in the middle of the cycle, so statement (ii) is wrong.
b) The luteal phase is the secretory phase; the follicular phase is the proliferative phase.

4. Select the mismatched pair : [1 Mark]
(A) Mega diversity countries in the world - 12
(B) Genetically different strains of rice in India - Less than 1000
(C) Robert Mays estimation on global species diversity - 7 million species worldwide
(D) Zoological parks - Ex-situ conservation

Answer: (B) Genetically different strains of rice in India - Less than 1000

Teacher's Note:
a) India has more than 50,000 genetically different strains of rice, not less than 1000.
b) Remember the 12 megadiversity countries and Robert May's figure of about 7 million species.

5. Pomato was produced by fusing protoplasts of [1 Mark]
(A) Tomato and Potato
(B) Pomegranate and Tomato
(C) Pomegranate and Potato
(D) Pomegranate, Potato and Tomato

Answer: (A) Tomato and Potato

Teacher's Note:
a) Pomato is a somatic hybrid made by fusing naked protoplasts of tomato and potato.
b) The name itself gives a clue: "Po" from potato and "mato" from tomato.

6. In an ecosystem, different species occupy different levels and vertical distribution of species is found. This is called _________. [1 Mark]
(A) Stratification
(B) Layering
(C) Fragmentation
(D) Population

Answer: (A) Stratification

Teacher's Note:
a) In a forest, trees, shrubs and herbs occupy the top, middle and bottom layers - this is stratification.
b) Fragmentation refers to breaking of habitats, not vertical layers.

7. Which of the following sacred groves is found in Meghalaya ? [1 Mark]
(A) Jaintia hills
(B) Bastar
(C) Chanda
(D) Sarguja

Answer: (A) Jaintia hills

Teacher's Note:
a) Khasi and Jaintia hills in Meghalaya are famous sacred groves.
b) Bastar, Chanda and Sarguja are sacred groves of Madhya Pradesh/Chhattisgarh region.

8. In the following figure, two ways of pairing of two homologous pairs of chromosomes are shown. Which of the following phenomena is expressed ? [1 Mark]

[Figure: Possibility 1 and Possibility 2 - two equally probable arrangements of two homologous chromosome pairs at Metaphase I give rise to different chromosome combinations. Each possibility leads to two Metaphase II cells (Possibility 1: A A with B B, and a a with b b; Possibility 2: A A with b b, and a a with B B), which form gametes labelled Combination 1 (AB), Combination 2 (ab), Combination 3 (Ab) and Combination 4 (aB).]

(A) Linkage of genes
(B) Independent assortment of genes
(C) Multiple alleles
(D) Incomplete dominance

Answer: (B) Independent assortment of genes

Teacher's Note:
a) Random orientation of homologous pairs at metaphase I gives four types of gametes in equal proportion.
b) This chromosomal behaviour explains Mendel's law of independent assortment.

9. Match Column-I with Column-II and choose the correct option : [1 Mark]
Column-I | Column-II
a. Biolistic gun | i. Bacterial cell
b. Chitinase | ii. Tumour inducing
c. Ti | iii. Animal cell
d. Ca⁺⁺ | iv. Fungal cell
  | v. Plant cell
Options : a | b | c | d
(A) i | ii | iii | iv
(B) ii | v | i | iii
(C) v | iv | ii | i
(D) v | i | iv | ii

Answer: (C) v | iv | ii | i (a-v, b-iv, c-ii, d-i)

Teacher's Note:
a) Biolistic gun is used for plant cells; chitinase dissolves the fungal cell wall.
b) Ti stands for tumour inducing plasmid; Ca⁺⁺ treatment makes bacterial cells competent.

10. The idea of use and disuse of organs for evolution of organism was proposed by [1 Mark]
(A) Charles Darwin
(B) Thomas Malthus
(C) Hugo De Vries
(D) Lamarck

Answer: (D) Lamarck

Teacher's Note:
a) Lamarck gave the example of the long neck of the giraffe to explain use and disuse.
b) Darwin proposed natural selection and Hugo de Vries proposed the mutation theory.

