18.Sol: All the statements are correct except
V as viruses with RNA as genetic material
mutate faster.
19.Sol: Because in RNA thymine is replaced
by Uracil
20.Sol: All the given statements are correct.
21.Sol: Non-coding strand & Antisense strand
31.Sol: (i) Genetic Code for GUU, proline
CCU
(ii) mRNA strand is transcribed from
DNA based on complimentary base
pairing
(iii) Stop codon-UAA.
- Sol: Methionine - glycine - histidine - lysine
- valine; in eukaryotes, the codon that codes
for methionine act as starting signal
40.Sol: The correct order of events for
synthesis of the lagging strand is primase adds
RNA primer, DNA polymerase III
creats a stretch, DNA polymerase I removes
the primer, and ligase seals the gaps.
41.Sol: Each codon contain 3 Nitrogen bases
thus
- Sol:
m Both introns and exons are present in
mRNA and tRNA does not possess exons.
m An intron is any nucleotide sequence within
a gene that is removed by RNA splicing
during maturation of the final RNA product.
m Prokaryotes can’t have introns, because they
have transcription coupled to translation.
m They don’t have time/space for that, since
intron splicing will stop the coupling.
m Eukaryotes evolved the nucleus, where
splicing can be done.
45.Sol:
m In prokaryotes, regulator genes often code
for repressor proteins.
m Repressor proteins bind to operators or
promoters, preventing RNA polymerase
from transcribing RNA.
m They are usually constantly expressed so
the cell always has a supply of repressor
molecules on hand.
m Inducers cause repressor proteins to change
shape or otherwise become unable to
bind DNA, allowing RNA polymerase to
continue transcription.
m Regulator genes can be located within an
operon, adjacent to it, or far away from it
49.Sol: The shorter DNA fragment move
faster towards the anode than the longer
fragments.- Sol: There are an estimated 19,000-
20,000 human protein-coding genes. The
estimate of the number of human genes has
been repeatedly revised down from initial
predictions of 100,000 or more as genome
sequence quality and gene finding methods
have improved, and could continue to drop
further.
52.Sol: The correct sequence is I, IV, II, III
54.Sol:1+3=4, one initiating code, -AUG and
3 stop codons UAA, UAG, UGA
57.Sol: In a 3.2 Kbp long piec of DNA =
3200bp.
If-820 adenine = 820 Thymine
according to chargaff’s rule
purines = pyrinidinies
A= T G = C
820 + 820 = 1640
3200 - 1640 = 1560
G + C = 1560
59.Sol: Statement II and V are correct
remaining statements are wrong.
60.Sol: Deoxyribose is the sugar found in the
backbone of DNA. C 1 of this sugar is linked
to the N 1 of a pyrimidine or the N 9 of a
purine base by a - N-glycosidic linkage.
61.Sol: 0.1mm=10^5 nm, diameter of DNA =2
n m. A x i a l r a t i o = = 5 0 , 0 0 0
(nm=nanometer)
62.Sol: Divide the total number of base pairs
by the size of the Okazaki fragment or
initiation sites.
63.Sol: If the one base is coding one amino
acid, then(12)^1 =12 but we have 96 amino acids
Suppose two basis form one codon, then (12)^2
=144.It is sufficient to code 96
or 20 amino acids and hence two, bases
from one codon.