BIOLOGY
CBSE
Sample Question Paper
Solution of Question Paper (2020)
SECTION ‘A’
1. (a) Medical Termination of Pregnancy (MTP) 1
Commonly Made Error
Many students fail to identify the correct act. Some are unaware of the concept of MTP (Medical
Termination Pregnancy).
OR
(a) GIFT (Gamete Intra Fallopian Transfer) 1
2. (c) Banana 1
3. (d) Down's syndrome 1
4. (c) Fish 1
5. (a) Lichen 1
OR
(d) In secondary succession soil is already their, hence the rate of succession is much faster. 1
SECTION ‘B’
6. Three Antipodals Two Polar nuclei One Egg Two Synergids Central Cell ½ × 4 Commonly Made Error Many students, instead of drawing the Embryo sac, drew the entire anatropous ovule. Some students
labelled the diagram incorrectly.
7. Synergids = n / haploid , egg = n / haploid , polar nuclei = n / haploid , antipodals = n / haploid //
all types of cell of female gametophyte are haploid / n = 2 ½ × 4
Time : 3 Hours Maximum Marks : 70
Answering Tip
Students should understand the concept of Haploid (n) and Diploid (2n). While studying plant
reproduction, at every step, understand the basic concepts of alternation of generation, viz., megasporogenesis, microsporogenesis, fertilization, embryonic development and endosperm formation.
Understand the location of haploid and diploid nuclei in embryo sac.
8. A unit of eight molecules of positively charged histones, negatively charged DNA , wrapped around
the histone octamer, contains 200 bp of DNA helix. ½ × 4
// In lieu of the above explanation the following diagram along with the following statement can be considered.
H1 histone
Histone octamer DNA
Core of histone molecules
DNA is negatively charged, histone is positively charged, 200 bp of DNA helix. ½ × 3
Diagram ½
9. Can grow easily on materials like waste water from potato processing plants / straw / molasses /
animal manure / sewage , reducing environmental pollution. ½ + ½
It serves as food rich in protein / carbohydrates / fats / vitamins / minerals (Any two) ½ + ½ OR
Thorns of Bougainvillea and tendrils of Cucurbita represent homologous organs (divergent evolution).
Example in animals: Forelimbs of whales, bats and cheetah. 1 + 1
Answering Tip
Homologous and Analogous organs should be discussed with respect to their origin and functions.
Merely giving a list of these organs does not help in proper learning.
10. For maintaining continuous culture system in bioreactors, they are continuously and regularly
fed with growth culture medium steadily, maintaining appropriate temperature, pH and other parameters within the bioreactors.
From the other end of bioreactor, product generated as a result of microbial action is continuously or periodically withdrawn in specified amounts.
This type of culturing method produces a larger biomass leading to higher yields of desired protein.
2 11. Ways by which GMO's have been useful for enhanced crop output are :
(i) Genetically modified crops are more tolerant to abiotic stresses (i.e. cold, drought, salt, heat).
(ii) They have reduced reliance on chemical pesticides (pest-resistant crops).
(iii) They have helped to reduce post-harvest losses.
(iv) They have increased efficiency of mineral usage by plants (this prevents early exhaustion of
fertility of soil).
12. (a) CFC adds to degradation of ozone layer in the stratosphere. (b) Ozone shields the earth by absorbing UV radiation from sun. (c) UV rays are highly injurious to organisms.
(d) To prevent degradation of ozone layer (in stratosphere) / to maintain a balance between
production and degradation of ozone in stratosphere. ½ × 4
SECTION ‘C’
13. (a) Difference between spermatogenesis and spermiogenesis :
Spermatogenesis is a process of formation of haploid spermatozoa from germinal cells while
spermiogenesis is a process of differentiation of spermatozoa into a spermatid. Here, a spermatid forms a single spermatozoon.
(b) Mitochondria provide energy for the movement of sperm. 2+1 Answering Tip
Understand the differences between spermatogenesis and spermiogenesis in tabular form for easy
retention. 14. Epidermis Endothecium Middle layers Microspore mother cells Tapetum
(Any four of the labels) ½ × 4 = 2
Tapetum nourishes the developing pollen grains 1
Commonly Made Error
Don't forget to answer the 2nd part of the question.
OR
After implantation, interdigitation of maternal & foetal tissues results in formation of structural and
functional unit between embryo & maternal body called placenta. 1 It facilitates supply of oxygen and nutrients to the embryo, removal of carbon dioxide and excretory
material, also acts as an endocrine tissue and produces hormones like hCG, hPL, Estrogen,
Progesterone, Relaxin. (Any two) 1+1
Commonly Made Error
Many students forget to mention the endocrine and protective functions of the placenta.
15. Gain or loss of a chromosome 1 (i) Down's Syndrome : Additional copy of 21st chromosome/ trisomy of 21 ½ + ½ (ii) Klinefelter’s Syndrome : Presence of an additional copy of X chromosome leading to XXY // Turner’s Syndrome : Absence of one of the X chromosome i.e. 45 with XO ½ + ½ 16. Transformation is the technique. Avery, MacLeod and McCarty identified the biochemical nature of
transforming principle of DNA.
