AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

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Students get through AP Inter 2nd Year Botany Important Questions 10th Lesson Molecular Basis of Inheritance which are most likely to be asked in the exam.

AP Inter 2nd Year Botany Important Questions 10th Lesson Molecular Basis of Inheritance

Very Short Answer Questions

Question 1.
What is the function of histones in DN A packaging?
Answer:

  1. The function of histones is to package DNA into structural units called nucleosomes.
  2. Histones are main protein in chromatin, which is component of cell nucleus.
  3. DNA wraps around histones and plays a role in gene regulation.

Question 2.
Distinguish between heterochromatin and euchromatin. Which of the two istranscriptionally active? [TS MAY-22] [AP 20]
Answer:

Heterochromatin Euchromatin
1)  The chromatin that is more densely packed and stains dark is called Heterochromatin.
2)   Heterochromatin is inactive.		 1) The chromatin that is loosely packed and stains light is called Euchromatin.
2)  Euchromatin is transcriptionally active.

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance
Question 3.
Who proved that DNA is genetic material? What is the organism they worked on? [TS MAR-19] [AP MAY-17]
Answer:

  1. Alfred Hershey and Martha chase proved that DNA is the genetic material.
  2. They worked with viruses that infect bacteria called bacterio phages.

Question 4.
What is the function of DNA polymerase?
Answer:

  1. DNA polymerase is a highly efficient enzyme which catalyse polymerisation of a large number of nucleotides in a very short time.
  2. It also catalyse the reaction with a high degree of accuracy.

Question 5.
What are the components of a nucleotide? [AP,TSMAY-22] [TS1V1AR-20] [APMAR-17]
Answer:
A nucleotide has 3 components, Nitrogenbase, pentose sugar and phosphate molecule.

Question 6.
Write the structure of chromatin.
Answer:

  1. Chromatin is a combination of DNA and histone.
  2. It is present in nucleus of the cell and is concerned with heredity and inheritance.
  3. Light stained loosely packed chromatin is called Euchromatin.
  4. Dark stained densely packed chromatin is heterochromatin.

Question 7.
What is the cause of the discontinuous synthesis of DNA on one of its parental strands? What happens to these short stretches of synthesised DNA?
Answer:

  1. The two strands of DNA cannot be separated throughout the entire length in one stretch due to very high energy requirement.
  2. The replication occur as two antiparallel strands of DNA acts as templates for the synthesis of new strands.
  3. So one new strand is synthesised continuously while the other new strand is synthesised discontinuously as the opening of fork continues.
  4. The fragments of discontinuously synthesised new strand are called ‘Okazaki’ fragments and are joined together by ‘DNA ligase’.

Question 8.
Given below is the sequence of coding strand of DNA in a transcription Unit.
5’AATGCAGCTATTAGG-3′
Write the sequence of (a) its complementary strand (b) the mRNA
Answer:
a) 3′-TTACGTCGATAATCC-5′.
b) 3′-UUACGUCGAUAAUCC-5′

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 9.
In a nucleus, the member of ribonucleoside triphosphates is lOtimes the number of deoxyribose nucleoside triphosphates , but only deoxyribose nucleoside are added during DNA replication. Suggest a mechanism
Answer:

  1. The synthesis of identical two daughter DNA molecules from a parental DNA is called replication.
  2. During replication, each new strand of DNA is synthesised on each template strand of DNA due to polymerisation of deoxyribose nucleotides only by DNA polymerase enzyme.

Question 10.
Name any three viruses which have RNA as the genetic material. |AP MAR-16|
Answer:

  1. Influenza virus
  2. Polio virus
  3. HIV

Question 11.
Write the sequence of nucleotides after single base insertion and deletion in the give-name gene. THE CAT ATE THE FAT RAT.
Answer:

  1. If we insert a letter B in between CAT and ATE and rearrange the statement, the sequence is “THE CAT BAT ETH EFA TRA T”.
  2. If we remove one letter in CAT, the sequence is ” THE CAA TET HEF ATR AT”.

Question 12.
Why was DNA chosen over RNA as genetic material in the majority of the organisms.
Answer:

  1. DNA is chemically less reactive and structurally more stable when compared to RNA due to absence of 2′-OH group in its nucleotides.
  2. Presence of thymine in place of uracil also confers additional stability to DNA therefore DNA is better genetic material than RNA.

Question 13.
What are the components of a transcription unit? [TS MAR, MAY-17] [AP MAR-16, 19]
Answer:
The components of the transcription unit are a) A promoter b) The structural gene and c) A Terminator.

Question 14.
What is the difference between exons and introns? [TS MAY-22] [TS MAR-15]
Answer:

  1. Exons: These are coding sequences . They appear in mature or processed RNA.
  2. Introns: These are non-coding sequences. They do not appear in mature or processed RNA.

Question 15.
What is meant by capping and tailing? [AP MAR-17] [TS M-16]
Answer:

  1. Capping: Adding of an unusual nucleotide (methyl guanosine triphosphate) to the 5′-end of hnRNA is called Capping.
  2. Tailing: Adding of adenylate residues (200-300) to the 3′-end in a template is called tailing.

Question 16.
What is meant by point mutation? Given an example. [TS MAR-19]
Answer:

  1. The mutation that occurs in a single base pair of DNA mole is called point mutation
  2. Ex: Sickle cell anemia.

Question 17.
Define peptide bond why are proteins referred to as peptide chains?
Answer:

  1. Peptide bond: The bond present between carboxyl group of aminoacid and amino groups of adjacent amino acid by dehydration is called a peptide bond.
  2. Proteins are composed of long chains of aminoacids which are connected by peptide bonds, hence referred as polypeptide chains.

Question 18.
What is meant by charging of tRNA?
Answer:
tRNAnswer: Activation of aminoacids, in the presence of ATP and linking to their cognate tRNA, is known as charging of tRNA.

Question 19.
What is a regulator and a promoter?
Answer:

  1. Regulator: It is a region of DNA involved in controlling the expression of one or more structural genes.
  2. Promoter: It is a region of DNA where RNA polymerase binds and initiates transcription.

