CBSE Class 12 Biology Question Paper 2026 Solved Code 57-1-3

SECTION A

Question Nos. 1 to 16 are Multiple Choice Type Questions, carrying 1 mark each. Choose the best option.

1. Perisperm present in seed of black pepper is remnant of [1 Mark]
(A) Ovary
(B) Nucellus
(C) Anther
(D) Megaspore

Answer: (B) Nucellus

Teacher's Note:
a) Perisperm is the residual, persistent nucellus seen in seeds of black pepper and beet.
b) Do not confuse perisperm (from nucellus) with endosperm (from the primary endosperm nucleus).

2. Interferons are most effective in making non-infected cells resistant against the spread of which of the following diseases ? [1 Mark]
(A) Malaria
(B) Ascariasis
(C) Pneumonia
(D) Cancer

Answer: (D) Cancer

Teacher's Note:
a) Interferons are proteins secreted by virus-infected cells that protect nearby non-infected cells.
b) Alpha-interferon is used in cancer therapy to activate the immune system and destroy the tumour.

3. 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 (Meghalaya) have sacred groves protected by tribal communities.
b) NCERT lists Sarguja, Chanda and Bastar as sacred grove areas of Madhya Pradesh, not Meghalaya.

4. 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) Vertical distribution of species at different levels is called stratification, e.g. trees, shrubs and herbs in a forest.
b) Stratification is one of the structural features of an ecosystem.

5. 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 replicate autonomously using their own origin of replication (ori), not under chromosomal control.
b) Read the question carefully: it asks for the incorrect statement.

6. 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.

7. Select the option that shows correct statements :
(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 : [1 Mark]
(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 peak in the middle of the cycle, so statement (ii) is wrong.
b) The follicular phase is the proliferative phase; the luteal phase is the secretory phase.

8. Which one of the following evidences does not support Darwin's theory of natural selection ? [1 Mark]
(A) Branching Descent
(B) Small & directional variations
(C) Appearance of new traits by mutations
(D) Survival of the fittest

Answer: (C) Appearance of new traits by mutations

Teacher's Note:
a) Mutation as a source of sudden new traits was proposed by Hugo de Vries, not Darwin.
b) Darwin's key ideas were branching descent, small directional variations and natural selection.

9. Expression of a particular gene helps to identify transformants. What is this gene known as ? [1 Mark]
(A) Plasmid
(B) Structural gene
(C) Selectable marker
(D) Intron

Answer: (C) Selectable marker

Teacher's Note:
a) Selectable markers, such as antibiotic resistance genes, help pick out transformants from non-transformants.
b) Examples: genes for resistance to ampicillin, tetracycline or kanamycin in E. coli.

10. A DNA molecule is 140 base pairs long. It has 20% thymine. How many guanine bases are present in the molecule ? [1 Mark]
(A) 64
(B) 84
(C) 32
(D) 56

Answer: (B) 84

Teacher's Note:
a) 140 base pairs means 280 bases; T = A = 20% each, so G + C = 60% and G = 30% of 280 = 84.
b) Common mistake: taking 140 as the total number of bases, which gives the wrong option.

11. 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, so "less than 1000" is wrong.
b) Remember also about 1,000 varieties of mango in India.

12. 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 protoplasts of tomato and potato.
b) The cell wall is first removed to obtain naked protoplasts before fusion.

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) : Fossils provide direct evidence for evolution.
Reason (R) : Fossils help to trace evolutionary relationships among organisms over time. [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) Fossils in rock layers of different ages show how life forms changed over time.
b) Because they trace relationships through time, fossils are direct evidence of evolution.

14. 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, not the grazing food chain.
b) The grazing and detritus food chains are linked at some levels, so R is true.

15. 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 (2.4 grams per litre), not alpha lactaglobulin.
b) Watch for small changes in protein names in Assertion-Reason questions.

16. 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 digest organic matter aerobically, consuming oxygen.
b) Anaerobic digestion happens later in anaerobic sludge digesters, not by flocs.