11. Which of the following statements about plasmids is incorrect ? [1 Mark]
(A) Plasmids have the ability to replicate within the bacterial cell.
(B) Their replication is controlled by chromosomal DNA.
(C) They are autonomously replicating circular extra-chromosomal DNA.
(D) They often carry antibiotic resistant genes.

Answer: (B) Their replication is controlled by chromosomal DNA.

Teacher's Note:
a) Plasmids have their own origin of replication (ori), so they replicate independently of chromosomal DNA.
b) Read "incorrect" carefully in such questions before choosing.

12. Which connective tissue connects ovary to pelvic wall and uterus ? [1 Mark]
(A) Tendons
(B) Ligaments
(C) Cartilage
(D) Bone

Answer: (B) Ligaments

Teacher's Note:
a) Each ovary is held to the pelvic wall and uterus by ligaments.
b) Tendons join muscle to bone, while ligaments join bone to bone or hold organs in place.

For Questions number 13 to 16, two statements are given - one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) given below :
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

13. Assertion (A) : The population of melanized moths increased in industrial areas after Industrial Revolution.
Reason (R) : In Industrial environment lichen covered trees were replaced by soot-covered trees offering better camouflage to dark coloured moths. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).

Teacher's Note:
a) Dark moths were hidden on soot-covered trunks, so predators picked out the white moths.
b) Industrial melanism is a classic example of natural selection.

14. Assertion (A) : The milk produced by transgenic cow 'Rosie' was nutritionally more balanced product for human babies than natural cow milk.
Reason (R) : It was human protein enriched milk containing human alpha lactaglobulin. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (C) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) Rosie's milk contained human alpha-lactalbumin, not alpha lactaglobulin, so the Reason is false.
b) Learn the exact protein name, as examiners often change one word in such statements.

15. Assertion (A) : Flocs are masses of bacteria associated with fungal filaments in secondary treatment of sewage.
Reason (R) : Flocs help in digestion of solid waste by anaerobic respiration. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (C) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) Flocs are formed in aeration tanks and consume organic matter aerobically, reducing the BOD.
b) Anaerobic digestion occurs later in the anaerobic sludge digesters, not by flocs.

16. Assertion (A) : In terrestrial ecosystem much larger fraction of energy flows through grazing food chain.
Reason (R) : Grazing food chain may be connected to Detritus food chain at some levels. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (D) Assertion (A) is false, but Reason (R) is true.

Teacher's Note:
a) In terrestrial ecosystems, a much larger fraction of energy flows through the detritus food chain.
b) In aquatic ecosystems, the grazing food chain is the major conduit for energy flow.

SECTION B

17. (a) Define Humoral immune response.
(b) Name any two types of antibodies found to give humoral immune response in humans. [2 Marks]

Answer:
1. (a) Humoral immune response is the immune response in which B-lymphocytes produce an army of proteins (antibodies) into the blood in response to pathogens, to fight them.
2. (b) Two types of antibodies: IgA and IgG (IgE and IgM are also correct).

Teacher's Note:
a) It is called "humoral" because antibodies are found in body fluids (blood).
b) Mention B-lymphocytes clearly; T-lymphocytes give cell-mediated immunity.

18. (a) (i) How does a farmer use dormancy of seed to his advantage ?
(ii) Differentiate between pea seed and castor seed. [2 Marks]

Answer:
1. (i) Dormancy allows the farmer to store dry (dehydrated) seeds for a long time and use them to raise crops in the next season; seeds can also be stored as food or for commercial purposes.
2. (ii) Pea seed is non-endospermic (non-albuminous) because the endosperm is fully consumed during embryo development, whereas castor seed is endospermic (albuminous) because some endosperm remains in the mature seed.

Teacher's Note:
a) Use the terms albuminous and non-albuminous for full marks in the difference.
b) One clear difference is enough for (ii); write it in a matched pair.