Question 20.
During DNA replication, why is it that the entire molecule does not open in one go? Explain replication fork.
Answer:

  1. DNA replication requires very high energy.
  2. It has to be fast and accurate as well. If entire molecule opens in one go, energy is also required in higher levels to reduce tension.
  3. So long DNA molecules of eukaryotes have a large number of replicons all of which work simultaneously.

Replication fork: Y-shaped configuration produced during unzipping of a part of DNA double helix during replication is called Replication fork.
or
It is a site on a DNA double helix where both unwinding of the helix and synthesis of daughter molecules occur.

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 21.
What is the function of the codon-AUG
Answer:
AUG has two functions:

  1. It acts as initiation codon of mRNA.
  2. It codes for the amino acid methionine.

Question 22.
Define stop codon. Write the codons. [TS MAR-19) [AP MAR-15,19|
Answer:

  1. The codons which terminates the protein synthesis are calied stop codons.
  2. They are UAA, UAG, UGA.
  3. They donot code for any amino acid.

Question 23.
What is the difference between the template strand and a coding strand in a DNA molecular?
Answer:

  1. During the transcription process, uncoiling of two strands of DNA takes place.
  2. The strand which acts as a complementary strand for the synthesis of mRNA is called template strand.
  3. The other strand of DNA which does not code for anything is called “coding strand”.

Question 24.
Write any two differences between DNA and RNA. [AP MAR-17]
Answer:

  1. DNA has deoxyribose sugar. DNA undergoes self replication.
  2. RNA has ribose sugar. RNA doesnot undergoes self replication.

Question 25.
In a typical DNA molecule, the proportion of Thymine is 30% of the N bases. Find out the percentages of other N bases. [TS M-16,22] [AP MAR-15]
Answer:
Adenine 30%; Guanine 20%; Cytosine 20% (Total 70%)

Question 26.
The proportion of nucleotides in a given nucleic acid are: Adenine 18%, Guanine 30%, Cytosine 42%, and Uracil 10%. Name the nucleic acid and mention the number of strands in it. [TS MAR-17]
Answer:
(a) RNA
(b) Only one strand

Question 27.
If the base sequence of a codon in mRNA is 5’AUG-3\ What is the sequence of t- RNA pairing with it?
Answer:
3’UAC-5′

Short Answer Questions

Question 1.
Define transformation in Griffith’s experiment. Discuss how it helps in the identification of DNA as genetic material.
Answer:

  1. Transformation: It is the genetic alteration of a cell resulting from the direct uptake and incorporation of genetic material from its surroundings through the cell membranes.
  2. In 1928, Fredrick Griffith conducted a series of experiments on a bacterium called streptococcus pneumoniae, which is responsible for pneumonia.
  3. During his experiment he found that the bacteria can change its physical form. He has grown the bacteria on a culture medium.
  4. He observed two strains of this bacterium, one forming smooth colonies with capsule (S-type) and the other forming rough colonies without capsule (R-type)
  5. The S-type cells are virulent while R-type cells are non-virulent.
  6. When live S-type cells are injected into mice, they suffered from pneumonia and died.
  7. When live R-type cells are injected into mice, the disease did not appear and the mice survived.
  8. When heat killed S-type cells were injected, the disease did not appear.
  9. When heat killed S-type were mixed with live R-type and injected into mice, the mice died of pneumonia and live S-type cells were isolated from the body of dead mice.
  10. He concluded that two R-strain cells have been transformed by the heat killed S-strain bacteria. 1 l)This must be due to the transfer of genetic material by the transforming principle.

Question 2.
Who revealed the biochemical nature of the transforming principle? How was it done.
Answer:

  1. Oswald Avery, Colin Macleod and Maclyin Mercarty worked to determine the biochemical nature of transforming principle in griffith’s experiment.
  2. They purified biochemical (proteins, DNA, RNA etc) from heat killed ‘S’ cells to see which one transform like R cells into S cells.
  3. They discovered DNA alone transforrried from S cell to R cell. They also discovered that proteases and RNases did not affect transformation.
  4. So the transforming substance is not protein or RNA. Digestion with DNase did inhibit transformation suggesting that the DNA caused the transformation.

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 3.
Discuss the significance of heavy isotope of nitrogen in Meselson and Stahl’s experiment.
Answer:
Matthew Meselson and Franklin Stahl performed experiments on E.coli and proved that DNA replicates semi-conservatively.

  1. They used heavy isotope of nitrogen called 15N.
  2. It is not radioactive and can be separated from 14N, based on densities only.
  3. Replication of DNA of E.coli will be completed within 20 minutes.
  4. They grew’E.coli in a medium containing 15NH4Cl. For many generations.
    AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance 4
  5. As a result 15N was incorporated into newly synthesized DNA.
  6. This heavy DNA mole could be distinguished from the normal DNA by centrifugation in Cesium Chloride (CsCl)
  7. Then they transformed the cells into a medium with nonnal 14NH4Cl.
  8. They took out the samples at various time intervals, extracted the DNAs and centrifuged them to measure their densities.
  9. Thus the DNA extracted from the culture one generation after the transfer from 15N to 14N medium (after 20 minutes) had a hybrid density.
  10. The DNA extracted after two generations (after 40 minutes) consisted of equal amounts of hybrid DNA and light DNA.
  11. It is now’ proved that DNA replicates semi conservatively.

Question 4.
Define a cisfron. Differentiate between monocistronic and polycistronic transcription unit with suitable examples.
Answer:
Cisfron: A segment of DNA coding for a polypeptide bond is called cisfron.

Monocistronic Polycistronic
1)  mRNA molecule contains the genetic information to translate only a single protein chain.
2)  Ex: Eukaryotes		 1) mRNA molecule contains several opening reading frames, each of which is translated into a polypeptide.
2) Ex: Prokaryotes (or) bacteria.

Question 5.
Recall the experiment done by Griffith in which DNA was speculated to be the genetic material. If RNA instead of DNA was the genetic material, would the heat killed strain of pneumococcus have transformed the R-strain into a virulent strain? Explain.
Answer:
No.

