Biology Times (^19)
March
Molecular basis of inheritance
1.Sol: Nucleosome is the smallest unit of DNA
packaging containing 200 nitrogen bases and four
types of histone proteins i.e. H2A, H2B, H3 and
H4. H1 type of histone is used in plugging.
2.Sol: The preferred molecule is glucose first, then
lactose is used by E.coli because ultimately lactose
is also broken down into galactose and
glucose. Permease is a carrier protein which helps
in facilitated diffusion. Lac operon is always
operational at low levels.
3.Sol: Answer can be (D) If the diagram (d) is GGU,
else no option is correct.
4.Sol: Endonucleases do not cut the DNA from free
ends. They cut from some where in the middle
depending on their restriction site.
5.Sol: Type of protenin.
6.Sol: Number of base pair
length of DNA = Number of base pairs nm
7.Sol: Lac operon use lactose as inducer hence its
an inducible operon
8.Sol: Presence of glucose inhibit the lac operon
9.Sol : No. of amino acid in one term in helix = 3.6
Pitch length for Helix = 5.4 Aº
The No. of Amino acid in polypeptide
Amino acid
= 18000 amino acid
10.Sol: Total no. of codon = 64
Functional codon = 61
Stop codon = 3
frequency of encountering stop codon
11.Sol: Histone is basic protein and it is rich inlysine
arginine
12.Sol: Molecular weight of polypeptide = 15.3 kda
Molecular weight of amino acid = 90 Da
No. of Amino acid in polypeptide90
One amino acid is coded by = 3 Nitrogen base
170 amino acid would be coded by
= 170 × 3 = 510 Nitrogen base
13.Sol: Melting temperature of DNA GC content
14.Sol: Cytosine = 15%
According to chargaff principle
A = T & G = C
A + T + G + C = 100
Thus A will be 35%
15.Sol: 5′- CGTACTA 3′
3′- GCATGAT 5′ 5′-TAGTACG-3
16.Sol: Total number of combination
Number of stop codens
So maximum number of unique amin oacids will
be 625 – 1 =624.
17.Sol: In Griffith experimentt mice died when
injected with combination heat killed s-strain +
Live R strain which resulted in transformation of
R II into S III form
18.Sol: As maximum number of GC pairs are present
in (iv).
19.Sol: Insertion of thiamine causes formation of
stop codon.
20.Sol:Watson and crick model of DNA is double
helix model of DNA, it is a B form of DNA
having a spiral length of 34Aº and diameter of 20
Aº.
21.Sol: The clover leaf model is the 2-D model.
Given by Holley. Its 3-D model is a L-shaped
model
22.Sol: The direction of RNA sequence is also from
5’-3’. The sequence of sense strand is
5’-GTTCATCG-3’
We know the sequence of m-RNA is similar to
sense strand. Only uracil is present instead of
Thymine. Hence the m-RNA sequence will be
5’-GUUCAUCG-3’
23.Sol: Given sequence is 5’-
AGCATATGATCGTTTCTCTGCTTTGAACT-3’
After Transcription the m-RNA sequence will be-
5’-
24.Sol: If the genetic code been quadruplet, the gene
that codes for the protein would have been longer
in size by 25 %.
25.Sol:Transcription is conversion of DNA
into RNA. Hence the sequence containing T
(Thymine) in DNA will be replaced by U (Uracil)
in the RNA ( transcripted sequence). Hence the
transcripted copy of 5’-ATGTATCTCAAT-3’
will be 5’-AUUGAGAUACAU-3’.
26.Sol: Puffs in Polytene chromosome of Drosophila
melanogaster salivary gland represents
transcriptionally active genes.
singke
(singke)
#1