166113
The central dogma is not applicable in the case of
1 retroviruses
2 all prokaryotes
3 all animal viruses
4 all plant viruses
Explanation:
The central dogma is not applicable in the case of retroviruses.
Manipal-2011
Molecular Basis of Inheritance and
166115
Total number of complete spirals occurring in a segment of DNA having 100 nucleotides will be
1 5
2 6
3 3
4 4
Explanation:
Number of nucleotide bases of a spiral of dsDNA molecule is twenty (10 base pairs) So, 1 spiral \(=20\) nucleotide 5 spiral \(=20 \times 5\) 5 Spiral \(=100\) nucleotide
MHT CET 5.10.2020 Shift-I
Molecular Basis of Inheritance and
166117
Which one of the following is NOT needed to construct solenoid fibre?
1 Nucleosome
2 DNA molecule
3 RNA molecule
4 Basic proteins histones
Explanation:
The solenoid model represents the organization of the nucleosome with the \(30 \mathrm{~nm}\) wide chromatin fibre. The term solenoid itself defines the winding of the DNA helix. In eukaryotes, the DNA is compressed into chromatin fibre after the association of DNA-linking histone proteins. Nucleosome and Solenoid Model
MHT CET 5.10.2020 Shift-I
Molecular Basis of Inheritance and
166119
If the total amount of adenine and thymine in a double-stranded DNA is \(45 \%\), the amount of guanine in this DNA will be
1 \(22.5 \%\)
2 \(27.5 \%\)
3 \(45 \%\)
4 \(55 \%\)
Explanation:
The total amount of \(\mathrm{T}\) and \(\mathrm{A}=45 \%\) then, total amount of \(\mathrm{G}\) and \(\mathrm{C}\) will be equal to \(100 \%-45 \%=\) \(55 \%\) Also, \(\% \mathrm{G}=\% \mathrm{C}\), Therefore \(\% \mathrm{G}=\% \mathrm{C}\) \(=\frac{55 \%}{2}=27.5 \%\) so, the percentage of guanine in DNA will be equal to \(27.5 \%\).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Molecular Basis of Inheritance and
166113
The central dogma is not applicable in the case of
1 retroviruses
2 all prokaryotes
3 all animal viruses
4 all plant viruses
Explanation:
The central dogma is not applicable in the case of retroviruses.
Manipal-2011
Molecular Basis of Inheritance and
166115
Total number of complete spirals occurring in a segment of DNA having 100 nucleotides will be
1 5
2 6
3 3
4 4
Explanation:
Number of nucleotide bases of a spiral of dsDNA molecule is twenty (10 base pairs) So, 1 spiral \(=20\) nucleotide 5 spiral \(=20 \times 5\) 5 Spiral \(=100\) nucleotide
MHT CET 5.10.2020 Shift-I
Molecular Basis of Inheritance and
166117
Which one of the following is NOT needed to construct solenoid fibre?
1 Nucleosome
2 DNA molecule
3 RNA molecule
4 Basic proteins histones
Explanation:
The solenoid model represents the organization of the nucleosome with the \(30 \mathrm{~nm}\) wide chromatin fibre. The term solenoid itself defines the winding of the DNA helix. In eukaryotes, the DNA is compressed into chromatin fibre after the association of DNA-linking histone proteins. Nucleosome and Solenoid Model
MHT CET 5.10.2020 Shift-I
Molecular Basis of Inheritance and
166119
If the total amount of adenine and thymine in a double-stranded DNA is \(45 \%\), the amount of guanine in this DNA will be
1 \(22.5 \%\)
2 \(27.5 \%\)
3 \(45 \%\)
4 \(55 \%\)
Explanation:
The total amount of \(\mathrm{T}\) and \(\mathrm{A}=45 \%\) then, total amount of \(\mathrm{G}\) and \(\mathrm{C}\) will be equal to \(100 \%-45 \%=\) \(55 \%\) Also, \(\% \mathrm{G}=\% \mathrm{C}\), Therefore \(\% \mathrm{G}=\% \mathrm{C}\) \(=\frac{55 \%}{2}=27.5 \%\) so, the percentage of guanine in DNA will be equal to \(27.5 \%\).
