Class 12 Sample Paper 2023 Solutions Biology

Class 12 Sample Paper 2023 Solution

Practice Questions – Marking Scheme
Session 2022-23
CLASS XII
BIOLOGY (044)

Q.No Question Marks
SECTION A
1 (b) inhibiting ovulation and implantation 1
2 (c) only Q, R and S 1
3 (b) 30% 1
4 (b) only Q 1
5 (c) reduced levels of oxygen in the blood 1
6 (c) only P and R 1
7 (d) dead organic matter 1
8 (a) only P 1
9 (c) only Q and R 1
10 (b) only P and R 1
11 (b) Q 1
12 (b) productivity 1

Question No. 13 to 16 consist of two statements – Assertion (A) and Reason (R).
Answer these questions selecting the appropriate option given below:
A. Both A and R are true and R is the correct explanation of A
B. Both A and R are true and R is not the correct explanation of A
C. A is true but R is false
D. A is False but R is true

13 (c) A is true, but R is false. 1
14 (d) A is false, but R is true. 1
15 (a) Both A and R are true, and R is the correct explanation for A. 1
16 (d) A is false, but R is true. 1
SECTION B
17 (a) 0.5 marks each for the following: 2

- sterile females with rudimentary ovaries
- short stature and underdeveloped feminine character

- category: chromosomal disorders/aneuploidy
- cause: failure of segregation of chromatids during cell division cycle
18 1 mark each for the following: 2

(a) Well P contains the uncut vector whereas well Q contains the vector
cut by a restriction enzyme.

(b) The vector in well Q has been cut by a restriction enzyme that has two

sites whereas the vector in well R has been cut either by different
enzymes or by one enzyme that has more than two sites.
19 (a) 0.5 marks each for the following: 2

- metastasis
- Pepsin is produced mainly by the stomach cells, so the tumor may have
metastasised from the stomach to the liver.

(b) malignant tumors [0.5 marks]

(c) They will disrupt the normal functioning of the liver cells. [0.5 marks]

[Accept any other valid answer]
20 (a) bone marrow OR preserved embryonic stem cells from Reena's 2
umbilical cord [1 mark]

- Stem cells are capable of differentiating into all types of tissues and
organs.
- Replacing the damaged nerve cells with healthy nerve cells that have
differentiated from the stem cells can help in treating the diseased
condition.
21 1 mark each for: 2

- dry weight
- Water content, as part of fresh (or wet) weight can be different due to
seasonal or ecosystem variations.

OR

1 mark each for the following:

- Very few decomposers can break down chitin present in the hard outer
coverings, so the decomposition would be slow.
- Soil submergence in marshy areas would lead to anaerobic
decomposition which is slower than aerobic decomposition.

[Accept any other valid answer]
SECTION C
22 (a) 0.5 marks for each of the following: 3

- Menstruation results in breakdown of the endometrial wall of the uterus.
- In case of fertilisation, the foetus needs the endometrial wall as the
womb and so in case of pregnancy continues to sustain and menstruation
does not happen.

(b) 1 mark for each correct answer as follows:

- rupture of Graafian follicle
- release of ovum

23 (a) inbreeding depression [1 mark] 3

(b) 1 mark each for correct name and explanation:

- self incompatibility
- The pollen of a plant is not allowed to germinate on the stigma of the
same flower or on a different flower of the same plant due to pollen-pistil
interaction.
24 (a) that DNA is the genetic material and not proteins [1 mark] 3

(b)

- X: 35S amount in supernatant [0.5 marks]
- Y: 32P amount in supernatant [0.5 marks]

- Reason: When agitated in a blender, the bacteriophages separated from
the cells. The supernatant so formed had a high content of 35S but there
was also some 32P which may have not yet transferred their genetic
material to bacterial cells. [1 mark]
25 (a) 0.5 marks for each of the following steps: 3