OR

(b) Identify the stage of follicle where primary oocyte undergoes first meiotic division.
Also mention the products of this division. [2 Marks]

Answer:
1. The primary oocyte completes its first meiotic division in the tertiary follicle in the ovary.
2. The products are a large haploid secondary oocyte and a tiny first polar body.

Teacher's Note:
a) The division is unequal, so most of the cytoplasm stays with the secondary oocyte.
b) Do not write "Graafian follicle" or "ovum" as the product; the scheme expects tertiary follicle and secondary oocyte.

19. Study the sequence of bases in a DNA molecule that is shown below and answer the questions.
5' - GAATTC - 3'
3' - CTTAAG - 5'
(a) What are such sequences called ? Name the enzyme used that recognises such nucleotide sequences.
(b) How do these enzymes function ? [2 Marks]

Answer:
1. (a) These are palindromic nucleotide sequences. They are recognised by restriction endonucleases, here EcoRI.
2. (b) Restriction endonucleases cut both strands at a specific recognition site, a little away from the centre of the palindrome, between the same two bases on opposite strands; EcoRI cuts between G and A, producing sticky ends on each strand.

Teacher's Note:
a) A palindrome reads the same on both strands when read in the same direction (5' to 3').
b) Mention "sticky ends" as they help in joining DNA with DNA ligase.

20. (a) Analyse the following two cases of sex determination in different organisms :
Case I : Males have (XO) sex chromosomes and females have two copies of same sex chromosome (XX).
Case II : Females have two different sex chromosomes (ZW) and males have two copies of same sex chromosome (ZZ).
Identify the type of heterogamety in each case, giving one example of each. [2 Marks]

Answer:
1. Case I: Male heterogamety, as males produce two types of gametes (with X and without X). Example: grasshopper (many insects).
2. Case II: Female heterogamety, as females produce two types of eggs (Z and W). Example: birds such as hen.

Teacher's Note:
a) The sex that produces two different types of gametes is the heterogametic sex.
b) XY type (humans, Drosophila) is also male heterogamety; ZW type is found in birds.

OR

(b) (i) Differentiate between a DNA and RNA nucleotide.
(ii) Name the two types of bonds present within a single nucleotide. [2 Marks]

Answer:
1. (i) A DNA nucleotide has deoxyribose sugar (bases A, G, C, T), whereas an RNA nucleotide has ribose sugar (bases A, G, C, U).
2. (ii) N-glycosidic linkage (between sugar and nitrogen base) and phosphoester linkage (between sugar and phosphate).

Teacher's Note:
a) Phosphodiester bonds join two nucleotides; they are not present within a single nucleotide.
b) Name the two parts each bond joins to earn full marks.

21. (a) Given below is a pyramid found in an ecosystem, where each bar represents the standing crop available in the trophic level.
(i) Identify the kind of pyramid and with the help of an example explain the conditions where this kind of pyramid is possible in nature.
(ii) Write any two limitations of ecological pyramids. [2 Marks]

[Figure: Two horizontal bars stacked one above the other; the lower, shorter bar is labelled PP and the upper, longer bar is labelled PC.]

Answer:
1. (i) It is an inverted pyramid of biomass. It is seen in aquatic ecosystems, where a small standing crop of phytoplankton supports a large standing crop of zooplankton or fish.
2. (ii) Limitations: it does not take into account the same species belonging to two or more trophic levels; it does not accommodate a food web.

Teacher's Note:
a) PP stands for primary producers and PC for primary consumers; PC being larger makes it inverted.
b) Another limitation: saprophytes are given no place in ecological pyramids.

OR

(b) In an ecosystem there was loss of biodiversity due to some project in that area.
(i) How will biodiversity be affected ? (2 points)
(ii) List two major causes of loss of biodiversity. [2 Marks]

Answer:
1. (i) There will be a decline in plant production and lowered resistance of the ecosystem to environmental perturbations such as drought.
2. (ii) Habitat loss and fragmentation; alien species invasion.