166113
The central dogma is not applicable in the case of
1 retroviruses
2 all prokaryotes
3 all animal viruses
4 all plant viruses
Explanation:
The central dogma is not applicable in the case of retroviruses.
Manipal-2011
Molecular Basis of Inheritance and
166115
Total number of complete spirals occurring in a segment of DNA having 100 nucleotides will be
1 5
2 6
3 3
4 4
Explanation:
Number of nucleotide bases of a spiral of dsDNA molecule is twenty (10 base pairs) So, 1 spiral \(=20\) nucleotide 5 spiral \(=20 \times 5\) 5 Spiral \(=100\) nucleotide
MHT CET 5.10.2020 Shift-I
Molecular Basis of Inheritance and
166117
Which one of the following is NOT needed to construct solenoid fibre?
1 Nucleosome
2 DNA molecule
3 RNA molecule
4 Basic proteins histones
Explanation:
The solenoid model represents the organization of the nucleosome with the \(30 \mathrm{~nm}\) wide chromatin fibre. The term solenoid itself defines the winding of the DNA helix. In eukaryotes, the DNA is compressed into chromatin fibre after the association of DNA-linking histone proteins. Nucleosome and Solenoid Model
MHT CET 5.10.2020 Shift-I
Molecular Basis of Inheritance and
166119
If the total amount of adenine and thymine in a double-stranded DNA is \(45 \%\), the amount of guanine in this DNA will be
1 \(22.5 \%\)
2 \(27.5 \%\)
3 \(45 \%\)
4 \(55 \%\)
Explanation:
The total amount of \(\mathrm{T}\) and \(\mathrm{A}=45 \%\) then, total amount of \(\mathrm{G}\) and \(\mathrm{C}\) will be equal to \(100 \%-45 \%=\) \(55 \%\) Also, \(\% \mathrm{G}=\% \mathrm{C}\), Therefore \(\% \mathrm{G}=\% \mathrm{C}\) \(=\frac{55 \%}{2}=27.5 \%\) so, the percentage of guanine in DNA will be equal to \(27.5 \%\).
166113
The central dogma is not applicable in the case of
1 retroviruses
2 all prokaryotes
3 all animal viruses
4 all plant viruses
Explanation:
The central dogma is not applicable in the case of retroviruses.
Manipal-2011
Molecular Basis of Inheritance and
166115
Total number of complete spirals occurring in a segment of DNA having 100 nucleotides will be
1 5
2 6
3 3
4 4
Explanation:
Number of nucleotide bases of a spiral of dsDNA molecule is twenty (10 base pairs) So, 1 spiral \(=20\) nucleotide 5 spiral \(=20 \times 5\) 5 Spiral \(=100\) nucleotide
MHT CET 5.10.2020 Shift-I
Molecular Basis of Inheritance and
166117
Which one of the following is NOT needed to construct solenoid fibre?
1 Nucleosome
2 DNA molecule
3 RNA molecule
4 Basic proteins histones
Explanation:
The solenoid model represents the organization of the nucleosome with the \(30 \mathrm{~nm}\) wide chromatin fibre. The term solenoid itself defines the winding of the DNA helix. In eukaryotes, the DNA is compressed into chromatin fibre after the association of DNA-linking histone proteins. Nucleosome and Solenoid Model
MHT CET 5.10.2020 Shift-I
Molecular Basis of Inheritance and
166119
If the total amount of adenine and thymine in a double-stranded DNA is \(45 \%\), the amount of guanine in this DNA will be
1 \(22.5 \%\)
2 \(27.5 \%\)
3 \(45 \%\)
4 \(55 \%\)
Explanation:
The total amount of \(\mathrm{T}\) and \(\mathrm{A}=45 \%\) then, total amount of \(\mathrm{G}\) and \(\mathrm{C}\) will be equal to \(100 \%-45 \%=\) \(55 \%\) Also, \(\% \mathrm{G}=\% \mathrm{C}\), Therefore \(\% \mathrm{G}=\% \mathrm{C}\) \(=\frac{55 \%}{2}=27.5 \%\) so, the percentage of guanine in DNA will be equal to \(27.5 \%\).
