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KENDRIYA VIDYALAYA, NAD, VISAKHAPATNAM–9 MONTHLY TEST FOR SEPTEMBER: 2019–2020

MARKING SCHEME CLASS: XII MAX. MARKS: 70 SUBJECT: BIOLOGY TIME: 03 Hours

SECTION–A

1) a–b = 3′–5′ c–d = 5′–3′

OR a – Phosphate b – Adenine or Guanine

½ + ½

½ + ½

2) Wind, light weight/enormous number/non–sticky ½ + ½ 3) Amphibians – outside the body / in external medium = ½

Reptiles – inside the body

½ + ½ 4) Forelimbs of horse/cow/dog/cat ….

Wings of bird/bat. Flippers of dolphins/whale/seal…. (Any two)

OR It indicates gene migration / gene flow / genetic drift / mutation / genetic recombination / natural selection leading to evolution.

½ + ½

½ + ½

5) Ans. 4 x 106 = 1

Calculation: Length of DNA= No. of bps (base pairs) x 0.34 x 10–9

1.36 x 10–3 = No. of bps (base pairs) x 0.34 x 10–9

No. of bps (base pairs) = �.�� ��

�

�.� ���� = 4 x 106

1

SECTION–B

6) Advantage–self–pollination assured/seed production assured = 1 Disadvantage–Least variations/leading to inbreeding depression=1

1+1

7)

General Instructions: 1. All questions are compulsory. 2. The question paper consists of four sections, A, B, C and D. 3. Internal choice is given in all the sections. A student has to attempt only one of the

alternatives in such questions. 4. Section–A contains 5 questions of 1 mark each. 5. Section–B contains 7 questions of 2 marks each. 6. Section–C is of 12 questions of 3 marks each. 7. Section–D has 3 questions of 5 marks each. 8. Wherever necessary, the diagrams drawn should be neat and properly labelled.

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=½x4= 2 [2 Marks]

8) Albuminous – (with residual) endosperm is not completely used up during embryonic development e.g., wheat/maize/ castor / sunflower. Non albuminous – (without residual) endosperm is completely consumed during embryonic development e.g. pea/ groundnut. = ½×4 (If endosperm present or absent written =½ mark).

½×4

9) Ans.

IAi IBi

A Group B Group Possible blood groups: A = ½ B = ½ Alleles – IA, IB, I – (All three = 1 Any two = ½ Any one = 0) [½+½+1=2 marks]

Mother Father

O Group AB Group ii IAIB

i IA IB Gametes

Progeny

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10) VNTR–Variable Number Tandem Repeats = 1 Probe–is labelled / radioactive (single stranded hybridise DNA fragments) = 1

// VNTR are Variable Number Tandem Repeats = 1, VNTR are short repetitive nucleotide sequences of an individual = ½, Probes are strands of DNA with specific nucleotide sequences = ½,

// Probes are strands of DNA complementary to VNTR = ½, VNTR are strands of DNA of specific individual = ½ [1+½+½= 2 marks]

OR Zosterophyllum, fern, Gingo, Gnetals

1+1 //

1+½+½ //

½+½

½x 4=2

11) a–Hydrogen bonds. b–Purines, i.e., adenine or guanine. c–Pentose sugar (deoxyribose sugar). d—5′ end.

½ x 4

12) Ans. (a) Binding of inducer / lactose to the repressor = 1

(b) In the absence of inducer lactose / when repressor binds with the operator = ½

(c) β–galactosidase = ½ [1+½+½=2 Marks]

OR Ans. (a) Repressor=½, Lactose (inducer) binds with the repressor molecule=½

(b) Z gene =½, (c) When all the lactose molecules are consumed / repressor becomes free to bind with operator = ½ [1+½+½=2 Marks]

SECTION–C

13) (a)

Geitonogamy Xenogamy Transfer of pollen grains from

anther to stigma of another flower of the same plant

Transfer of pollen grains from anther to stigma of a different plant of the same species.

(b) Characters of progeny in geitonogamy are same as parents/no variation/

introduces homozygosity (pure lines)/low rate of variation can cause inbreeding depression = ½

Characters of progeny in Xenogamy are different from parents/ variation is observed/ genetically different from parent/ no inbreeding depression = ½

1+1

½

½

14) Pollen release and stigma receptivity are non-synchronized, either the pollen is released before the stigma becomes receptive / stigma becomes receptive before the release of pollen = ½ × 2

Anther and stigma are placed at different positions, pollen cannot come in contact with stigma of the same flower = ½ × 2

Self-incompatibility, prevents self-pollen from fertilising the ovules = ½ × 2

½+½

½+½

½+½

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15)