Teacher's Note:
a) Remember the "Evil Quartet": habitat loss, over-exploitation, alien species invasion and co-extinction.
b) Increased variability in water use and pest and disease cycles is also an accepted effect.

SECTION C

22. (a) Do all pollen grains remain viable for the same length of time ? Support your answer with two suitable examples.
(b) How are pollen grains stored in pollen banks ? [3 Marks]

Answer:
1. (a) No, the period of pollen viability differs from plant to plant.
2. In cereals such as rice and wheat, pollen grains lose viability within 30 minutes of their release.
3. In some members of Rosaceae, Leguminosae and Solanaceae, pollen grains remain viable for months.
4. (b) Pollen grains are stored by cryopreservation in liquid nitrogen at \( -196^{\circ} \)C.

Teacher's Note:
a) Start with a clear "No" - it carries marks on its own.
b) Always write the temperature \( -196^{\circ} \)C with liquid nitrogen for the storage part.

23. (a) What is pedigree analysis ? Mention its importance in human genetics. (2 pts.)
(b) Analyse the following pedigree and write the
(i) Pattern of inheritance.
(ii) Give one example of disease showing such an inheritance pattern. [3 Marks]

[Figure: Pedigree chart. Generation I: an unaffected male (square) and an unaffected female (circle). Generation II (their children): an unaffected male, an affected male (shaded square), an unaffected female married to an unaffected male, an affected female (shaded circle) and an unaffected male. Generation III (children of the unaffected female and her unaffected husband): one affected male and two unaffected females.]

Answer:
1. (a) Pedigree analysis is the analysis of inheritance of a trait in several generations of a family.
2. Importance: it helps to trace the inheritance of a trait, an abnormality or a disease in a family, and to predict its chances in future generations.
3. (b) (i) Autosomal recessive inheritance (unaffected parents have affected sons and daughters).
4. (ii) Example: Sickle cell anaemia (thalassemia or phenylketonuria are also correct).

Teacher's Note:
a) Two unaffected parents with an affected child means the trait is recessive.
b) An affected daughter from unaffected parents rules out X-linked recessive inheritance.

24. (a) How is Hardy-Weinberg expression (p² + 2pq + q²) = 1 derived ?
(b) List any two factors that disturb the genetic equilibrium. [3 Marks]

Answer:
1. (a) According to the Hardy-Weinberg principle, the sum of all allelic frequencies in a population is 1. If the alleles A and a have frequencies p and q, then \( p + q = 1 \).
2. The frequency of AA is \( p^2 \), because the chance that allele A (frequency p) appears on both chromosomes of a diploid individual is the product of the probabilities, \( p \times p \). Similarly, the frequency of aa is \( q^2 \).
3. The frequency of Aa is \( 2pq \). Hence \( (p + q)^2 = p^2 + 2pq + q^2 = 1 \).
4. (b) Factors: gene migration (gene flow) and genetic drift (mutation, genetic recombination and natural selection are also correct).

Teacher's Note:
a) Each step (p + q = 1, \( p^2 \), \( q^2 \), 2pq) carries separate marks, so write all of them.
b) Explain why Aa is 2pq: the heterozygote can arise in two ways (A from father or from mother).

25. Draw a neat diagram of megasporangium of an angiosperm and label any six parts. [3 Marks]

[Figure: Diagram of an anatropous ovule (megasporangium) showing hilum, funicle, micropyle, micropylar pole, outer integument, inner integument, nucellus, embryo sac and chalazal pole.]

Answer: Draw an inverted (anatropous) ovule attached to the placenta by a stalk. Label any six of the following parts:
1. Funicle - the stalk attaching the ovule to the placenta.
2. Hilum - the point where the body of the ovule joins the funicle.
3. Micropyle and micropylar pole - the small opening at the tip of the integuments.
4. Outer integument and inner integument - the protective layers around the nucellus.
5. Nucellus - the mass of cells inside the integuments.
6. Embryo sac - the female gametophyte inside the nucellus; chalazal pole - the basal end opposite the micropyle.

