Plus Two Zoology Chapter Wise Previous Questions Chapter 4 Molecular Basis of Inheritance are part of Plus Two Zoology Chapter Wise Previous Year Questions and Answers. Here we have given Plus Two Zoology Chapter Wise Previous Questions Chapter 4 Molecular Basis of Inheritance.
Kerala Plus Two Zoology Chapter Wise Previous Questions and Answers Chapter 4 Molecular Basis of Inheritance
Expressed sequences in the gene are called [March-2018]
DNA is tightly packed structure and is found as units called nucleosomes. [March-2018]
a. Explain the concept of nudeosomes.
b. Differentiate between euchromation and hetero chromatin.
a. The negatively charged DNA is wrapped around positively charged histones octamer to form a structure called nucleosomes. Nucleosome constitute the repeating unit of a structure in nucleus called chromatin, thread like stained bodies seen in nucleus.
b. The region of chromatin, which is loosely packed and stains light arc euchromatin, which are transcriptenally alive. Densely packed and transcriptionally inactive chromatin, which stains dark arc called as heterochromatin.
Identify the disadvantages of RNA over DNA as a genetic material and explain it. [March-2018]
RNA. Compared to DNA, RNA acts as the genetic material in some viruses, and RNA, Uracil is present (less stable compared to thymine) RNA is single stranted molecule.
a. In lac Opero lactose act as inducer molecule. Evaluate the statement and explain it.
b. Observe the diagram of lac Operon and identify labelled parts. A, B, Cand D. [March-2018]
In the presence of an inducer, such as lactose or allolactose, the repressor is inactivated by interaction with inducer. This allows RNA polymerase access to the promoter and transcription proceeds.
b. A – Repressor
B – P-galactosidase
C – Permease
D – Transacitylase
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) [March-2017]
2. (b) and (c)
Examine the diagram of mRNA given below. Mark the 5’ and 3’ ends of the mRNA by giving reasons.
Polyadenylation. It is the addition of a poly (A) tail to a messenger RNA (Adding adenosine to 3′ end). The poly (A) tail consists of multiple adenosine monophosphates, it is a stretch of RNA that has only adenine bases. Polyadenylation is part of the process that produces mature messenger RNA (mRNA) for translation.
A small fragment of skin of a different person was extracted from the nails of a murdered person. This fragment of skin led the crime investigators to the murderer. Based on this incident answer the following questions:
a. What technique was used by the investigators?
b. What is the procedure involved in this technique?
In an E. coli culture lactose is used as food instead of glucose. If so, answer the following questions:
a. How do the bacteria respond to I the above situation at genetic level?
b. If lactose is removed from the medium what will happen? [March-2017]
a. DNA Finger printing
b. Geneticis, Francois Jacob and Jacque Monod were the first to describe a transcriptionally regulated system of gene expression. In bacteria, Polycistronic structural gene is regulated by a common promoter and regulatory genes called op
eron. Eg., lac operon, trp operon, ara operon, his operon, val operon.
components of lac operon
I It consists of
1. Regulatory genes ( gene)
It codes for repressor.
2. Structural genes (z, y, a)
- z-gene – Codes for beta-galactosidase ( β -gal).
It responsible for the hydrolysis of the disaccharide, lactose into it’s monomeric units, galactose and glucose.
- y – gene – Codes for permease. Which increases permeability of the cell to beta-galactosidases (β-gal).
- a – gene – it encodes a transacety¡ase.
Read carefully the sequence of codons in the mRNA unit and answer the questions.
a. What change is needed in the first codon to start the translation process ?
b. If translation starts by that change, till which codon it can be continuous ? Why ? [March-2016]
a. For initiation, the ribosome binds to the mRNA at the start codon that is recogonised only by the initiator to tRNA.
b. A translational unit in mRNA is the sequence of RNA that is flanked by the start codon (AUG) and the stop codon and codes for a polypeptide. An mRNA also has some additional sequences that are not translated.
Explain transcription. A transcription unit in DNA is defined primarily by three regions. Write the names of any two regions.[March-2016]
The three regions in the DNA:
1. A promoter
2. The structural gene
3. A terminator
Schematic representation of DNA fingerprints are shown below: [Hints : C is a sample taken from a crime scene, A and B from two suspected individuals.
a. Which one of the suspected individual may involved in the crime?
b. Write any other use of DNA fingerprinting.
DNA I rum individual A DNA Hum individual [March-2016]
a. Individual B
b. It is used as a powerful florensic tool, used in the diagonosis of genetic diseases, also used in the determination of phylogenetic status of animals.
Results of a famous experiment are given in the figure. Answer the questions.
a. Identify the experiment.
b. Which property of the DNA is proved by this experiment ? [March-2016]
a. Meselson and Stahl’s experiment
b. The experiment proved that the DNA in chromosomes replicate semiconservatively.
a. The steps in DNA fingerprinting are given below. Complete the flowchart (A and B).
b. Mention the applications of DNA fingerprinting.
a. A – Digestion of DNA by restriction endonuclease.
B – Hybridisation using labelled VNTR probe
b. It is used as a forensic tool to the solve the problems of paternity, rape, murder etc.
It is used to determine population and genetic diversity. Used in the diagnosis of genetic diseases.
“Prediction of the sequence of amino acids from the nucleotide sequence in mRNA is very easy, but the exact prediction of the nucleotide sequence in mRNA from the sequence of amino acids coded by mRNA is difficult.”
a. Which properties of the genetic code is the reason for the above condition? Explain.
b. Which are the stop codons in DNA replication? [March-2014]
a. Degeneracy of specificity, unambigious
b. UAA, UAG, UGA
Diagrammatic representation of’Central dogma’ is given below. Observe the diagram carefully and re-draw it making appropriate corrections.[March-2015,2014,2013]
Complete the table and copy. [Model-2014]
|Tanin and Baltimore||Reverse trascription|
|Watson and Crick||(a)|
|(b)||DNA finger printing|
|Harshey and Chase||(c)|
a. DNA helical structure
b. Alec Jeffreys
c. Blender experiment
d. Jacob Monod
A transcriptional unit is given below. Observe it and answer the questions.
a. How can you identify the coding strand?
b. Write the sequence of RNA formed from this unit.
c. What would happen if both strands of the DNA act as templates for transcription? [March-2012]
a. The strand with polarity 5’-3’is called coding strand.
- If both strands act as template,they would code for RNA molecule with different sequences.
- If they code for proteins, different amino acids will be formed. So more complications.
- If two RNA molecular produced at same time they would be complementary to each other and would form double strand ed RNA. So translation affected.
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