They worked to determine the biochemical nature of ‘transforming principle’ in Griffith’s experiment. They purified biochemicals (proteins, DNA, RNA etc.) from the heat killed S cells to see which ones
RNases did not affect transformation. Digestion with DNase inhibited transformation, suggesting that the DNA caused the transformation. Thus, they concluded that DNA is the hereditary material.
OR
Blood related autosomal Mendelian disorder is sickle cell anaemia. It is called Mendelian disorders. It is mainly due to alteration or mutation in a single gene.
The defective haemoglobin undergoes polymerisation under low oxygen tension and changes the shape of RBC from biconcave cells to sickle-shaped elongated cells. The disease is controlled by a single pair of alleles, HbA and HbS. Of the three possible genotypes, only individual homozygous
for HbS show the disease. Heterozygous individuals (HbAHbS) are carriers. 3
17. (a) 5' 3' 3' 5' 5' 3' Continuous synthesis Template DNA (parental strands) Discontinuous synthesis Newly synthesised strands 5' 3' 4 labellings = ½ × 4
(b) 2 strands are antiparallel, DNA polymerase acts only in one direction i.e. 5’ → 3’ ½ + ½ Commonly Made Error
Students often fail to label correct polarity of parental strand, continuous strand and discontinuous
strand. Answering Tip
Practice diagram with correct polarity.
18. (a) Continuous inbreeding especially close inbreeding reduces fertility, and even productivity /
yield. ½ + ½
(b) Helps in accumulation of superior genes / elimination of less desirable genes, increases
homozygosity, pure lines, true breeding, helps to restore fertility, helps to increase yield / productivity, produces more milk per lactation, produces superior progeny, produces disease
resistant breeds. (Any two) 1 + 1
19. Flocs - Aerobic microbes consume the major part of the organic matter in the effluent, significantly
reduces BOD. 1 + 1
Activated sludge - Small part of activated sludge is used as inoculum and pumped back to aeration
tank / pumped into anaerobic sludge digesters where microbes or bacteria grow anaerobically to
produce CH4 or H2S or CO2 or biogas. 1
20. Habitat loss and fragmentation = ½ , Habitat loss from tropical rainforest / The Amazon rain forest is
being cut and cleared / for raising cattle / for conversion to grasslands / for cultivating soyabeans / large habitats are broken up into small fragments due to human activities / mammals and birds requiring large territories are badly affected leading to decline in population = ½
Over exploitation = ½ , when ‘need’ turns ‘greed’ lead to over exploitation of natural resources /
Steller’s sea cow / passenger pigeon were over exploited / marine fish populations around the world are over exploited / endangering existence of commercially important species = ½
Alien species invasions = ½ , when introduced unintentionally or deliberately for any purpose some
of them turn invasive and decline indigenous species / carrot grass / Parthenium / African cat fish /
Co-extinctions = ½ , when a species becomes extinct the plant or animal species associated with it (an obligate way) become extinct / when a host species becomes extinct (its unique assemblage of) parasites meets the same fate / extinction of any member in plant pollinator mutualism leads to
extinction of other = ½ (Any three named and explained) = 1 × 3
21. Competition Predation Commensalism
Differences Both species lose in their
interaction. Only one species benefits in their interaction. One species is benefitted and the other is neither benefitted nor harmed in their interaction.
Similarity Both the interacting
species live together. Both the interacting species live together. Both the interacting species live together.
1+1+1
SECTION ‘D’
22. (a) Gases are methane, ammonia, hydrogen and water vapour. ½ (b) In flask A, electric discharge is created using electrodes. ½
(c) B is amino acids and organic acids. 1
(d) They concluded that life could have originated from pre-existing non-living organic molecules
and their formation was preceded by chemical evolution. 1
23. (a) Patient should ensure that the donor’s blood has been screened for HIV and other pathogens
before transfusion. 1
(b) If this advice is not followed, he might contract AIDS. It is a fatal disease in which the immune
system of the patient gets weakened. It is caused by a virus called HIV, which is a retrovirus, which attacks helper T-cells of the body and greatly reduce their number. These helper T-cells are responsible for stimulating the antibody production by B-cells. Thus, reduction in their number results in the loss of natural defence of our body. ½ × 4 = 2
Retro virus
Viral RNA core Virus infects normal cell Viral protein coat Plasma membrane Animal cell
Viral RNA is introduced into cell Cytoplasm Viral DNA is incorporated
into host genome New viral RNA is by the infected cellproduced
DNA Nucleus New viruses are produced Viral DNA is produced by reverse transcriptase New viruses
can infect other cells
24. (a) The recombinant colonies can be differentiated from non-recombinant colonies by their inability
The recombinants do not produce any colour while the non-recombinants produce a blue colour
with chromogenic substrate in the medium.