Teacher's Note:
a) Each correct label carries ½ mark, so label six parts clearly with straight lines.
b) Show the embryo sac inside the nucellus near the micropylar end.

26. (a) Mention the scientific name of the source plant and the part from which opioids are extracted.
(b) How are morphine and heroin related ? Mention the effect each one of them has on human body. [3 Marks]

Answer:
1. (a) Opioids are obtained from the latex of the poppy plant, Papaver somniferum.
2. (b) Heroin (smack) is formed by acetylation of morphine (diacetylmorphine).
3. Morphine is a very effective sedative and pain killer, useful for patients who have undergone surgery; heroin is a depressant and slows down body functions.

Teacher's Note:
a) Write the scientific name correctly, with the genus starting with a capital letter.
b) The keyword "acetylation" carries 1 mark in part (b).

27. A farmer grew two 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 that corn borers attacked only in field A. To control it, spores of Bt were sprayed on field A.
(a) Name the gene in the spores responsible for control of pest.
(b) What effect will the spores of Bt have on insect pest ?
(c) How has corn plants of field B developed resistance against this pest ? Explain. [3 Marks]

Answer:
1. (a) The gene is cryIAb, which controls the corn borer.
2. (b) The Bt spores, mixed with water and sprayed on the plants, are eaten by the insect larvae. In the gut of the larvae, the toxin is released and activated, and the insect is killed.
3. (c) The GM corn plants of field B carry and express the Bt (cry) gene, so they produce Bt toxin themselves; when the pest eats these plants, the toxin kills it.

Teacher's Note:
a) cryIAb controls corn borer, while cryIAc and cryIIAb control cotton bollworms.
b) The toxin is inactive (protoxin) in the bacterium and becomes active only in the alkaline gut of the insect.

28. Observe the given picture carefully. A mixture of DNA with fragments ranging from 100 base pairs to 1800 base pairs were separated by electrophoresis on agarose gel with the following arrangement :
(a) What result will be obtained in staining with ethidium bromide ? Explain with reasons.
(b) The above setup was modified as shown below and a band with 100 base pairs was obtained at X.
What changes were made to the previous design to get a band at 'X'. Why did the band appear at 'X' ? [3 Marks]

[Figure: First set-up: a gel with a row of five wells (labelled Wells, one marked by an arrow) near the top; the positive terminal (+) is at the top next to the wells and the negative terminal (-) is at the bottom, with a downward arrow from + to -. Second (modified) set-up: the negative terminal (-) is at the top next to the wells and the positive terminal (+) is at the bottom; DNA bands are seen below some wells, and the band farthest from the wells is marked X.]

Answer:
1. (a) No bands will be seen in the agarose gel, because DNA fragments are negatively charged and will not move towards the negative end (cathode); they remain at the anode side near the wells.
2. (b) The positions of the positive terminal (anode) and the negative terminal (cathode) were interchanged, so the wells are now at the cathode end.
3. The band appeared at X because the 100 bp fragment is the smallest fragment, so it moved the farthest towards the anode through the sieving gel.

Teacher's Note:
a) DNA always moves towards the anode (positive electrode) as it is negatively charged.
b) In the gel, smaller fragments move faster and farther; larger fragments stay close to the wells.

SECTION D

Question Nos. 29 and 30 are case-based questions. Each question has sub-sections with internal choice in one sub-section.

29. The diagram below shows the distribution of two barnacle species, Chthamalus and Balanus on a rocky sea shore.
When Balanus is experimentally removed, Chthamalus expands its range in lower intertidal zone.

[Figure: Barnacles labelled Balanus balanoides and Chthamalus stellatus shown above a rocky sea shore. The shore between Highest tides and Lowest tides is divided into three zones: Upper intertidal zone - Balanus dessicates in this zone, allowing Chthamalus to thrive; Middle intertidal zone - Where it can survive, Balanus outcompetes Chthamalus for space; Lower intertidal zone - Balanus is subject to predation in the lower tidal zone.]