  1. Avery, MacLeod and Mccarty worked to determine the biochemical nature of transforming principle.
  2. They purified biochemical (proteins, DNA, RNA etc) from the heat killed ‘S’ cells to see which one could transform ‘R’ cells into ‘S’ cells.
  3. They discovered that DNA alone from ‘S’cell caused ’R” cell to become transformed.
  4. They also discovered that protein digesting enzymes (proteases) and RNA digesting enzymes (RNAases) did not affect transformation. So transforming was not a protein or RNA but only DNA.

Question 6.
There is only one possible sequence of amino acid when deduced from a given nucleotides. But a multiple nucleotide sequence can be deduced from a single amino acid sequence. Explain this phenomenon.
Answer:

  1. Some aminoacids are coded by more than one codon.
  2. From an amino acid sequence, multiple nucleotide sequence will be obtained.
  3. For example Met-Ile can have the following nucleotide sequence.
    (i) AUG-AUU (ii) AUG-AUC (iii) AUG-AUA nucleotide sequences will code for Met-Ile.

Question 7.
A single base mutation in a gene may not always result in loss or gain of function. Do you think the statement is correct? Defend your answer.
Answer:

  1. No, the statement is not correct.
  2. A classical example of point mutation is a change of single base pair in the gene for beta globin chain (human haemoglobin)
  3. It result in the change of amino acid residue glutamate to valine.
  4. It results in a diseased condition called “Sickle cell anemia”.
  5. Thus a change in a single nucleotide in a codon alters the amino acid in a polypeptide chain.
  6. Consider a statement that is made up of the following words each having three letter like a genetic code.
    RAM HAS RED CAP
  7. If we insert letter B in between HAS and RED and rearrange the statement, it would be read as follows
    RAM HAS BRE DCA P
  8. Similarly if we delete letter R, it would be read as follows AMH ASE DCA P.
  9. It is concluded that insertion or deletion of one or two bases change the reading frame from the point of insertion or delation.
  10. It can disrupt the protein structure and its function.

Question 8.
A low level expression of lacoperon occurs all Ihe time. Can you explain the logic behind this phenomenon.
Answer:

  1. Lactose is the substrate for the enzyme beta-galactosidase and it regulates switching on and off of the operon. Hence it is termed as inducer.
  2. In the absence of glucose, if lactose is provided in the growth medium of the bacteria, it is transported into cells through the action of permease.
  3. A very low level of expression of lac operon has to be present in cell all the time, otherwise lactose cannot enter the cells. The lactose then induces the operon.
  4. The repressor of the operon is synthesised all the time from the i-gene.
  5. The repressor protein binds to operator region of the operon and prevents RNA-polymerase from transcribing the operon.
  6. In the presence of lactose, the repressor is inactivated by interaction with the inducer. This allow’s RNA polymerase to access the promotor and transcription proceeds.

Question 9.
What back ground information did Watson & Crick have for developing a model of DNA? What was their contribution?
Answer:

  1. DNA as an acidic substance present in nucleus was first identified by Frederick Meisher in 1869 and he named it as ‘Nuclein’.
  2. X-ray diffraction data produced by Mauris Wilkins and Rosalind Franklin. Erwin charge off states that the purines & Pyramidines exist in 1:1 ratio.
  3. Based on the above information James watson and Francis crick proposed a very simple but famous double helix model for the structure of DNA.

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 10. What are the functions of (i) Methylated guanosine cap (ii) Poly-A ‘tail’ in a mature on RNA.
Answer:

  1. Methylated guanosine cap- it is an unusual nucleotide added to the 5′ end of hnRNA.
  2. Adenylate residues are added at 3-end in a template-independent manner to hnRNA.

Question 11.
How many types of RNA polymerases exist in cells? Write their names and functions. [TS MAR-17]
Answer:
There are three RNA polymerases in the nucleus (in addition to the RNA polymerase found in the organelles). They are
RNA Polymerase I : It transcribes rRNAs (28S, 18S and 5.8S)
RNA Polymerase II: It transcribes the precursor of mRNA, the heterogeneous nuclear RNA (hnRNA).
RNA Polymerase III: It is responsible for transcription of tRNA, 5srRNA and snRNAs (Small nuclear RNAs)

Question 12.
Write briefly about DNA polymerase.
Answer:

  1. DNA polymerase is a highly efficient enzyme as it has to catalyse polymerisation of large number of nucleotides in a very short time.
  2. They are essential for DNA replication, DNA repair, genetic recombination, reverse transcription and the generation of antibody diversity.
  3. They are widely used in molecular biology laboratories notably for PCR, DNA sequencing and molecular cleaning.

Question 13.
What are the contributions of George Gamow. H.GKhorana, Marshall Nirenbergin deciphering the genetic code? [TS M AR-19]
Answer:

  1. George Gamow is a physicist.
  2. He stated that there are only 4 bases to code for 20 amino acids.
  3. The code should constitute 2 combinations of these 4 bases only.
  4. Each code should contain three nucleotides (Tripletcodons)
  5. A permutation and combination of 43 (4 x 4 x 4) would generate by codons.
  6. Har Gobind Khorana developed a chemical method in synthesizing RNA moles with defined combinations of bases (homopolymers such as UUU and copolymers such as UUC, CCA)
  7. Marshall Nirenberg made cell free system for protein synthesis.

Question 14.
On the diagram of the secondary structure of tRNA shown below label the location of the following parts:
a) Anticodon b) Acceptor stem c) Acceptor stem d) 5′ end e) 3′ end
Answer:
AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance 5

Question 15.
If there are 2.9 × 109 complete turns in DNA molecule. Estimate the length of the molecule (1A°=108cm)
Answer:
1 turn of DNA = 3.4 nm = 3.4 x 10-9 m
Number of turns = 2.9 × 109
The length of DNA molecule = 2.9 × 109 x 3.4 × 10-9 = 0.986 మీ.