Correct diagram with: 1 labelling = ½, 2 labellings = 1, 3 labellings = 2, 4 labellings = 3 [3 marks]

OR

= ½ x 6

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16)

1 x 3

17) (a) Non- medicated (e.g. lippes loop) , phagocytosis of sperms = ½ + ½ Copper releasing IUDs (CuT , Cu7, Multiload 375) , suppress sperm

motility / reduces fertilizing capacity of sperm= ½ + ½ Hormone releasing IUDs (Progestasert , LNG - 20) , makes uterus

unsuitable for implantation / cervix hostile to sperms = ½ + ½ (Any two) (b) Advantages of Saheli Non -steroidal /once a week / high contraceptive value / less side effects (Any two) = ½ + ½

3

18)

(Correct diagram with labelling of four wall layers) Function Epidermis, Endothecium, Middle layers – protection and dehiscence = ½ Tapetum – nourishment of developing pollen grains = ½

3

19) (a) Nucloesomes = ½ (b) a – Histone octamer = ½, b – DNA = ½, c – H1 histone = ½ (c) In bacteria DNA in nucleoid, is organised in large loops held by proteins= ½ + ½ [½+1½+1 = 3 Marks]

½ ½ x 3

½+½

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20) Ans. XhY X XX h Father is haemophilic Mother is carrier = 1 = 1 XhXh = 1 Possibility of the daughter being haemophilic.

OR DNA sequences which are repeated many a times, show a high degree of polymorphism, and form a bulk of DNA in a genome, called satellite DNA=½x3 DNA from every tissue from an individual shows the same degree of polymorphism and is heritable. Hence very useful in DNA finger printing=½x3

1 x 3

1½

1½

21) DNA is packaged in the cell in the following manner: (a) As Nucleosomes consists of Histone octamer around which the positively

charged DNA is wrapped around to form a nucleosome. A typical nucleosome contains 200bp of DNA helix.

(b) Repeated units of nucleosomes then form chromatin (in a nucleus). The nucleosomes represent the “Beads on String” structure” as seen in electron microscopic picture.

(c) These are then further coiled and condensed at metaphase stage to form chromosomes.

(d) For packaging of chromatin at higher level, non–histone proteins are required.

½x6=3

22) (i) RNA polymerase II = ½ (ii) Has (nonfunctional) introns = ½

(Methyl guanosine tri–phosphate is added to 5′ end) capping, tailing (Poly A tail at 3′ end added), splicing (introns are removed and exons are joined) = ½ x 3 = 1½ Nucleus = ½ [½+2½=3 Marks]

½

(½ x5 = 2½ )

23) a) From the original seed eating features, many other forms with altered beaks arose, enabling

them to become insectivorous and vegetarian finches=½ x 3 b) The process of evolution of different species in a given geographical area starting from one

point, and literally radiating to other areas of geography (habitats), is called adaptive radiation = ½ x 3 [1½+1½ =3 Marks]

OR (a) Australia = ½

(b) Adaptive Radiation = ½ The process of evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography (habitat) is called adaptive radiation = 1

(c) (Convergent evolution / Adaptive Convergence) Organisms coming from different stock, evolved similar features and adapted to same habitat = 1 [½+½+1+1 = 3 Marks]

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24)

(a) Homo erectus Homo habilis

(i) Brain capacity 900 cc Brain capacity 650 – 800 cc (ii) (Probably) ate meat (Probably) did not eat meat

1+1 (b) Silurian Carboniferous Jurassic = 1

(No mark to be awarded if all the three are not in proper sequence) OR

(a) He could not get desired results because: i) O2 was used instead of H2 ii) Temperature maintained was 800c instead of 8000. 1

(b) It was concluded that life could have come from pre–existing non-living organic molecules and their formation was preceded by chemical evolution 1

(c) He observed formation of Amino acids when in a closed flask CH4, H2, NH3 and water vapour were heated at 8000 C in presence of electric discharge. Analysis of meteorite content also reveals similar compounds indicating that similar process are occurring elsewhere in space / Chemical evolution. Urey & Miller proved that life originated abiogenetically whereas theory of spontaneous generation emphasized that units of life called spores were transferred to different planets including Earth. 1

SECTION–D

25) (a) (i) To detect chromosomal disorders / sex determination (legally banned)

/ detect genetic disorder / Karyotyping = 1 (ii) To prevent pregnancy / means of natural contraception = 1 (iii) To assist an infertile couple to have children by tranferring the zygote

