MODEL PAPER DISUCSSION CHAPTER 4:- MOLECULAR BASIS OF INHERITANCE

 

UNIT TEST

TOPIC: MOLECULAR BASIS OF INHERITANCE

MARKS:-25                                                         TIME:-1.30HRS.

SECTION-A

QN.1) SELECT AND WRITE THE CORRECT ANSWER: -                   (04)

            i) The terminator codons are

                  a) UAA, UAG, UGA                b) AUG, UAG, UGA

                  c) UAC, AUG, UAG                         d) DCC, UAA, CAC     

            ii) The main step involved in the process of translation is

                  a) Activation of amino acid

                  b) Initiation of polypeptide chain synthesis

                  c) Termination of polypeptide chain formation

                  d) Transfer of polypeptide chain to tRNA               

            iii) Exons help in synthesis of

                  a) Proteins                      b) fatty acids

                  c) Fats                         d) glycogen

    iv) The octamer core of nucleosome contains two molecules each of the following histones 

            a) H2A, H2B, H3 and H4

                  b) H1, H2A, H2B, H3

                  c) H1, H2A, H2B, H4

          d) H1, H2B, H3, H4  

 

QN.2) ANSWER THE FOLLOWING: -                                 (03)

        i) How many base pairs of DNA helix are wound over on nucleosome approximately?

            Ans:-Approximately 200 base pairs.

             ii) Name the enzyme which nicks DNA strands temporarily at point ‘O’.

            Ans:-Enzyme endonuclease

             iii) What is the number of estimated base pairs in the bacterium E.coli?

            Ans:-4.6 million base pairs.

SECTION-B

ATTEMPT ANY FOUR: -                                                     (08)

QN.3) Give any two applications of genomics.

Ans: - 1) Used in agriculture to develop transgenic crops having more desirable characters.

        2) Genetic markers developed in genomics have applications in forensic analysis.

        3) Genomics can lead to introduce new gene in microbes to produce enzymes, therapeutic proteins and even biofuels.

        4) Helps on treatment of genetic disorders through gene therapy.

        5) Improvement of livestock is also possible.

QN.4) Give the role of b-galactoside permease and b-galactoside acetyltransferase.

Ans:-1) b-galactoside permease- permits lactose molecules to enter into the cell.

  2) b-galactoside acetyltransferase- transfers acetyl group from acetyl CO-A to galactoside.

QN.5) Draw a well labelled diagram of transcription unit.

 

QN.6) Identify ‘A’ and ‘B’ in the following figure,

 

QN.7) Differentiate between leading strand and lagging strand (any two points).

Leading strand	Lagging strand
1) Continuously growing strand is   called as leading strand	1)  The discontinuously growing     strand which is in the form of     Okazaki fragments is called as     lagging strand.
2) Synthesis of leading strand   proceeds towards the replication   fork.	2)  Synthesis of lagging strand     proceeds away from replication     fork.
3) Only one RNA primer is required.	3)  Large number of RNA primers are    required.
4) Okazaki fragments are not   produced.	4)  Okazaki fragments are produced.
5) Activity of DNA ligase is not   observed	5)  Activity of DNA ligase  observed to join DNA fragments
6)  Leading strand is synthesized over leading templet.	6) Lagging strand is synthesized over lagging templet.

 

QN.8) Match the following columns and write the correct answer,

Column(I)	Column(II)
a) UUU	i) Leu
b) AUG	ii) Isoleu
c) AUU	iii) Phe
d) UUA	iv) Met

 

SECTION-C

ATTEMPT ANY TWO: -                                                      (06)

QN.9) Explain any three properties of genetic code.

Ans:-1) Genetic code is triplet code

        2) Distinct polarity

        3) Non-overlapping

        4) Commaless

        5) Degeneracy

        6) Universal

        7) Non-ambiguous

(FOR EXPLAINATION WATCH THE FULL VIDEO)

QN.10) Give an account of Hershey-Chase experiment that proved, ’DNA is the genetic material’.