Question 16.
Draw the schematic/diagrammatic presentation of the lac operon. [TS MAY-22] [APMAR-15] [TS MAR-16]
Answer:
AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance 6

Question 17.
What are the differences between DNA and RNA. [TS MAY-17,22] |AP,TS 20]
Answer:
DNA

  1. DNA stands for Deoxyribo Nucleic Acid.
  2. DNA is double stranded Heliptical structure.
  3. DNA is stable under alkaline condition.
  4. DNA contains the sugar Deoxyribose
  5. DNA is made up of more than 4 million nucleotides.
  6. DNA undergoes self replication.
  7. DNA is genetic material.
  8. DNA does not participate directly in protein synthesis.
  9. The base pairing is A=T and G=C

RNA

  1. RNA stands for Ribo Nucleic Acid.
  2. RNA is single-stranded Helix.
  3. RNA is unstable under alkaline condition.
  4. RNA contains the sugar Ribose.
  5. RNA is made up of 75-2000 nucleotides.
  6. RNA does not undergo self-replication.
  7. RNA is non-genetic material.
  8. RNA participates directly in protein synthesis.
  9. Metabolically RNA is of three types.
  10. The base pairing is A=U and G=C

Question 18.
Write the important features of Genetic code? [AP,TS MAY-22] [AP MAR-16,17] [TS MAR-15]
Answer:
The important features of genetic code:

  1. Genetic code is a set of instructions that direct the translation of DNA into 20 amino acids.
  2. Genetic code consists of 64 triplets of Nucleotides. Each triplet is called a codon.
  3. 61 codons code for amino acids. 3 codons donot code for any amino acids, hence they are called stop codons.
  4. One codon codes for only one amino acid, hence it is unambiguous and specific.
  5. Some amino acids are coded by more than one codon, hence the code is degenerate.
  6. The codon is read in mRNA in a contiguous fashion. There are no punctuations.
  7. The code is nearly universal.
  8. Ex: From bacteria to human, UUU would code for phenylalanine (phe)

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 19.
Describe sequential steps in the replication of DNA molecule?
Answer:

  1. The double helix is unwound by helicase enzyme exposing two single strand DNAs.
  2. This creates a replication fork on to which replication occurs on each fork.
  3. The single stranded binding proteins cover the DNA strands preventing them from annealing into a double strand.
  4. The original strand is used as a template strand to synthesise to DNA strand in 5→3 direction with help of an extension formed by RNA primer.
  5. DNA Polymerase can synthesise the strand only in 5→3. Thus one strand in synthesised in short bursts, joined together later on RNase H and DNA polymerase recognise.
  6. RNA polymers that are bound to DNA template and removes the primers by RNA hydrolysis.
  7. The gaps formed due to hydrolysis of RNA are filled by DNA polymerase.
  8. The DNA pieces or nicks filled by deoxyribonucleotides and joined together by the enzyme ligase, thus completing the replication process.

Question 20.
Give a diagrammatic presentation of the process of transcription in a bacterial cell.
Answer:
AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance 7

Question 21.
Write briefly on nucleosomes [TS MAY-22] [AP MAR-19] [AP MAY-17]
Answer:
AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance 8

  1. Nucleosome is a bead like structure of chromosomes.
  2. It consists of eight histone molecules (histone octamer) and a DNA segment of about 150 base pairs.
  3. The negatively charged DNA is wrapped around the positively charged histone octamer to form the structure of the nucleosome.
  4. Nucleosome helps to fold DNA into a compact form.
  5. DNA and basic histone proteins constitute chromatin.
  6. Nucleosomes constitute the repeating unit of a structure in nucleus called chromatin.
  7. The nucleosomes in chromatin are seen as ‘beads-on-string’ when viewed under electron microscope.

Long Answer Questions

Question 1.
Given account Of the Hershey and Chase experiment What did it conclusively prove? If both DNA and proteins contained phosphorus and sulphur do you think the result would have been the same.
Answer:

  1. Alfred Hershey and Martha Chase (1952) worked with viruses that infect bacteria, bacteriophages to prove that DNA is the genetic material.
  2. Genetically bacteriophage attacks bacteria and manufactures more virus particles.
    AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance 9
  3. Hershey and Chase worked to discover whether it was protein or DNA from the viruses that entered the bacteria.
  4. They grew some viruses on a medium that contained radioactive phosphorous and some other on a medium that contained radio active sulphur.
  5. Viruses grown in presence of radioactive phosphorous contained radioactive DNA but not radioactive protein because DNA contains phosphorous but protein does not.
  6. Similarly viruses grown on radioactive sulphur contained radio active protein but not radioactive DNA because DNA doesnot contain sulphur.
  7. Bacteria which were infected with virus that had radio active DNA were radio active indicating that DNA was the material that passed from the virus to the bacteria.
  8. Bacteria that were infected with virus that had radio active proteins were not radioactive.
  9. Therefore DNA is the genetic material that is passed from virus to bacteria.
  10. If both DNA and proteins contained phosphorous and sulphur, the result would have been different from result given by Hershey and Chase.

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 2.
Give an account of post-transcriptional modification of a eukaryotic mRNA.
Answer:
The primary transcript is extensively modified after transcriptions in cell nucleus. This includes
1) 5 “Capping” this is the first of processing reactions for hnRNA,. The cap is a 7-methyl guanosine attached “backward” through a triphosphate linkage to 5 terminal end of the mRNA. Additional methylation of 7-methyl guanosine “Cap” through unusual 5’→ 5′ triphosphate linkage appears to facilitate the initiation of translation and helps to stabilise mRNA.

2) Addition of a Poly-A tail: Most mRNA have a chain of 40 to 200 adenine nucleotides attached to 3-end . This poly-A-tail is not transcribed from the DNA but it is added after transcription. These tails help to stabilize mRNAs and facilitate their exit from nucleus. After mRNA enters cytosol, poly-A tail is gradually shortened.

3) Splicing – Removal of inlrons: Maturation of eukaryotic mRNA may involve removal of RNA sequences that do not code for protein from primary transcript.
The remaining coding sequences the exons are spliced together to form mature mRNA.
After all these transcriptional modifications m-RNA is transported out of nucleus for translation.

