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QUESTIONS
1 Score Questions
1. Which of the following combinations do not apply to DNA?
(a) Deoxyribose, Guanine
(b) Ribose, Adenine
(c) Deoxyribose, Uracil
(d) Guanine, Thymine
(1) (a) and (b)
(2) (b) and (c)
(3) (c) and (d)
(4) (a) and (d)
2. In DNA, the proportion of A is equal to T and the
proportion of G is equal to C. This rule is known as ………
3. Biochemical characterization experiment proved that transforming
principle is
(a) RNA
(b) DNA
(c) Protein
(d) Lipid
4. During DNA replication, one strand
is synthesised in small stretches called …………
5. Note the relationship between first two words and fill up the
fourth place.
UGG: Tryptophan
AUG:
…………………
6. Expand the abbreviation VNTR.
2 Score Questions
7. In a DNA segment, cytosine is 22%. Find
out the percentage of adenine, thymine and guanine.
8. Identify the structure given below. Label
A, B and C.
9. Nucleosomes constitute the repeating unit to
form chromatin.
a. Name the two forms of chromatin.
b. Mention any two differences.
10. Following are the first two steps in
Griffith’s transformation experiment.
i. S-strain → inject into mice → mice
live
ii. R-strain → inject into mice → mice
die
a. If there is any mistake correct it.
b. Write the remaining steps.
11. Blender experiment (Hershey and
Chase experiment) proved that DNA is the genetic material.
a.
Name the organisms used for this
experiment.
b.
What are the steps involved in this
experiment?
12. DNA is the better genetic material than RNA. Do
you agree with this statement? Substantiate.
13. The flow of genetic information is
shown below:
Name the processes a, b, c and d.
14. Results of a famous experiment is given in the
figure.
a. Identify the experiment.
b. Which property of DNA is proved by experiment?
15. The diagram of an mRNA is given below. Copy it and
mark the 5’ end 3’ end of mRNA by giving reason.
16. Read the sequence of codon in the
mRNA unit.
a. What changes is needed in the first
codon to start the translation process?
b. If translation starts by that
change, till which codon it can be continues?
17. Observe the diagram and answer the question.
a. Identify the process and its site.
b. What is the role of mRNA and tRNA in this process?
18. Human
Genome Project (HGP) was the first mega project for the sequencing of
nucleotides and mapping of all the genes in human genome.
a. What are the two main approaches in the
methodologies of HGP?
b. Expand BAC & YAC.
19. Schematic representation of DNA
finger prints is given.
a. Which one of the suspected
individuals may involve in the crime?
b. Write any other use of DNA fingerprint.
3 Score Questions
20. Observe the diagram and answer the
question.
a. What is meant by replication fork?
b. What is the difference in the
replication process in strand A and strand B?
c. What is the role of DNA ligase
in the replication process in B strand?
21. A transcription unit is given below.
Observe it and answer the question.
a. How can you identify the coding
strand?
b. Write the sequence of RNA formed
from this unit.
c. What would happened if both strand
of DNA act as template for transcription?
22. Given below is the diagrammatic representation of
first stage of a process in a bacterium.
a. Identify the process.
b. Name the enzyme catalyses this process.
c. What are the additional complexities in eukaryotes
in this process?
23. With the help of the figure given, explain the
processing of hnRNA to mRNA in eukaryotes.
24. Observe the figure of mRNA.
a. Find the start codon and stop codon.
b. How many amino acids will be present
in the protein translated from this mRNA?
c. The additional sequences that are
not translated in the mRNA is called ................
25. “Prediction of the sequence of amino
acids from the nucleotide sequence in mRNA is very easy, but the exact
prediction of nucleotide sequence in mRNA from the sequence of amino acids
coded by mRNA is difficult”
a. Which property of genetic code is
the reason for the above condition? Explain.
b. Which are the stop codons in DNA translation?
26. A DNA sequence for coding a peptide
is given below:
3’-CAAGTAAATTGAGGACTC-5’
|
Codons |
Amino acids |
|
UUA |
Leu |
|
CCU |
Pro |
|
CAU |
His |
|
ACU |
Thr |
|
GUU |
Val |
|
GAG |
Glu |
a. Write the sequence of complementary
mRNA.
b. Find out the amino acids sequence of
peptide chain using the codon given in the hints.
c. Coding region of a gene consists of
450 nucleotide base pairs. Avoiding stop codons & introns, how many amino
acids would the corresponding polypeptide chain contain? Justify your answer.
27. In an E. coli culture,
lactose is used as food instead of glucose.
a.
How do the bacteria respond to this situation
at genetic level?
b.
If lactose is removed what will
happen?
28. Lac operon in the absence of inducer
(Lactose) is given.
a. What is ‘P’?
b. Name the enzyme produced by the
structural gene ‘Z’,’Y’, and ‘A’.
c. Redraw the diagram in the presence
of an Inducer.
29. a) The steps in DNA Finger printing
are given below. Complete the flow chart (A and B).
b) Mention the application of DNA
finger printing.
30. A small fragment of a skin of
different person was extracted from nails of a murdered person. This led the
crime investigators to the murder.
a. What technique was used by the
investigators?
b. What is the procedure involved in
this technique?
1. 2. (b) and (c)
2. Erwin Chargaff’s rule.
3. (b) DNA
4. Okazaki fragments.
5. Methionine.
6. Variable Number Tandem Repeats.
7. According to Erwin Chargaff’s rule, in DNA, proportion of A is equal to T and the proportion of G is equal to
C.
If cytosine (C) = 22%, then guanine (G)= 22%. (C+G = 44%.)
So A+T = 100 – 44 = 56%.
Adenine (A) = 28% and Thymine (T) =
28%.