(a) Identify and define the ecological phenomenon demonstrated by this observation. [1 Mark]

Answer:
1. The phenomenon is competitive release.
2. A species whose distribution is restricted to a small geographical area because of a competitively superior species expands its range when the superior competitor is removed.

Teacher's Note:
a) The name (competitive release) and its definition carry ½ mark each.
b) This is Connell's classic field experiment on the rocky coast of Scotland.

OR

(a) State the principle that explains the elimination of one species. [1 Mark]

Answer: Gause's competitive exclusion principle states that two closely related species competing for the same resources cannot co-exist indefinitely, and the competitively inferior one will be eliminated eventually.

Teacher's Note:
a) Name the principle (Gause's competitive exclusion principle) and then state it.
b) Use the words "closely related", "cannot co-exist indefinitely" and "inferior one eliminated".

(b) Two different species can compete for the same resource. Give another example of it. [1 Mark]

Answer: In some shallow South American lakes, visiting flamingos and resident fishes compete for their common food, the zooplankton.

Teacher's Note:
a) This example shows that even totally unrelated species can compete for the same resource.
b) Name both species and the common resource to get the full mark.

(c) How do species avoid competition in nature ? Explain with an example. [2 Marks]

Answer:
1. Species avoid competition by resource partitioning, that is, by choosing different times for feeding or different foraging patterns.
2. Example: MacArthur showed that five closely related species of warblers living on the same tree could avoid competition and co-exist due to behavioural differences in their foraging activities.

Teacher's Note:
a) The term "resource partitioning" is the key value point.
b) The warbler example must mention "behavioural differences in foraging".

30. A student performed some crosses in plants and represented the result in the form of bar graphs as shown below. Each graph displays the phenotypic proportion of the progeny. Study the graphs and answer the questions :

[Figure: Bar graph of Number of Plants (y-axis, 0 to 600) against Crosses (x-axis: I, II, III), with black bars for Tall and white bars for Dwarf. Cross I: Tall 500, no Dwarf. Cross II: Tall 375, Dwarf 125. Cross III: Tall 250, Dwarf 250.]

(a) What can you infer about the genotype of parents in crosses I and II ? [1 Mark]

Answer:
1. Cross I: the parents are TT and tt (TT and Tt, or TT and TT, also give only tall plants).
2. Cross II: both parents are heterozygous, Tt and Tt (giving 3 tall : 1 dwarf).

Teacher's Note:
a) All-tall progeny means at least one parent is homozygous dominant (TT).
b) A 3 : 1 ratio (375 : 125) always points to a Tt x Tt cross.

OR

(a) Which genetic cross is represented by these crosses ? [1 Mark]

Answer: These crosses represent a monohybrid cross, as only one character (height - tall or dwarf) is studied.

Teacher's Note:
a) Only one pair of contrasting traits is involved, so it is a monohybrid cross.
b) A dihybrid cross would involve two characters and a 9 : 3 : 3 : 1 ratio.

(b) Looking at bar graph of cross III, identify the type of cross performed and its importance in genetics. [2 Marks]

Answer:
1. Cross III is a test cross (Tt x tt), giving tall and dwarf plants in a 1 : 1 ratio (250 : 250).
2. Importance: it is done to find the genotype of an organism showing the dominant phenotype, by crossing it with a homozygous recessive parent.

Teacher's Note:
a) A 1 : 1 ratio in a test cross shows the dominant parent is heterozygous.
b) If all progeny were tall, the dominant parent would be homozygous (TT).

(c) What conclusion can you draw from the results of bar graphs of crosses I and II ? Name the genetic principle illustrated. [1 Mark]

Answer: The results show that the tall trait is dominant over the dwarf trait. The principle illustrated is the Law of Dominance.

Teacher's Note:
a) In cross I, dwarf does not appear at all, and in cross II tall appears in 3 out of 4 plants.
b) Give both the conclusion and the name of the law for the full mark.

