Answer key of Teacher selection test
(PGT BIOLOGY)
1. c) Higher number of parental types
2. c) Kary Mullis
3. b) lymphocytes
4. b) Chromosome 1 and X
5. c)only P, Q and R
6. c) It is single-stranded DNA
7. c) Diener
8. d) 80 sperms and 10 ovum
9. c) Ureotelic
10. a bacterium producing a large amount of carbon dioxide
11. Space travelers use Chlorella and Spirulina algae as food supplements
12. Gelidium, Gracilaria,
13. the X-shaped points of contact between non-sister chromatids of homologous chromosomes,
where crossing over (exchange of genetic material) occurs.
14. Echinodermata
15. Lysosomes
16. The partial pressure of oxygen (pO2) in the alveoli is around 104 mmHg, while in deoxygenated
blood, it's about 40 mmHg.
17. Theodor Wilhelm Engelmann (1843-1909) conducted experiments on photosynthesis using the
green alga Cladophora
18. 2 molecules.
19. Ribulose 1,5 bi phosphate
20. Ribulose-1,5-bisphosphate carboxylase/oxygenase
21. (i) Translational unit in mRNA is the sequence of RNA that is flanked by the start codon (AUG)
and the stop codon (UAA) and codes for a polypeptide/ AUG AUC UCG UAA.
1 (ii) Untranslated regions (UTR). The UTRs are present at both 5' -end (before start codon) and
at 3' -end (after stop codon). They are required for an efficient translation process.
22.
23. Phenylketonuria (PKU) is a rare, inherited metabolic disorder where the body cannot properly
break down the amino acid phenylalanine, leading to its buildup and potentially causing brain
damage if left untreated.
24. The Test Cross:
A test cross involves crossing an individual with an unknown genotype (in this case, the
violet-flowered plant) with a homozygous recessive individual (a plant with white flowers,
genotype 'pp').
25. Coitus withdrawl, lactational amenorrhoea, periodic absteenence (description of any 2)
26. This DNA segment possesses 750 pyrimidine bases (cytosine and thymine).
Here's the breakdown:
● Total nucleotides: 1500
● Guanine (G): 410
● Cytosine (C): Since G pairs with C, C = G = 410
● G + C: 410 + 410 = 820
● A + T: 1500 - 820 = 680
● Thymine (T): Since A pairs with T, T = A = 680/2 = 340
● Pyrimidines (C + T): 410 + 340 = 750
27. a) i- Cryptorchidism of both testes means infertility. - Spermatogenesis cannot happen at the
normal internal body temperature.
(a) (ii) 0.5 each mark for answering and giving reason:
- A male with cryptorchidism of one testis might be able to produce sperm. - Spermatogenesis
can occur in the descended testis as its temperature will be lower than the normal internal body
temperature.
(b) (i) scrotum (ii) 34.5 °C - 35 °C
28. Continuous self-pollination leads to inbreeding depression
a) Self-sterility: A gene prevents pollen grains from germinating in its stigma. b) Heterostyly: The
stigma and anthers are located at different levels.
c) Protandry: The male reproductive part (stamen) matures before the female part (carpel).
d) Protogyny: The female reproductive part matures before the male part.
( Candidate should write any three devices)
29. (a) Chain A
(b) A point mutation in hemoglobin beta gene
and HbsHbs condition lead due to Glutamic acid being replaced by Valine at the sixth position of
the beta chain.
( c) At low Oxygen tension, the mutant hemoglobin undergoes polymerisation and become
sickle-shaped.
30. a) Morphine b)
Papaversomniferum
c) Acts as depressant and slow down body functions
31.
32.