Question 3.
Discuss the process of translation in detail.
Answer:
Translation: It is the processing of polymerisation of amino acids to form a polypeptide chain.
The triplet sequence of bases in mRNA designs the order and sequence of aminoacids in the polypeptide chain. This process involves three steps.
1) Initiation 2) Elongation 3) Termination

  1. During the initiation, tRNA gets charged when the aminoacid bind by using ATR
  2. The start codon AUG present on mRNA is recognised only by the charged tRNA.
  3. The ribosome acts as an actual site for the process of translation.
  4. It contains two separate sites in the larger subunit for the attachment of subsequent amino acid.
  5. The smaller subunit of ribosome binds to mRNA at the initiation codon (AUG) followed by the larger sub unit.
  6. During the elongation process, the ribosome moves one codon downstream along with mRNA so as to leave space for binding of another charged tRNA.
  7. The amino acid brought by tRNA gets linked w ith the previous amino acid through a peptide bond.
  8. This process results in formation of a polypeptide chain.
  9. When the ribosomes reach one of the stop codons (UAA, UAG and UGA), the process of translation gets terminated.
  10. The polypeptide chain is released and ribosomes gets detached from mRNA.

Question 4.
Write briefly about Griffith’s experiments on S. pneumoniae bacteria. What was his conclusion.?
Answer:

  1. Transformation: It is the genetic alteration of a cell resulting from the direct uptake and incorporation of genetic material from its surroundings through the cell membranes.
  2. In 1928, Fredrick Griffith conducted a series of experiments on a bacterium called streptococcus pneumoniae, which is responsible for pneumonia.
  3. During his experiment he found that the bacteria can change its physical form. He has grown the bacteria on a culture medium.
  4. He observed two strains of this bacterium, one forming smooth colonies with capsule (S-type) and the other forming rough colonies without capsule (R-type)
  5. The S-type cells are virulent while R-type cells are non-virulent.
  6. When live S-type cells are injected into mice, they suffered from pneumonia and died.
  7. When live R-type cells are injected into mice, the disease didnot appear and the mice survived.
  8. When heat killed S-type cells w ere injected, the disease did not appear.
  9. When heat killed S-type were mixed w ith live R-type and injected into mice, the mice died of pneumonia and live S-type cells w’ere isolated from the body of dead mice.
  10. He concluded that two R-strain cells have been transformed by the heat killed S-strain bacteria.
  11. This must be due to the transfer of genetic material by the transforming principle.
  12. Stability of genetic material is proved by Griffith’s transforming principle.
  13. Heat, which killed the bacteria, did not destroy the properties of the genetic material.
  14. Further, 2′-OH group present in every nucleotide in RNA is a reactive group.
  15. It makes RNA labile and easily degradable.
  16. So DNA is a better genetic material than RNA.
  17. If RNA is used in the experiment, the heat killed strain of pseudomonas could not transform the R-strain into virulent strain.

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 5.
Define an operon. giving an example. Explain w hat is an inducible operon.
Answer:

  1. It is a group of closely placed structural genes and regulatory elements (DNA sequences) such as a regulator, a promoter and an operator which functions in a co-ordinated manner- Qperon.
  2. Lac operon is a segment of DNA that is made up of three adjacent structural genes, namely an operator gene, a promoter gene and a regulator gene.
    AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance 10
  3. It works in a coordinated manner to metabolise lactose to glucose and galactose.
  4. In Lac Operon , lactose act as an inducer.
  5. It binds to the repressor and inactivates it once the lactose binds to the repressor , RNA polymerise binds to the promotor region.
  6. Hence three structural genes express their product and respective enzymes are produced .
  7. These enzymes act on lactose or metabolise it into glucose and galactose .After some time, when the level of inducer decreases, it causes the synthesis of the repressor from regulator gene.
  8. The repressor binds to the operator gene and prevents RNA polymerase from transcribing the operon Hence transcription is stopped.
  9. Repressoi binds to the operator region (o) and prevents RNA polymerase from transcribing the operon.

Question 6.
Give the salient features of double helix structure of DNA.
Answer:
The salient features of Double-helix structure of DNA proposed by Watson & Crick

  1. It is made of two poly nucleotide chains where the backbone is sugar phosphate, and the bases project inside. Nucleotide is deoxyribose sugar, nitrogen base and a phosphate molecule.
  2. The two chains have anti-parallel polarity. This means that if one chain has the polarity 5’→3′, the other has 3’→5′
    AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance 11
  3. The bases in two strands are paired through hydrogen bonds forming base pairs. Adenine forms two hydrogen bonds with Thymine.
    Guanine forms three hydrogen bonds with Cytosine. This generates approximately uniform distance 20A° between two strands of helix.
    (Purines- Adenine & Guanine Pyramidines – Thymine & Cytosine, Always a purine pairs with a pyramidine) –
  4. The two chains are coiled in a right handed fashion. The pitch of helix is 3.4nm and there are roughly lObp in each turn. So the distance between two successive base pairs is equal to 0.34nm.
  5. The plane of one base pair stacks over the other in a double helix. This confers stability of helical structure in addition to hvdrouen bonds.

Question 7.
Replication was allowed to take place in presence of radioactive Deo-oxy nucleotide precursors in EXolil that was a mutant for DNA ligase. Explain how the newly synthesised radioactive DNA will be when purified?
Answer:

  1. In the long DNA molecule, the replication occurs within a small opening of DNA helix, called replication fork.
  2. On one strand (the template with polarity 3’—5′) the replication is continuous, but on the other hand, the template with other polarity, it would be discontinuous .
  3. The discontinuously synthesized fragments will be are generally joined by DNA ligase.
  4. But E.Coli is mutant for DNA ligase, these Okazaki fragments will not be joined on lagging strand.
  5. When the strand of DNA is denatured and purified by Density Gradient Centrifugation, both high molecular fragments (only leading strand) and low molecular weight fragments (on lagging strand) are formed .
  6. The radioactivity will be obtained in two zones high molecular zone and low molecular zone.