SECTION B

17. A health officer visiting two different regions reported the following symptoms in patients :
Region A - Sustained high fever with stomach pain, constipation and loss of appetite.
Region B - Inflammation in the lower limbs and genital organs.
Name the diseases likely affecting the patients in regions A and B. Write the scientific name of the causative organisms in both the cases. [2 Marks]

Answer:
1. Region A: The disease is Typhoid, caused by the bacterium Salmonella typhi (Salmonella typhimurium as given in the marking scheme).
2. Region B: The disease is Filariasis (Elephantiasis), caused by the filarial worm Wuchereria bancrofti (or Wuchereria malayi).

Teacher's Note:
a) Four value points of half mark each: two disease names and two causative organisms.
b) Sustained high fever with constipation points to typhoid; swelling of limbs and genital organs points to filariasis.
c) Write scientific names correctly, with genus starting with a capital letter.

18. (a) (i) How does a farmer use dormancy of seed to his advantage ? [1 Mark]
(ii) Differentiate between pea seed and castor seed. [1 Mark]

Answer:
(i) Dormancy allows seeds to be dried and stored safely, so the farmer can use them to raise crops in the next season (stored seeds can also be used as food or sold commercially).
(ii) Pea seed is a non-endospermic (non-albuminous) seed, as its endosperm is fully consumed during embryo development. Castor seed is an endospermic (albuminous) seed, as some endosperm remains in the mature seed.

Teacher's Note:
a) Link dormancy to storage of seeds for the next sowing season.
b) For (ii), one correct difference with both seeds named is enough for 1 mark.

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 of the ovary.
2. The products of this unequal division are a large haploid secondary oocyte and a tiny first polar body.

Teacher's Note:
a) One mark for "tertiary follicle" and half mark each for the two products.
b) Do not write "ovum" as a product; the second meiotic division is completed only after sperm entry.

19. Write any two advantages of producing plants by micro-propagation. [2 Marks]

Answer:
1. A large number of plants can be produced in a very short time.
2. The plants produced are genetically identical to the original plant (somaclones).

Teacher's Note:
a) Another accepted advantage is obtaining disease-free plants from virus-infected plants using meristem culture.
b) Give only two points, as the question asks for any two.

20. (a) Mendel published his work on inheritance in 1865, but it remained unnoticed for over three decades.
State the reasons why his work was not recognised during his life time. (4 points) [2 Marks]

Answer:
1. Communication was not easy in those days, so his work was not widely publicised.
2. His idea of genes (factors) as stable, discrete units controlling traits, and of alleles that do not blend with each other, was not accepted by his contemporaries.
3. His use of mathematics (statistics) to explain biological phenomena was new and unacceptable to biologists of that time.
4. He could not provide any physical proof for the existence of factors or say what they were made of.

Teacher's Note:
a) Each reason carries half mark, so write exactly four clear points.
b) Keywords: poor communication, non-blending factors, use of mathematics, no physical proof.

OR

(b) What is adaptive radiation ? Give one example from any habitat. [2 Marks]

Answer:
1. Adaptive radiation is the process of evolution of different species in a given geographical area, starting from a point and literally radiating to other areas of geography (habitats).
2. Example: Darwin's finches of the Galapagos Islands. From the original seed-eating finches, other forms with altered beaks arose, enabling them to become insectivorous and vegetarian. (Australian marsupials evolving from a common ancestral stock is another example.)

Teacher's Note:
a) One mark for the definition and one mark for a correct example.
b) Mention the change in beak type if you give Darwin's finches as the example.

21. (a) (i) What is primary productivity of an ecosystem and how is it expressed ? [1 Mark]
(ii) Explain the equation given below :
NPP = GPP - R [1 Mark]

Answer:
(i) Primary productivity is the amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis. It is expressed in terms of weight (g m^-2 yr^-1) or energy (kcal m^-2 yr^-1).
(ii) Net Primary Productivity (NPP) is equal to Gross Primary Productivity (GPP) minus the respiratory losses (R) of the plants. NPP is the biomass available for consumption by heterotrophs.