/ early embryo/ embryo at eight blastomere stage into fallopian tube=1 (b) A poster made on RCH - Any relevant slogan or sketch made should be

awarded marks e.g. Hum Do Hamare Do, Do Boond Zindagi Ke , Beti Bachao Beti Padhao, Stop STD, Gender selection and detection is

punishable, (Any other relevant theme) = 2 [3 + 2 = 5 marks]

OR (a)

(i) A diploid egg is formed without reduction division which develops into embryo without fertilization = 1

(ii) Some cells of the nucellus (which are diploid in nature) start dividing and develop into embryo = 1

(b) Advantage: No segregation of characters in hybrid progeny / Apomictic hybrid can be used to grow crop year after year / economical as ordinary hybrid seeds are costly = 1 Disadvantage: Cannot control deleterious genetic mutation / it reduces genetic diversity from parents to offspring plants due to lack of variations (in asexual reproduction) / lack ability to adapt to changing environment= 1

(c) Hybrid seeds are costly as farmers have to purchase seeds year after year /production of hybrid seeds is a technical and expensive method to be done under controlled conditions = 1 [2 + 2+ 1 = 5 marks]

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26) (a)

It is less reactive due to absence of 2′–OH group//RNA has 2′–OH group – More reactive, ///

In the nucleotide of RNA the 2′–OH group is a reactive group and this makes RNA labile and degradable, while DNA is chemically less reactive and structurally more stable.

DNA does not act as catalyst, RNA is a catalyst// RNA is also now known to be catalytic, hence reactive.

In DNA, the presence of thymine at the place of uracil also confers additional stability to DNA.

Stability as one of the properties of genetic material was very evident in Griffith’s ‘transforming principle’ itself that heat, which killed the bacteria, at least did not destroy some of the properties of genetic material. This now can easily be explained in light of the DNA that the two strands being complementary if separated by heating come together, when appropriate conditions are provided.

Therefore, DNA chemically is less reactive and structurally more stable when compared to RNA. Therefore, among the two nucleic acids, the DNA is a better genetic material. (Any two) =1+1=2

(b) “Unambiguous” as one codon codes for only one amino acid. “Degenerate” because some amino acids are coded by more than one codon.

“Universal” because codons code for the specific amino acid no matter what the organism is. ///

Ans. Unambiguous–One codon codes for one amino acid =½, e.g. AUG (Methionine) =½ Universal–Codon and its corresponding amino acid are the same in all organisms=½ Example: - Bacteria to human UUU codes for phenylalanine (phe) = ½ Degenerate–Some amino acids are coded by more than one codon = ½ Example: - UUU and UUC code for phenylalanine (phe) = ½ [1 + 1 + 1 = 3 Marks] [2+3=5 Marks]

OR Ans.

Labelling – Polarity, promoter, structural gene, template strand, coding strand, terminator

(6 labels x ½ = 3) Promoter–provides binding site for RNA polymerase/initiates transcription process = ½ Structural gene – codes for the enzymes. = ½ Template strand – codes for mRNA. = ½ Terminator – ends the transcription process = ½ (½x4=2) [3+2=5 Marks]

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27) Ans. The aim of the experiment done by Messelson and Stahl is to prove that DNA

replication is semiconservative. =1 They grew E. coli, in 15NH4Cl for many generations to get 15N incorporated into DNA, Then the cells are transferred into 14NH4Cl. The extracted DNA are centrifuged in CsCl and measured to get their densities, DNA extracted from the culture after one generation (20 minutes), showed intermediate hybrid density, DNA extracted after two generations (40minutes) showed light DNA, and hybrid DNA = ½ × 8 = 4

//

A correctly labelled diagrammatic representation in lieu of the explanation of experiment=4 The conclusion of the experiment is that DNA replication is semi–conservative=1 [5 marks]

OR Ans.

(a) S Strain Inject into mice Mice die = ½ R Strain Inject into mice Mice live = ½

S Strain Inject into mice Mice live = ½ (heat–killed)

S Strain (heat–killed) + Inject into mice Mice dead = ½ R Strain (live) He concluded that R Strain bacteria had somehow been transformed by the heat killed S Strain bacteria = ½ (½×5 = 2½)

(b) Oswald Avery, Colin Mac Leod, Maclyn Mc Carty (if all three mentioned=1, if two mentioned = ½, if one mentioned = 0)

(c) They concluded that DNA is the hereditary material = ½ By discovering that protein digesting enzymes (proteases) and / RNA digesting enzymes (RNAses) did not affect transformation so the transforming substance was not a protein or RNA, but digestion with DNAse did inhibit transformation therefore they concluded that DNA is the hereditary material // by purifying the biochemicals using enzymes like proteases, RNAses and DNAses = ½×3 [2½+1+1½ = 5 Marks]

Do Well and Excel

End of exam