Exercise

Question 1.
Croup the following as nitrogeneous bases and nucleosides: Adenine, Cytidine, Thy none, Cuanosine, IJraeil and Cytosine.
Answer:
Nitrogen bases are – adenine, thymine, uracil and cytocine
Nucleosides are -cytidine and Guanosine

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 2.
If a double stranded DINA has 20 percent of cytosine, calculate (he percent of adenine in the DINA
Answer:

  1. Adenine and Guanine are the purines.
  2. Thymine and Cytocine are the Pyrimidines.
  3. Adenine is bound to Thymine by two hydrogen bonds (A=T)
  4. Guanine is bond to Cytosine by three hydrogen bonds (G=C)
  5. Erwin Chargaff discovered that purines (A,G) and Pyrimidines (T,C) exist in 1:1 ratio.
  6. The base ratio may vary from species to species, but it is constant for a given species.
  7. According to this law if a double stranded DNA has 20% of cytocine, it would have 20% of Guanine. G+C= 40%
  8. The remaining 60% represents both A+T molecule.
  9. Since ‘A’ and T are in equal proportion, the percentage of Adenine is 30% and the percentage of Thymine is 30%.

Question 3.
If the sequence of one strand of DNA is written as follows:
Write down the sequence of the complementary strand in 3′-5’ direction.
5′-ATGC ATGC ATGC ATGC ATGC ATGC ATG C-3′
Answer:
1) The sequence of one strand of DNA is
5′- ATGC ATGC ATGC ATGC ATGC ATGC ATGC -3′

2) The sequence of its complementary strand is
3′- TACG TACG TACG TACG TACG TACG TACG -5’

Question 4.
If the sequence of the coding strand in a transcription unit is written as follows:
5′ – ATG C AT G C AT G C ATG C ATG CAT G C AT G C-3 ‘
Answer:
The sequence of the coding strand is a transcription unit is
3′-TACG TACG TACG TACG TACG TACG TACG -5’

The sequence of mRNA is
5’- AUGC AUGC AUGC AUGC AUGC AUGC AUGC -3’

In mRNA Thymine is replaced by Uracil.

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesize a semi¬conservative mode of DNA replication? Explain.
Answer:

  1. Watson and Crick proposed double-helical structure for DNA.
  2. The two strands of DNA are antiparallel and complementary to each other with respect to their base sequences.
    AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance 12
  3. This type of arrangement has led to the hypothesis that DNA replication in semiconservative.
  4. The scheme suggested that the two strands would separate and act as a template for the synthesis of new complementary strands.
  5. After the completion of replication, each DNA mole would have one parental and one newly synthesized strand.
  6. This scheme was termed as semiconservative DNA replication. semiconservative DNA replication

Question 6.
Depending upon the chcmnical nature of the template (DNA or RNA) and the nature of nucleic acids synthesized from it (DNA or RNA), list the types of nucleic acid polymerases.
Answer:
There are two types of polymerases.
a) DNA-dependent DNA Polymerase
b) DNA- dependent RNA Polymerase

  • The DNA dependent DNA polymerases uses a DNA template for synthesizing a new strand of DNA.
  • The DNA dependent RNA molymerases use a DNA template strand for synthesizing RNA.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiement while proving that DNA is the genetic material? ,
Answer:

  1. Hershey and chase grew some viruses on a medium containing radio active phosphorns and some others on a medium containing radio active sulfur.
  2. Viruses grown in radio active phosphorus contained radioactive DNA but not radio active protein.
  3. Because DNA contains phosphorus but protein doesnot contain phosphorus.
  4. Similarly viruses grown on radio active sulfur contain radio active protein sheath but not radio active DNA.
  5. Because DNA does not contain sulfur.

Question 8.
Differentiate between the followings:
a) mRNA and tRNA
b) Template strand and Coding strand
Answer:
a) Differences between mRNA and tRNA:

mRNA tRNA
1) It is a single-stranded unfolded polynucleotide 1 mole. 1) It is also a single-stranded but folded like a clover leaf.
2) It carries genetic information for the synthesis of a specific protein from DNA to ribosome in the form of triplet codons. 2) It has three distinct loops and a fourth indistinct loop which is considered as accessory arm along with a tail.

b) During the process of transcription, the DMA at the promotor site gives two single strands.

One strand is called as the template strand which is transcribed.

  • Where as the other strand is called coding strand which does not code for anything and is not transcribed.

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 9.
List two essential rolessof ribosome during translation.
Answer:

  1. Ribosome is the cellular factory responsible for the synthesis of proteins.
  2. The ribosome consists of structural RNAs and about 80 different proteins.
  3. It exists as two sub units, a large sub unit and a small sub unit.
  4. When small sub unit encounters an mRNA, the process of translation of mRNA to protein begins.
  5. There are two sites on large sub unit for the subsequent amino acids to bind and form peptide bond.
  6. The ribosome also acts as a catalyst for the formation of a peptide bond.

Question 10.
In the medium where E.coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
Answer:

  1. Lactose is a substance which acts as a inducer for the lac operon.
  2. The lactose binds to the repressor protein.
  3. As a result the repressor becomes inactive and cannot bind to the operator.
  4. Nil; allows RNA polymerase to access the promotor and transcription proceeds
  5. Production of enzymes continue and metabolism of lactose takes place.
  6. If the substance lactose is absent, repressor becomes active, binds the operator and prevents RNA polymerase from transcribing the operon and switches off the process.

Question 11.
Explain the function of (a) Promoter (b) tRNA (c) Exons
Answer:
a) Promotor: It is a region of DNA where RNA polymerase binds and initiates transcription.

b) tRNAnswer: Is an adaptor mole, that is used by all living organisms to bridge the genetic code in mRNA with the twenty amino acids in proteins, t RNAs carry amino acids to ribosomes, where they are linked into proteins.

c) Exons: In Eukaryotes, the monocistronic structural genes have interrupted coding sequences – the genes in eukaryotes are split. The coding sequences are called exons. The exons are interrupted by introns.

