- DNA is a polymer of nucleotides which are linked to
each other by 3’-5’phosphodiester bond. To prevent
polymerisation of nucleotides, which of the following
modifications would you choose?
(a) Replace purine with pyrimidines
(b) Remove/Replace 3’-OH group in deoxyribose
(c) Remove/Replace 2’-OH group with some other
group in deoxyribose
(d) Both (b) and (c) - The differences between mRNA and tRNA are that
(i) mRNA has more elaborated 3-dimensional
structure due to extensive base pairing.
(ii) tRNA has more elaborated 3-dimensional structure
due to extensive base pairing.
(iii) tRNA is usually smaller than mRNA.
(iv) mRNA bears anticodon but tRNA has codons.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i), (ii) and (iii)
(d) (i), (ii), (iii) and (iv)
True or False - The string-on-beads structure in chromatin is packaged
to form chromatin fibres, that are further condensed
at metaphase stage to form chromatids. - The negatively charged DNA is wrapped around
the positively charged histone octamer to form a
nucleotid. - The DNA dependent DNA polymerase catalyses
polymerisation only in one direction, that is
5 ′ → 3 ′. - The structural gene in a transcription unit is monocistronic
in prokaryotes and polycistronic in eukaryotes. - The coding strand has 5′ → 3 ′ polarity and does not
code for anything. - Many times in the eukaryotic cell, the translation can
begin much before the mRNA is fully transcribed. - The chemical method developed by Har Gobind
Khorana was instrumental in synthesising RNA
molecules with defined combinations of bases. - Untranslated regions or UTRs are present at 5′ end
(before start codon) only. - Chromosome 1 has the maximum number of genes,
2968, while chromosome Y has the least number of
genes, about 231. - Polymorphism arises due to mutations and is the
basis of genetic mapping of human genome and DNA
fingerprinting.
Match The Columns - Match Column-I with Column-II.
Column-I Column-II
A. Z gene (i) Clover leaf
B. tRNA (ii) 5 ′ → 3 ′
C. Promoter (iii) hnRNA
D. Leading strand (iv) Chaperones
E. Lagging strand (v) 3 ′ → 5 ′
F. Primary transcript (vi) TATA box
G. Helix formation (vii) b-gal - Match Column-I with Column-II. (There can be more
than one match for items in Column-I).
Column-I Column-II
A. Initiation codon (i) AUG
B. Phenylalanine (ii) UAA
C. Template strand (iii) UUU
D. Termination codon (iv) Minus strand
E. Nontemplate strand (v) Plus strand
F. Arginine (vi) GUG
G. Tr p operon (vii) UGA
(viii) AGG
(ix) Antisense strand
(x) CGU
(xi) UUC
(xii) Sense strand
(xiii) Corepressor
(xiv) Negative control
passage Based Questions
28.(A) Complete the given passage with appropriate words
or phrases.
Transcription requires (i) polymerase and (ii) factors in
eukaryotes. There are at least three RNA polymerases
in eukaryotes. (iii) is for transcribing tRNA, 5SRNA
and some snRNAs; (iv) is for transcribing mRNA and
snRNAs while (v) is for rRNA, except 5S rRNA.
(B) Read the passage and correct the errors, wherever
present.
Additional nucleotides are added to ends of RNAs
for specific functions — CCC segment in tRNA, cap
nucleotides at 3′ end of mRNA and poly-G segments
at 5′ end of mRNA. The cap is formed by modification
of ATP into 6-methyl adenosine or 6mA.
nora
(Nora)
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