(b) The enzyme α-galactosidase become inactivated on insertion of recombinant DNA, within the
coding sequence of enzyme. Thus, this method is called insertional inactivation. 2+1
SECTION ‘E’
25. According to Morgan and his group if genes were very tightly linked they showed very low
recombination = 1 (shown in cross A) = 1
If genes were loosely linked they showed very high recombination = 1 (shown in cross B) =1 The group used the frequency of recombination between gene pairs on the same chromosome as a
measure of distance between genes and ‘mapped’ their position on the chromosome = 1
Cross A Cross B + Parental y w w y
yellow, white wild type y+w+
m+
y+w+
y w
Wild type yellow, white w y + + White, mintature Wild type w w m m m+ w+ w+ m w w m
White, mintature Wild type + Cross A Cross B + Parental y w w y
yellow, white wild type y+w+
m+
y+w+
y w
Wild type yellow, white w y + + White, mintature Wild type w w m m m+ w+ w+ m w w m
White, mintature Wild type + Parental type (98.7%) Recombinant type (37.2%) Recombinant type (1.3%) Parental type (62.8%) Gametes w+ m+ w+ m m+ m w w
Wild type miniature
white White,miniature w+ w+ w w w m m m m m+ m+ m m w w w Wild type miniature
White,miniature white yellow,white yellow white Wild type w+ y+ y+ w w+ w y y
Wild type white
yellow yellow,white w+ w+ y y y y y y w w w w w w y+ y+ 2 + 2 + 1 OR
(a) (i) Before industrialization, white coloured lichen covered the trees in which white winged
moths camouflaged themselves from predators.
(iii) After industrialization, there were more dark winged moths in the same area i.e. proportion was
reversed.
(iv) Predators would spot a moth easily against a contrasting background.
(v) During post industrialization, tree trunks became dark due to industrial smoke and soot.
(vi) White winged moth did not survive due to detection by predators whereas dark winged
survived. ½ × 6
(b) The process of evolution of different species in a given geographical area starting from a point,
radiating to other areas of geography (habitats) is called adaptive radiation. Finches evolved in the same island from original seed eating features, many other altered beaks arose enabling
them to become insectivorous and vegetarian finches ½ × 4
26. (a) (i) a-S = CA2 1
(ii) b = log S = log C+ Z log A 1
Slope-Z (regression coefficient)
(iii) Value of Z = 1.15 (frugivorous birds) 1
(Normal value 0.6 to 1.2)
(b) (i) Animals are mobile- can avoid predator or unfavourable event.
(ii) Well developed nervous system to receive stimuli against external factors and respond to
them. 2
Answering Tip
Students are advised to answer the questions part by part.
OR
(i) Phosphorus in living systems occur in biological membranes/nucleic acids/cellular energy
transfer system/bones, teeth/in animal shells. (Any two) ½ + ½ = 1 (ii) The natural reservoir of phosphorus is rock which contains it in the form of phosphates. 1 (iii) Phosphorus cycle : The phosphates dissolved in soil solution are absorbed by the roots of higher
plants. The dead organic remains of animals and plants are decomposed by microorganism like bacteria to release phosphorus again in the soil from where it is again absorbed by plants. 3
Consumers Producers Detritus Soil solution Litter fall Decomposition Uptake Run off Weathering Rock minerals
Figure : A simplified model of phosphorus cycling in a terrestrial ecosystem
27. Plasmodium falciparum causes malignant malaria. It enters into human body in sporozoites form.
Life cycle of Malarial parasite in human body is as follows: The malarial parasite requires two hosts – human and female Anopheles mosquito to complete their life cycle.
(a) Life cycle of plasmodium starts with inoculation of sporozoites (infective stage) through the bite
(b) The parasite initially multiplied within the liver cells and then attack the red blood cells (RBCs)
resulting in their rupture.
(c) There is release of a toxic substance called hemozoin from the ruptured RBCs which is responsible
for the chill and high fever.
(d) From the infected human, the parasite enters into the body of Anopheles mosquito during biting
and sucking blood.
(e) Further development takes place in the body of Anopheles mosquitoes. (f) The female mosquito takes up gametocytes with the blood meal.
(g) Formation of gametes and fertilization takes place in the intestine of mosquito.
(h) The zygote develops further and forms thousands of sporozoites which migrates into the salivary
gland of mosquito. When the mosquito bite another human, sporozoites are injected. 1+1+3 OR
(a) Two limitations of traditional breeding that led to promotion of micro-propagation are: (i) It failed to fulfil the demand.
(ii) It failed to provide an efficient and fast crop improvement system. (b) Two advantages of micro-propagation are:
(i) It helps in the propagation of a large number of plants in a short span of time.
(ii) It leads to the production of healthier plantlets, which exhibit better disease resistant powers.
(c) Micro-propagation technology is adapted in the commercial production of many important food
plants such as tomato, apple and banana. 2+2+1