Question 12.
Briefly describe the following: (a) Transcription (b) Translation.
Answer:
a) Transcription: The formation of m-RNA from DNA is called transcription. It occurs in the nucleus.

b) Translation: The synthesis of polypeptide chain by binding specific amino acids in a sequence based on the codons of mRNA is called “translation”. It takes place on ribosomes.

Question 13.
How the polymerization of nucleotides can be prevented in a DNA molecule.
Answer:

  1. DNA polymerases on their own cannot initiate the process of replication.
  2. By removing primer DNA template the polymerisation can be prevented.
  3. In the absence of DNA polymerase the polymerisation can be prevented.

Question 14.
In an experiment, DNA is treated with a compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive base pairs increases from 0.34nm to 0.44 nm. Calculate the length of DNA double helix (which has 2xl09 bp) in the presence of saturating amount of this compound.
Answer:

  1. Distance between 2 consecutive base pairs is 0.44 nm (0.44 x 10-9nm)
  2. The length of DNA double,helix = 6.6×2 x 109bp x 0.44 x 10-9/bp=5.8 meters.

Question 15.
Recall the experiments done by Frederick Griffith. Where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat killed strain of Pneumococcus have transformed the R-strain into virulent strain? Explain.
Answer:

  1. Stability of genetic material is proved by Griffith’s transforming principle.
  2. Heat, which killed the bacteria, did not destroy the properties of the genetic material.
  3. Further, 2′-OH group present in every nucleotide in RNA is a reactive group.
  4. It makes RNA labile and easily degradable.
  5. So DNA is a better genetic material than RNA.
  6. If RNA is used in the experiment, the heat killed strain of pseudomonas could not transform the R-strain into virulent strain.

Question 16.
Your are repeating the Hershey-Chase experiment and are provided with two isotopes;
32P and 15N (in place of 35S in the original experiment). How do you expect your results to be different?
Answer:

  1. In the repeated experiment of Hershey-Chase, they have provided two isotopes of 32P and 15N in the place of 35S of original experiment.
  2. The experiment is to prove whether it is the protein or DNA that has entered the bacteria from virus.
  3. Finally it is proved that we cannot distinguish the difference between protein or DNA because 32P and 15N are present in both DNA and protein of bacteria which infected with virus.

Question 17.
Do you think that the alternate splicing of exons may enable a structural gene to code for several isoproteins from one and the same gene? If yes, how? If not, why so?
Answer:

  1. Yes, In Eukaryotes the genes are called split genes.
  2. Split gene contain two sections called exons and introns..
  3. Exons are the coding sequences of the gene.
  4. Introns are the non-coding or intervening sequences.
  5. Before translation, the introns are to be removed and extrons are reattached.
  6. Thus a copy of mRNA with continuous codon is produced.
  7. This process is called gene splicing.
  8. Each-gene will separately code for one protein and that genes donot overlap.
  9. So alternate splicing of exons may enable a structural gene to code for several isoproteins from one and the same gene.

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 18.
Can you recall what centrifugal force is, and think why a molecule with higher mass/ density would sediment faster?
Answer:

  1. Centrifugal force is the force that acts away from the centre of the circle.
  2. Higher mass/ density will be thrown towards out side because heavier particles experience more centrifugal force.
  3. So they would sedement faster.

Question 19.
Do Retroviruses follow central Dogma?, (live one example.
Answer:

  1. Francis Crick proposed the central dogma in molecular biology.
  2. It states that genetic information flows from DNA → RNA → Protein
  3. AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance 13
  4. Retro viruses doesnot follow central dogma.
  5. Retro viruses contain RNA as genetic material.
  6. In that genetic material information flows in the reverse direction, that is from RNA to DNA. Eg: HIV- which belong to family retroviridae.

Question 20.
If there are 2.9 x 10-9 complete turns in a DNA molecule estimate the length of the molecule. (I angstrom = I0~* cm)
Answer:
Distance between two consecutive base pairs is 0.34 nm (0.34 x 10-9nm)
Complete turns in DNA = 2.9 x 10-9
Length of DNA = Total number of turns x distance between two consecutive base pairs = 2.9 x 109 x 0.34 x 10-9=0.986m.

Multiple Choice Questions

Question 1.
The individual functional monomers of nucleic acids are
1. Nucleosides
2. Nitrogen bases
3. Nucleotides
4. Pentose monosaccharides
Answer:
3. Nucleotides

Question 2.
The most common genetic nucleic acid in majority of living organisms is
1. DNA
2. RNA
3. mRNA
4. tRNA
Answer:
1. DNA

Question 3.
The monomers present in the nucleic acid polymers are
1. Pentose sugars
2. Nucleosides
3. Nucleotides
4. heterocyclic molecules
Answer:
3. Nucleotides

Question 4.
The most abundant genetic material is
1. RNA
2. DNA
3. mRNA
4. rRNA
Answer:
2. DNA

Question 5.
DNA molecule is
1. Positively charged
2. Negatively charged
3. Exhibits positive charge during denaturation only
4. Does not show any charge
Answer:
2. Negatively charged

Question 6.
The alternate name for Thymine is
1. Thymidine
2. Uridine
3. 5-methyl uracil
4. Thymedylic acid
Answer:
3. 5-methyl uracil

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 7.
DNA acts as a template for the synthesis of
1. RNA only
2. DNA only
3. Both DNA and RNA
4. Proteins only
Answer:
3. Both DNA and RNA

Question 8.
Doubling of DNA generally occurs in
1. Mitosis
2. Interkinesis
3. Meiosis
4. Interphase
Answer:
4. Interphase

Question 9.
RNA does not differ from DNA in
1. Number of polynucleotide strands
2. Type of Pyrimidines
3. Type of Purines
4. Type of Sugar
Answer:
3. Type of Purines

Question 10.
Each sugar molecule in DNA is linked to
1. 2 Nitrogen bases and 1 phosphate
2. 2 Phosphates and 2 Nitrogen bases
3. 1 phosphate and 1 Nitrogen base ;
4. 1 phosphate and 2 Nitrogen bases
Answer:
3. 1 phosphate and 1 Nitrogen base ;