Teacher's Note:
a) Give both the definition and the units in (i) to earn the full mark.
b) Expand all three terms, NPP, GPP and R, in (ii).

OR

(b) Biodiversity conservation is crucial for maintaining ecological balance and ensuring the survival of future generations.
Describe four advanced ex-situ methods used for conservation of biodiversity. [2 Marks]

Answer:
1. Cryopreservation: gametes or pollen of threatened species are preserved in viable and fertile condition for long periods at \( -196^{\circ}C \) (in liquid nitrogen).
2. In vitro fertilisation: eggs can be fertilised in vitro to produce offspring of threatened species.
3. Tissue culture: plants can be propagated in large numbers by tissue culture methods.
4. Seed banks: seeds of different genetic strains of commercially important plants can be stored for long periods in seed banks.

Teacher's Note:
a) The question asks for advanced methods, so do not write only zoos or botanical gardens.
b) Each correct method carries half mark.

SECTION C

22. 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 :

[Figure: An agarose gel with a row of five wells at the top. The positive terminal (+) is at the top, next to the wells, and the negative terminal (-) is at the bottom. An arrow on the left side points from + down to -. A small arrow marks the middle well, and the label "Wells" points to the wells.]

(a) What result will be obtained in staining with ethidium bromide ? Explain with reasons. [1 Mark]

Answer: No bands will be seen in the agarose gel. DNA fragments are negatively charged, so they will not move towards the negative end (cathode) at the bottom; they remain at the positive end (anode) near the wells.

Teacher's Note:
a) Half mark for "no bands" and half mark for the reason based on the negative charge of DNA.
b) Remember: DNA always moves towards the anode (positive electrode).

(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' ? [2 Marks]

[Figure: The same gel with five wells at the top, but now the negative terminal (-) is at the top next to the wells and the positive terminal (+) is at the bottom. Dark DNA bands are seen below the first, third and fifth wells at different distances. The band below the third well that has moved farthest from the wells is labelled X.]

Answer:
1. The positions of the positive terminal (anode) and the negative terminal (cathode) were interchanged, so the wells are now at the cathode end and the anode is at the far end of the gel.
2. The band at X is the smallest DNA fragment (100 bp). Smaller fragments move faster through the pores of the agarose gel, so it moved the farthest towards the anode.

Teacher's Note:
a) One mark for the change in electrode position and one mark for the reason based on fragment size.
b) Agarose gel works as a sieve: the smaller the fragment, the farther it moves.

23. (a) Mention the scientific name of the source plant and the part from which opioids are extracted. [1 Mark]
(b) How are morphine and heroin related ? Mention the effect each one of them has on human body. [2 Marks]

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

Teacher's Note:
a) Write the scientific name correctly and mention "latex" as the part.
b) The keyword "acetylation" carries a full mark in part (b).

24. (a) Do all pollen grains remain viable for the same length of time ? Support your answer with two suitable examples. [2½ Marks]
(b) How are pollen grains stored in pollen banks ? [½ Mark]

Answer:
(a) 1. No, the period of viability of pollen grains is variable.
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.
(b) Pollen grains are stored by cryopreservation in liquid nitrogen at \( -196^{\circ}C \).

Teacher's Note:
a) Half mark for "No" and one mark for each correct example with its duration.
b) Always write the temperature \( -196^{\circ}C \) along with liquid nitrogen.

25. (a) What is pedigree analysis ? Mention its importance in human genetics. (2 pts.) [2 Marks]
(b) Analyse the following pedigree and write the
(i) Pattern of inheritance. [½ Mark]
(ii) Give one example of disease showing such an inheritance pattern. [½ Mark]

[Figure: A three-generation pedigree chart. Generation I: an unaffected male and an unaffected female. Generation II (their children): an unaffected male, an affected male (shaded square), an unaffected female who is married to an unaffected male from outside the family, an affected female (shaded circle), and an unaffected male. Generation III (children of the married unaffected couple): one affected male (shaded square) and two unaffected females.]