- q2 = 0.25, so q = 0.5

-p+q=1
So, p = 1 - 0.5 = 0.5

- So, population that consists of carriers (Ss) = 2pq = 2 x 0.5 x 0.5 = 0.5

- Carrier individuals in a population of 1000 individuals = 0.5 x 1000 =
500 individuals

(b) 0.5 marks for each of the following:

- No, it cannot be used.
- The Hardy-Weinberg principle takes into account only diploid
organisms/organisms with two alleles for a trait.
26 (a) 0.5 marks for each of the following: 3

- reverse transcription of the viral RNA to cDNA.
- this is the only process that is not normally carried out by the animal cell

(b) If the viral DNA continues to stay in the cytoplasm, it can get
degraded by enzymes in the cytoplasm. [1 mark]

(c) If the concentration of antigen/antibodies in the sample used is not
sufficient to give a positive result. [1 mark]

[Accept any other valid answer]

OR

1 mark each for the following:

(a) Anaerobic bacteria would not use the oxygen in the wastewater, thus
the BOD would not reduce.

(b) Anaerobic conditions need to be maintained as biogas-producing
bacteria are strict anaerobes.

(c) The slurry will mainly be rich in nitrogen and phosphorus as the
carbon in the sludge would be used in the formation of methane and
carbon dioxide.
27 (a) 0.5 marks for each of the following: 3

- As the bacteria starts to grow, it uses up the oxygen in the system.
- After a while, the oxygen gets used up and anaerobic respiration begins
which leads to the formation of lactic acid, causing a decrease in the pH.

(b) 0.5 marks for identification and 0.5 marks for the reason for any TWO
of the following:

- Oxygen delivery system: After the starter culture is added, oxygen is
also added which will need a delivery system.
- Stirrer: Once oxygen is added, the system would need to be mixed
thoroughly so that oxygen is available throughout the bioreactor.
- Sterilization unit: Milk needs to be sterilised before addition of the
starter culture to remove any other microorganisms already present in it.

[Accept any other valid answer]
28 (a) 1 mark for any ONE of the following reasons such as: 3

- low incident solar radiation results in low productivity
- extremely cold conditions do not favour survival of many species

(b) 1 mark for any ONE of the following reasons such as:

- greater competition between species
- greater climatic variations
- harsh climatic conditions for many species

(c) In cryopreservation, low temperature conditions are used to preserve
biological constructs. [1 mark]
SECTION D

29 (a) 0.5 marks each for the following: 4

- parent with purple kernel - Pp
- parent with yellow kernel – pp

(b) Complete answer:

P p
P PP Pp
p Pp pp

- 1 mark for correctly identifying genotypes of the parents in the cross
- 1 mark for drawing the correct Punnett square.

(c) 0.5 marks each for the following:

- genotypic ratio: 1:2:1
OR 1PP:2Pp:1pp

- phenotypic ratio: 3:1
OR 3 plants with purple kernel: 1 plant with yellow kernel

OR

(c) 0.5 marks each for the following:

- test cross with a plant having yellow kernels/homozygous recessive.
- The percentage of dominant and recessive phenotypes in the progeny of
a test cross can help identify the unknown genotype of the parent.
30 (a) 1 mark each for the following: 4

- active immunity
- Since he used matter from lesions, it is likely to be the antigen which
was then inoculated into the boy for protection against cowpox and
building a memory response against smallpox

- Vaccination can provide active immunity if antigenic proteins of a
pathogen or the inactivated/weakened pathogen (vaccine) are introduced
into the body.
- For example, the COVID-19 vaccine or polio vaccine contains an
antigen that when injected in the body initiates an antibody response.
- Vaccination can also provide passive immunity if pre-formed antibodies
or antitoxins are injected into an individual.
- For example, in case of snake bites, patients are injected with antibodies
against the snake venom.