8. Nucleosome. A= DNA, B= H1 histone,
C= Histone octamer.
9. (a) Euchromatin and Heterochromatin.
(b) Euchromatin: Loosely packed and transcriptionally active region
of chromatin. It stains light.
Heterochromatin: Densely
packed and inactive region of chromatin. It stains dark.
10. (a) i. S-strain → inject into mice →
mice die
ii. R-strain → inject into mice →
mice live
(b) S-strain (Heat
killed) → Inject into mice → Mice live
S-strain (Hk) + R-strain (live) → Inject into mice →
Mice die.
11. (a) E. coli & Bacteriophage.
(b) Infection bacteriophage with E.
coli.
Blending to remove the virus particles from the
bacteria.
Centrifugation to separate virus particles from bacterial
cells.
12. Yes. For the storage of genetic information, DNA is
better due to its stability.
13. (a) Replication. (b) Transcription.
(c) Translation
(d) Reverse transcription.
14. (a) Meselson & Stahl experiment.
(b) Semi-conservative model of DNA replication.
15. Here, one end of mRNA has polyadenylate residues (AAAAA...).
Polyadenylation (tailing) occurs at 5’ end.
16. (a) First codon should be AUG.
(b) Translation continues till 5th
codon (UUU) because after that the stops codon UGA comes. So translation is
stopped.
Read carefully the sequence of codon
in the mRNA unit and answer the question.
17. (a) Translation (Protein synthesis). Its site is
ribosome.
(b) mRNA:
Provide
template for translation.
tRNA: Brings amino acids for
protein synthesis and reads the genetic code.
18. (a) Expressed Sequence Tags (ESTs) & Sequence
annotation.
(b) BAC=
Bacterial Artificial Chromosomes.
YAC= Yeast Artificial
Chromosomes.
19. (a) Individual B.
(b) For the diagnosis of genetic diseases.
20. (a) Unwinding of
the DNA molecule at a point forms a ‘Y’-shaped structure called replication fork.
(b) In
strand A, continuous synthesis occurs. In strand B, discontinuous synthesis
occurs.
(c) It helps to join the Okazaki
fragments formed (due to discontinuous synthesis) together to form a new strand.
21. (a) Coding strand will be in 5’-3’
direction.
(b) 5’ – U C A G U A C A – 3’
(c) If 2 RNA
molecules are produced simultaneously, this would be complimentary to each
other. It forms a double stranded RNA and prevents translation
22. (a) Transcription.
(b) DNA-dependent
RNA polymerase.
(c) There are
3 RNA polymerases (RNA polymerase I, II & III). The
primary transcripts (hnRNA) contain exons and introns and are non-functional.
Hence introns must be removed.
23. Steps of processing of hnRNA to mRNA:
Splicing: From hnRNA, introns are removed (by the spliceosome)
and exons are spliced (joined) together.
Capping: Here, a nucleotide methyl guanosine triphosphate (cap) is added to the 5’ end of hnRNA.
Tailing
(Polyadenylation): Here, adenylate residues (200-300) are added at 3’-end.
24. (a) Start codon is AUG (3rd
codon). Stop codon is UAG.
(b) 4 amino acids.
(c) Untranslated Regions (UTRs).
25. (a) Degeneracy. An amino acid (except
methionine & tryptophan) is coded by more than one codon. They are called degenerate codons.
(b) UAA,
UAG & UGA.
26. (a) 5’-GUU CAU UUA ACU CCU GAG-3’
(b) Val – His – Leu – Thr – Pro – Glu
(c) 150 amino acids. A codon contains 3 bases. So in this gene, 150
codons are there. A codon codes for one amino acid. Thus 150 codons code for
150 amino acids.
27. (a) If lactose is
provided in the growth medium, it is transported into E. coli cells by the action of permease. Lactose
(inducer) binds with repressor protein. So repressor protein cannot bind to
operator region. The operator region becomes free and induces the RNA
polymerase to bind with promoter. Then transcription starts.
(b) If there is no lactose, lac operon remains switched off. The regulator gene synthesizes mRNA to
produce repressor protein. This
protein binds to the operator region and blocks RNA polymerase movement. So the structural genes
are not expressed.
28. (a) P is the promoter for structural
genes.
(b) z gene: b galactosidase, y gene: Permease
a gene: Transacetylase.
(c)
29. (a) A = Digestion of DNA by restriction
endonucleases. B= Hybridization using radioactive labelled
VNTR probe.
(b) Forensic tool to
solve paternity, rape, murder etc.
For the diagnosis of genetic diseases.
To determine phylogenetic status of animals.
To determine population and genetic diversities.
30. (a) DNA fingerprinting.
(b) Procedures:
i. Isolation of DNA.
ii. Digestion of DNA by restriction
endonucleases.
iii. Separation of DNA fragments by gel
electrophoresis.
iv. Transferring DNA fragments to synthetic
membranes such as nitrocellulose.
v. Hybridization using radioactive labelled VNTR probe.
vi. Detection of hybridized DNA by autoradiography.
Chapter-wise Sample Q & A
CLASS 11 (PLUS 1) BOTANY: QUESTIONS & ANSWERS
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CLASS 12 (PLUS 2) BOTANY: QUESTIONS & ANSWERS
CLASS 12 (PLUS 2) ZOOLOGY: QUESTIONS & ANSWERS
Chapter-wise Previous Year Q & A
CLASS 11 (PLUS 1) BOTANY: CHAPTER-WISE Q & A
CLASS 11 (PLUS 1) ZOOLOGY: CHAPTER-WISE Q & A
CLASS 12 (PLUS 2) BOTANY: CHAPTER-WISE Q & A