Answer:
(a) 1. Pedigree analysis is the analysis of the inheritance of traits over several generations of a family.
2. Importance: it helps to trace the inheritance of a trait, an abnormality or a disease in a family.
(b) (i) The pattern of inheritance is autosomal recessive.
(ii) Example: Sickle-cell anaemia (Thalassemia or Phenylketonuria are also correct).

Teacher's Note:
a) Unaffected parents having affected sons and daughters is the key sign of autosomal recessive inheritance.
b) Do not give haemophilia or colour blindness as examples; they are X-linked.

26. (a) How is Hardy-Weinberg expression \( (p^{2} + 2pq + q^{2}) = 1 \) derived ? [2 Marks]
(b) List any two factors that disturb the genetic equilibrium. [1 Mark]

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

Teacher's Note:
a) Begin with \( p + q = 1 \); each step of the derivation carries half mark.
b) Write only two factors in (b), as asked.

27. Suggest and describe a technique through which a healthy sugarcane plant can be obtained from virus infected sugarcane plant. [3 Marks]

Answer:
1. The technique is tissue culture (meristem culture).
2. The meristem (apical and axillary), which is free of virus, is taken from the infected plant and grown in a sterile nutrient medium in vitro.
3. The medium contains a carbon source such as sucrose, along with inorganic salts, vitamins and amino acids.
4. Growth regulators such as auxins and cytokinins are also added.
5. Thus a healthy, virus-free sugarcane plant is regenerated, which is genetically similar to the parent plant.

Teacher's Note:
a) Name the explant as "meristem"; it is free of virus even when the plant is infected.
b) Mention any two components of the medium and the growth regulators to get full marks.

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

Answer: The diagram shows an anatropous ovule (megasporangium) attached to the placenta by a stalk. Six labelled parts:
1. Hilum - the point where the body of the ovule joins the funicle.
2. Funicle - the stalk that attaches the ovule to the placenta.
3. Micropyle - the small opening at the tip of the integuments, at the micropylar pole.
4. Outer and inner integuments - the two protective layers around the nucellus.
5. Nucellus - the mass of cells enclosed by the integuments.
6. Embryo sac - the female gametophyte located inside the nucellus; the base of the ovule opposite the micropyle is the chalazal pole.

Teacher's Note:
a) Each correct label carries half mark; label any six clearly with straight lines.
b) Draw the ovule inverted (anatropous), with the micropyle close to the funicle.

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: A rocky sea shore divided into zones from "Lowest tides" at the bottom to "Highest tides" at the top, with pictures of the two barnacles labelled Balanus balanoides and Chthamalus stellatus. 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: The phenomenon is competitive release. A species whose distribution is restricted to a small geographical area because of the presence of a competitively superior species expands its distributional range when the superior competitor is removed.

Teacher's Note:
a) Half mark for the term "competitive release" and half mark for its definition.
b) Connell's barnacle experiment on the Scottish coast is the classic example of competitive release.

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) Half mark for naming Gause's principle and half mark for stating 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, zooplankton.

Teacher's Note:
a) This example shows that even totally unrelated species can compete for the same resource.
b) Mention the shared resource (zooplankton) clearly to get the 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 because of behavioural differences in their foraging activities.

Teacher's Note:
a) One mark for the term "resource partitioning" and one mark for a correct example.
b) The warblers example is the standard NCERT example; name the scientist if you can.

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: A bar graph with "Number of Plants" (0 to 600) on the y-axis and "Crosses" (I, II, III) on the x-axis. Black bars show Tall and white bars show Dwarf plants. Cross I: 500 Tall, no Dwarf. Cross II: 375 Tall and 125 Dwarf. Cross III: 250 Tall and 250 Dwarf.]