Ans: - 1) worked with bacteriophages that composed of DNA and proteins.

        2) Used radioactive phosphorus ³²P in the medium for some viruses and radioactive sulphur ³⁵S for some others.

        3) Viruses cultured on medium with radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but proteins do not.

        4) Similarly viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.

        5) These phages were allowed to infect E.coli bacteria containing normal ‘P’ and ‘S’. After infection viral coats were removed by centrifugation.

6) Bacteria infected with viruses containing radioactive DNA were radioactive and those infected with containing radioactive sulphur were not radioactive.

7) This indicates that protein did not enter the bacteria. Hence, DNA is the genetic material.

QN.11) State any three aims of human genome project.

Ans: - 1) Mapping the entire human genome at the level of nucleotide sequence.

        2) To store the information collected from the project in databases.

        3) To develop tools and techniques for analysis of the data.

        4) Transfer the related technologies to the private sectors such as industries.

        5) Taking care of the legal, ethical and social issues which may arise from the project.

 

SECTION-D                   

ATTEMPT ANY ONE                                                         (04)

QN.12) Explain various steps of DNA fingerprinting in sequence. Draw a flow sheet diagram for the same.

Ans:-Technique of DNA fingerprinting involves following steps.

    (A) Extraction of DNA:

    DNA molecule from sample is extracted which may be blood stain, semen, saliva, etc.

    (B) Amplification:

    Many copies of DNA are made by polymerase chain reaction (PCR) if the content of DNA is less.

    (C) Restriction digestion / Fragmentation:

    It is cutting of DNA into fragments containing VNTRs at specific site with the help of restriction endonucleases.

    As a result DNA fragments of different lengths are obtained which contain VNTR.

    (D) Electrophoresis:

    1.  Broken DNA fragments are separated and arranged according to their
  length and electric charge on ‘Agarose polymer Gel slab’.

    2.  It is a method of separation of charged molecules by applying electric field.

    3.  The DNA molecule have negative charge. Therefore, they move towards anode.

4. The DNA fragments move through the gel column and the rate of movement depends upon length of fragment.

    E)  Splitting or Conversion of ds-DNA:

      By using alkaline chemicals ds-DNA can be made single stranded.

    (F) Southern Blotting (Development by Southern).

    1.  Transfer of separated DNA sequences from gel on nitrocellulose or nylon membrane which shows number of ss-DNA fragments.

    2.  These fragments get permanently fixed on membrane.

      3.  ss-DNA fragments are hybridized with already available DNA probes or markers (radiolabeled).

    4.  In autoradiography probe hybridizes with those ss-DNA strands which contain VNTR.

    5.  This nylon membrane is exposed to x-ray film to mark the places where DNA probe bound to DNA fragments.

6. The hybridized regions are marked as dark bands on x-ray film called as autoradiograph which represents DNA fingerprint.

QN.13) Describe the experiment that proves, ‘DNA replication is semiconservative’.

Ans: - 1) semiconservative nature of DNA replication was experimentally proved by Messelson and Stahl using equilibrium density gradient centrifugation technique.

        2) Cultured E.coli in the medium containing ¹⁴N (light nitrogen) and obtained equilibrium density gradient band by using CsCl₂.recorded position of this band.

        3) These cells were transferred to ¹⁵N (heavy isotopic nitrogen) medium and allowed to replicate for several generations.

        At equilibrium point density gradient bands were obtained and position was recorded.

        4) Heavy DNA can be distinguished from light DNA on the basis of density gradient.

        5) Now these cells were cultured on medium containing ¹⁴N (normal nitrogen). After first generation density gradient band for ¹⁴N¹⁵N was obtained and its position was recorded. Two such bands were obtained after second generation, one at ¹⁴N and other at ¹⁵N position.

        6) This clearly proved that replication is semiconservative.