[Accept any other valid examples]

OR

(b) 1 mark each for the following:

- passive immunization
- Passive immunization involves injecting an individual with pre-formed
antibodies rather than the antigen itself which is responsible for the
generation of a memory response.
SECTION E
31 (a) 0.5 marks each for the following: 5

- The viral RNA is isolated and converted to a dsDNA structure.
- The dsDNA so formed is inserted into an appropriate
vector (Agrobacterium or viral vector).
- The vectors are then introduced into the host plant where it transcribes
the mRNA for this gene.
- Whenever the virus infects the plant and injects its RNA into the host
plant, the host plant transcribes the viral RNA.
- The mRNA produced from the vector binds to the transcribed viral
RNA.
- This prevents the viral RNA from being transcribed and therefore
survives in the host plant without infecting the plant.

(b) 1 mark each for the following:

- RNAi or RNA interference
- RNAi helps in in-vitro silencing of a gene/set of genes so that they lose
their function.

OR

(a) 1 mark each for each of the following step explained in a diagram:

- From the amino acid sequence, the gene for growth hormone needs to be
synthesized chemically.
- The gene of interest is cut using a restriction enzyme and the same
restriction enzyme is used to cut the vector within the lac gene.
- The gene obtained is inserted into the vector using a ligase.
- These are transformed into E.coli cells/host cells for production.

- In recombinants, since the lac gene is inactivated, after insertion of the
gene of interest, it does not produce the β-galactosidase enzyme which
results in colourless colonies when a chromogenic substrate is added.

- In non-recombinants, since the lac gene is still active, it produced the
enzyme β-galactosidase which results in blue colonies when a
chromogenic substrate is added.

32 (a) 1 mark for the following: 5

- AUG UAC GGC AUG ACA UGG -

- yes
- since the mRNA begins with a start codon

[No marks are to be awarded if the reason is not mentioned]

(c) 0.5 marks each for the following:

- MET-TYR-GLY-MET-THR-TRP
- tRNA sequence: UAC

(d) Only methionine will remain in the amino acid sequence as the second
codon will get converted to a stop codon. [1 mark]

[No marks to be awarded if stop codon is not mentioned]

(e) 0.5 marks each for the following:

- point mutation
- guanine is getting converted to another base, thymine, resulting in the
loss of the gene

OR

(a) 1 mark for each of the following:

- man R
- The percentage of similarity of DNA fragments between the child and
the men is highest in man R.
(b) DNA profiling/DNA fingerprinting
(c) 1 mark each for the following:

- Each individual has a unique pattern of mini-satellites or repeated DNA
called VNTRs.
- This can be used to identify an individual using DNA from any body
part/fluid.
(d)

- twins/monozygotic twins [0.5 marks]
- The DNA profiles of men Q and S are the same which is possible only if
they come from the same zygote/if they are identical twins. [1 mark]
33 (a) 0.5 marks for each correct name: 5

- embryo
- endosperm

(b) 1 mark for each correct answer:
- The embryo is represented by the entire pea seed.
- The endosperm is consumed by the developing embryo and cannot be
identified as such.

(c) 1 mark each for mentioning the following:
- The pollen grain needs to be transferred to the stigma. Hence, the
smaller size of the male gametophyte makes movement easier.
- The ovule develops into the seed and supports the growing embryo. The
supporting cells provide nourishment to the growing embryo.

OR

(a) 0.5 marks each for the following:

- After 1st meiotic division - 23 chromosomes
- Reason - 1st meiosis/reductional division results in halving of the
chromosome content in the primary spermatocytes.

- After 2nd meiotic division - 23 chromosomes
- Reason - The 2nd meiotic division is similar to mitosis/equational
division where there is no reduction in the number of chromosomes.

[Accept any other valid answer]
(b) (i) formation of the sperm from secondary spermatocytes does not
happen
OR
differentiation of the spermatocyte into head, neck, tail and middle piece
does not happen
(b) (ii) low sperm count
(c)

- degenerate/degrade [0.5 marks]
- Since the secondary oocyte and the ovum retain bulk of the cytoplasm
that contains the nutrients for survival, the polar bodies are not likely to
survive for long and therefore degenerate. [1 mark]