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

Answer: In cross I, all progeny are tall, so the parents are TT and tt (or TT and Tt, or TT and TT). In cross II, tall and dwarf appear in a 3 : 1 ratio, so both parents are heterozygous, Tt and Tt.

Teacher's Note:
a) Half mark for each cross; a 3 : 1 ratio always means both parents are heterozygous.
b) Use T for the dominant (tall) allele and t for the recessive (dwarf) allele.

OR

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

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

Teacher's Note:
a) One pair of contrasting traits means a monohybrid cross.
b) Do not write dihybrid; that needs two characters.

(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, as it gives tall and dwarf plants in a 1 : 1 ratio (Tt crossed with tt).
2. Importance: a test cross is done to find the genotype of an organism showing the dominant phenotype, that is, whether it is homozygous (TT) or heterozygous (Tt).

Teacher's Note:
a) One mark for naming the test cross and one mark for its importance.
b) In a test cross the plant is always crossed with the homozygous recessive parent.

(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 genetic principle illustrated is the Law of Dominance.

Teacher's Note:
a) Half mark for the conclusion and half mark for naming the law.
b) In cross I, dwarf is hidden completely, which shows dominance.

SECTION E

31. (a) (i) Name the protozoan species that is responsible for causing the most serious and even fatal malarial disease. [½ Mark]
(ii) Name the host in which the parasite completes its sexual stages and explain the changes taking place. [2 Marks]
(iii) How does the parasite damage the human body after entering the blood stream ? [1½ Marks]
(iv) Suggest two effective preventive measures to control the spread of this disease in endemic regions. [1 Mark]

Answer:
(i) Plasmodium falciparum.
(ii) 1. The sexual stages are completed in the female Anopheles mosquito.
2. The mosquito picks up gametocytes when it bites an infected person.
3. Fertilisation and development take place in the gut of the mosquito.
4. Sporozoites formed in the gut migrate to the salivary glands of the mosquito, ready to infect a new host.
(iii) After entering the blood stream, the parasites first reach the liver, where they multiply, and then enter red blood cells (RBCs), where they reproduce asexually. The RBCs rupture and release a toxic substance, haemozoin, which causes chills and high fever that recur every three to four days.
(iv) 1. Avoid stagnation of water in and around residential areas and clean household coolers regularly.
2. Introduce larvivorous fishes such as Gambusia in ponds, which feed on mosquito larvae (use of mosquito nets is also effective).

Teacher's Note:
a) Write the species name, falciparum, not just Plasmodium, for part (i).
b) The keyword "haemozoin" is essential in part (iii).
c) In (ii), mention gametocytes, fertilisation in the gut and sporozoites in the salivary glands.

OR

(b) (i) What are bio-fertilizers ? [1 Mark]
(ii) Name the different types of microorganisms used as bio-fertilizers in organic farming and explain how each contributes to soil fertility. [3 Marks]
(iii) Write two advantages of using bio-fertilizers over chemical fertilizers. [1 Mark]

Answer:
(i) Bio-fertilizers are organisms that enrich the nutrient quality of the soil. The main sources are bacteria, fungi and cyanobacteria.
(ii) 1. Bacteria: Rhizobium forms a symbiotic association with the roots of leguminous plants and fixes atmospheric nitrogen into organic forms used by the plant. Azotobacter and Azospirillum fix atmospheric nitrogen while free-living in the soil.
2. Fungi: fungi of the genus Glomus form mycorrhiza with plant roots. They absorb phosphorus from the soil and pass it to the plant; such plants also show resistance to root-borne pathogens, tolerance to salinity and drought, and better growth.
3. Cyanobacteria: Anabaena, Nostoc and Oscillatoria are autotrophic microbes that fix atmospheric nitrogen; they are important bio-fertilizers in paddy fields.
(iii) 1. Bio-fertilizers do not pollute the environment (soil and ground water).
2. They conserve the beneficial soil microbes and do not harm human health.

Teacher's Note:
a) For each group in (ii), name the organism and state its role to get one full mark.
b) Mycorrhiza is linked with phosphorus absorption, while Rhizobium and cyanobacteria are linked with nitrogen fixation.