SECTION E

31. (a) Trace the events from fertilization till implantation of blastocyst in human female. Also mention the site where fertilization takes place. [5 Marks]

Answer:
1. Fertilization takes place in the ampullary region (ampullary-isthmic junction) of the fallopian tube.
2. After the sperm enters the secondary oocyte, the oocyte completes meiotic division II to form a haploid ovum (ootid) and a second polar body.
3. The haploid nucleus of the sperm fuses with that of the ovum to form a diploid zygote.
4. The zygote moves through the isthmus of the oviduct towards the uterus and undergoes mitotic divisions called cleavage, forming blastomeres.
5. The embryo with 8 to 16 blastomeres is called a morula; it divides further and changes into a blastocyst.
6. The blastomeres in the blastocyst are arranged into an outer layer called trophoblast and an inner group of cells called the inner cell mass.
7. The trophoblast layer gets attached to the endometrium of the uterus.
8. The uterine cells divide rapidly and cover the blastocyst, so it becomes embedded in the endometrium. This is called implantation.

Teacher's Note:
a) Write the events in the correct sequence; each correct step carries ½ mark.
b) The site of fertilization (ampullary region) carries 1 mark on its own - do not forget it.
c) Remember that the inner cell mass later differentiates into the embryo.

OR

(b) (i) Case studies of some couples are given below. They were not able to have kids even though the parents were not taking any precautions. They also do not wish to adopt a child or take the help of donors. Study these cases and suggest an appropriate Assisted Reproductive Technology that could help them.
Case I - Female - Normal Reports
Male - Normal sperms but no connection between epididymis and vas deferens.
Case II - Female - Normal eggs but Fallopian tube blocked.
Male - Normal sperms. Male reports are normal.
Case III - Female - Normal Reports.
Male - Low sperm count with less motility.
(ii) Copper releasing intrauterine devices are very effective and popular contraceptive devices. Name any two copper releasing IUDs. Write two reasons that make them effective contraceptives. [5 Marks]

Answer:
1. (i) Case I: In vitro fertilization followed by zygote intra fallopian transfer (IVF-ZIFT), using sperms collected directly from the male (IUT or IUI/artificial insemination are also accepted).
2. Case II: In vitro fertilization followed by intra uterine transfer (IVF-IUT) of the embryo, since the blocked fallopian tube cannot be used.
3. Case III: Intra cytoplasmic sperm injection (ICSI), in which a sperm is directly injected into the ovum (artificial insemination or IUI is also accepted).
4. (ii) Copper releasing IUDs: CuT and Cu7 (Multiload 375 is also correct).
5. They are effective because they increase phagocytosis of sperms within the uterus, and the Cu ions released suppress sperm motility and the fertilising capacity of sperms.

Teacher's Note:
a) In Case II, the zygote cannot be placed in the blocked tube, so it must be transferred into the uterus.
b) Do not name Lippes loop, as it is a non-medicated IUD, and do not name Progestasert or LNG-20, which are hormone-releasing IUDs.

32. (a) (i) Watson and Crick's discovery of double helical structure was based on which two findings. Also mention the name of the scientists associated with these findings.
(ii) Write down the salient features of double helix structure of DNA (any three points). [5 Marks]

Answer:
1. (i) X-ray diffraction studies of DNA, done by Maurice Wilkins and Rosalind Franklin.
2. Chargaff's rule, given by Erwin Chargaff: in double stranded DNA, the ratios between adenine and thymine and between guanine and cytosine are constant and equal to one.
3. (ii) DNA is made of two polynucleotide chains; the backbone is formed by sugar and phosphate, and the bases project inside.
4. The two chains have antiparallel polarity, and the bases of the two strands are paired by hydrogen bonds: adenine forms two hydrogen bonds with thymine, and guanine forms three hydrogen bonds with cytosine.
5. The two chains are coiled in a right-handed fashion; the pitch of the helix is 3.4 nm with about 10 bp in each turn, and the distance between two base pairs is about 0.34 nm.

Teacher's Note:
a) Link each finding with the correct scientist to get both halves of the mark.
b) Another feature: the plane of one base pair stacks over the other in the double helix.
c) Quote numerical values (3.4 nm, 10 bp, 0.34 nm) exactly.

OR

(b) Given below is a stretch of DNA showing the coding strand of structural gene of transcription unit.
5' - ATG ACC GTA TTT TCT GTA GTG CCC GTA CTT CAG GCA TTA 3'
(i) Write the corresponding template strand and m-RNA strand that will be transcribed along with its polarity.
(ii) If GUA of transcribed mRNA is an intron, then depict the sequence involved in formation of mRNA / mature / processed hnRNA strand :
(1) In a bacterium
(2) In humans
(iii) How many amino acids the resulting polypeptide will have after the process of translation in humans ? [5 Marks]

Answer:
1. (i) Template strand: 3' - TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC CGT AAT - 5'
2. mRNA: 5' - AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA - 3'
3. (ii) (1) In a bacterium (no introns, no splicing): 5' - AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA - 3'
4. (2) In humans (introns GUA removed by splicing, cap added at 5' end and poly A tail at 3' end): 5' - m⁷Gppp AUG ACC UUU UCU GUG CCC CUU CAG GCA UUA - poly A tail 3'
5. (iii) The resulting polypeptide in humans will have 10 amino acids.

Teacher's Note:
a) The mRNA has the same sequence as the coding strand, with U in place of T.
b) Always write the polarity (5' and 3') on every strand; marks are lost without it.
c) After removing three GUA introns, 10 codons remain, so 10 amino acids are coded.

33. (a) (i) Name the protozoan species that is responsible for causing the most serious and even fatal malarial disease.
(ii) Name the host in which the parasite completes its sexual stages and explain the changes taking place.
(iii) How does the parasite damage the human body after entering the blood stream ?
(iv) Suggest two effective preventive measures to control the spread of this disease in endemic regions. [5 Marks]

Answer:
1. (i) Plasmodium falciparum.
2. (ii) The sexual stages are completed in the female Anopheles mosquito. The mosquito picks up gametocytes while biting an infected person; fertilization and development take place in the mosquito's gut; the sporozoites formed then migrate to the salivary glands of the mosquito.
3. (iii) In the blood, the parasites first enter the liver cells and then the red blood cells, where they multiply asexually. The RBCs rupture and release a toxic substance, haemozoin, which causes chills and high recurring fever.
4. (iv) Preventive measures: avoid stagnation of water in and around residential areas, and use mosquito nets (introducing larvivorous fish like Gambusia in ponds is also effective).

Teacher's Note:
a) The mosquito is the primary (definitive) host because sexual reproduction occurs in it.
b) Name the toxin haemozoin correctly; it is responsible for the chill and fever.

OR

(b) (i) What are bio-fertilizers ?
(ii) Name the different types of microorganisms used as bio-fertilizers in organic farming and explain how each contributes to soil fertility.
(iii) Write two advantages of using bio-fertilizers over chemical fertilizers. [5 Marks]

Answer:
1. (i) Bio-fertilizers are organisms that enrich the nutrient quality of the soil, for example bacteria, fungi and cyanobacteria.
2. (ii) Bacteria: Rhizobium forms a symbiotic association with the root nodules of leguminous plants and fixes atmospheric nitrogen into organic forms used by the plant; Azotobacter and Azospirillum fix nitrogen while free-living in the soil.
3. Fungi: fungi of the genus Glomus form mycorrhiza with plant roots and help to absorb phosphorus from the soil and pass it to the plant; such plants also resist root-borne pathogens and tolerate salinity and drought.
4. Cyanobacteria: Anabaena, Nostoc and Oscillatoria fix atmospheric nitrogen and serve as important bio-fertilizers in paddy fields.
5. (iii) Bio-fertilizers do not pollute the environment (soil and ground water), and they conserve the beneficial soil microbes.

Teacher's Note:
a) For each type, name one organism and write its contribution; both carry marks.
b) Remember: bacteria and cyanobacteria fix nitrogen, while mycorrhizal fungi help in phosphorus uptake.

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