32. (a) Describe how the lac operon operates both in the presence and absence of an inducer in E.coli. [5 Marks]

Answer:
In the presence of inducer:
1. The regulator gene (i gene) codes for a repressor protein.
2. Lactose acts as the inducer. It binds with the repressor and inactivates it, so the repressor cannot bind to the operator.
3. The operon has three structural genes, z, y and a.
4. RNA polymerase binds to the promoter and starts transcription of the structural genes.
5. The enzymes beta-galactosidase (from z), permease (from y) and transacetylase (from a) are produced, which help in the metabolism of lactose.
In the absence of inducer:
6. The repressor protein made by the i gene binds to the operator region of the operon.
7. This prevents RNA polymerase from transcribing the structural genes z, y and a, so the enzymes are not produced.
(A labelled diagram showing the i gene, promoter, operator and structural genes z, y, a, with the repressor bound to the inducer in one case and to the operator in the other, may be drawn in place of the written steps.)

Teacher's Note:
a) Three marks are for the presence of inducer and two marks for the absence of inducer.
b) Name all three enzymes with their genes; this is a common place to lose marks.
c) Lac operon is an example of negative regulation, because the repressor switches the operon off.

OR

(b) (i) Differentiate between Mendelian and Chromosomal disorders. [3 Marks]
(ii) Which disorder is known as "Inborn Error of Metabolism" ? Mention its cause and any one symptom. Name the harmful compounds formed. [2 Marks]

Answer:
(i) 1. Cause: Mendelian disorders are caused by alteration or mutation in a single gene, while chromosomal disorders are caused by absence, excess or abnormal arrangement of one or more chromosomes.
2. Study: the inheritance pattern of Mendelian disorders can be traced by pedigree analysis, while chromosomal disorders cannot be studied by pedigree analysis.
(ii) 1. Phenylketonuria is known as an inborn error of metabolism.
2. Cause: a mutation in the single gene that codes for the enzyme phenylalanine hydroxylase.
3. Symptom: mental retardation (hair loss or reduced skin pigmentation are also seen).
4. Harmful compounds formed: phenylpyruvic acid and other derivatives of phenylalanine, which accumulate in the brain.

Teacher's Note:
a) In (i), each difference carries one and a half marks, so explain both sides of each point.
b) In (ii), each of the four points (name, cause, symptom, compound) carries half mark.

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

Teacher's Note:
a) Eight steps carry half mark each and the site of fertilisation carries one mark.
b) Write the steps in correct order: zygote, cleavage, morula, blastocyst, implantation.
c) Do not write uterus as the site of fertilisation; it is the ampulla of the fallopian tube.

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. [1 Mark]
Case II - Female - Normal eggs but Fallopian tube blocked.
Male - Normal sperms. Male reports are normal. [1 Mark]
Case III - Female - Normal Reports.
Male - Low sperm count with less motility. [1 Mark]
(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. [2 Marks]

Answer:
(i) Case I: In vitro fertilisation followed by zygote intra fallopian transfer (ZIFT) or intra uterine transfer (IUT), using sperms collected from the male (intra uterine insemination or artificial insemination is also accepted).
Case II: In vitro fertilisation followed by intra uterine transfer (IUT), as the blocked fallopian tube cannot be used.
Case III: Intra cytoplasmic sperm injection (ICSI), in which a sperm is directly injected into the ovum (artificial insemination or intra uterine insemination is also accepted).
(ii) 1. Two copper releasing IUDs: CuT and Cu7 (Multiload 375 is another).
2. They increase phagocytosis of sperms within the uterus.
3. The Cu ions released suppress sperm motility and the fertilising capacity of sperms.

Teacher's Note:
a) In Case II, avoid ZIFT or GIFT, because the fallopian tube is blocked.
b) Do not name LNG-20 or Progestasert in (ii); they are hormone releasing IUDs.

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