Question 11.
DNA is absent in
1. Chloroplasts
2. Mitochondria
3. Plasmids
4. Ribosomes
Answer:
4. Ribosomes

Question 12.
The radio active isotope used to trace out nucleic acid metabolism is
1. 35C
2. 32P
3. 14C
4. 18O
Answer:
2. 32P

Question 13.
Denaturation of DNA is applicable to
1. mRNA
2. DNA o’f ψ x 174 – Bacteriophage
3. RNA of TMV
4. DNA of Bacteriophage Lambda(X)
Answer:
4. DNA of Bacteriophage Lambda(X)

Question 14.
A correct statement with reference to the property of DNA is
1. It should be able to replicate
2. It should evolve fastly
3. It should be biochemically unstable
4. It should be able to mutate quickly
Answer:
1. It should be able to replicate

Question 15.
In Eukaryotes, a single mRNA is synthesised from
1. Two structural genes
2. Many structural genes
3. A single structural gene
4. Three structural genes
Answer:
3. A single structural gene

Question 16.
The time taken to complete the process of DNA replication in E.Coli is
1. 38 minutes
2. 80 Minutes
3. 40 minutes
4. 20 minutes
Answer:
1. 38 minutes

Question 17.
The length of DNA double helix in typical human cell is
1. 3.2 meters
2. 4.2 meters
3. 2.2 meters
4. 1.2 meters
Answer:
3. 2.2 meters

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 18.
DNA strands are antiparallel because of
1. Presence of disulphide bonds
2. Presence of phosphodiester bonds
3. H-bonds
4. Peptide bonds
Answer:
3. H-bonds

Question 19.
One turn of the helix in a (3-form DNA is approximately
1. 2.0 nm
2. 0.34 nm
3. 3.4nm
4. 20 nm
Answer:
3. 3.4nm

Question 20.
The process of transfer of genetic information from one strand of DNA to RNA is called
1. Translation
2. Transcription
3. Transformation
4. Polyadenylation
Answer:
2. Transcription

Question 21.
The strand of DNA that is involved in transcription is called
1. Template strand
2. Coding strand
3. Promoter strand
4. Terminator strands
Answer:
1. Template strand

Question 22.
The strand of DNA that is not involved in RNA synthesis is called
1. Antisence strand
2. Code complement
3. Coding-strand
4. Template strand
Answer:
3. Coding-strand

Question 23.
In bacteria, both transcription and translation occur respectively in
1. Nucleus and cytoplasm
2. Nucleoid and nucleus
3. Cytoplasm and cytoplasm
4. Cytoplasm and genophore ;
Answer:
3. Cytoplasm and cytoplasm

Question 24.
Identify the mismatch
1. Transcription – DNA dependent DNA polymerase
2. rRNA – catalytic role during translation
3. tRNA – Genetic code reader
4. mRNA – template for protein synthesis
Answer:
1. Transcription – DNA dependent DNA polymerase

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 25.
The primary transcript is longer than the functional hnRNA in
1. Prokaryotes
2. Eukaryotes
3. Prokaryotic animals
4. Cyanobacteria
Answer:
2. Eukaryotes

Question 26.
Transcription of snRNA is catalysed by
1) Prochromatin
2) Euchromatin
3) Heterochromatin
4) Chromatogram
Answer:
2) Euchromatin

Question 27.
Introns are present in
1. Prokaryotes only
2. Eukaryotes only
3. Both 1 & 2
4. Bacteria only
Answer:
2. Eukaryotes only

Question 28.
Exons are present in
1. only in Eukaryotes
2. both in prokaryotes and eukaryotes
3. only in prokaryotes
4. only in eukaryotic plants
Answer:
2. both in prokaryotes and eukaryotes

Question 29.
Split genes are found in
1. Eukaryotes
2. Bacteria
3. Cyanobacteria
4. All Monerans
Answer:
1. Eukaryotes

Question 30.
Genes are chemically composed of
1. Histone proteins
2. Glycoproteins
3. Lipoproteins
4. Polynucleotides
Answer:
4. Polynucleotides

Question 31.
The synthesis of ’repressor mRNA’ is carried out by
1. ’i’ gene
2. ’O’ gene
3. ‘P’ gene
4. ‘Z’ gene
Answer:
1. ’i’ gene

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 32.
Number of nucleotides in the DNA of Bacteriophage lamda.
1) 48502
2) 48520
3) 97004
4) 97040
Answer:
1) 48502

Question 33.
Transcriptionlly active chromatin is called
1) Prochromatin
2) Euchromatin
3) Heterochromatin
4) Chromatogram
Answer:
2) Euchromatin

Question 34.
DNA differs from RNA in
A) Sugar
B) Phosphate
C) Purines
D) Pyramidines
1) ABCD
2) ABC
3) A & C
4) A & D
Answer:
4) A & D

Question 35.
Nucleotide differs from Nucleoside in
1) Having phosphate
2) Lacking 2 N base
3) Having sugar
4) Lacking phosphate
Answer:
1) Having phosphate

Question 36.
The first proof that DNA is the genetic material was provided through
1) Bacterial transformation
2) Transfection
3) Bacterial translocation
4) Phage infection
Answer:
4) Phage infection

Question 37.
Griffith conducted his transformation experiments on
1) Escherichia coli
2) Streptococcus pyogenes
3) Streptococcus pneumoniae
4) Streptomyces griseus
Answer:
3) Streptococcus pneumoniae

Question 38.
O.T Avery, Mae Leod and MC Carty identified ……. the genetic material active factor/transformation principle
1) DNA
2) RNA
3) Polysaccharide
4) Proteins
Answer:
1) DNA

Question 39.
Each histone core is wrapped by number of turns of DNA are
1) 2
2) 3
3) 4
4) 5
Answer:
1) 2

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 40.
Uracil differ from Thymine in
1) Having methyl group
2) Lacking methyl group
3) Having NH2
4) Having two C-N rings
Answer:
2) Lacking methyl group

Question 41.
The haploid content of human DNA is
1) 5386 nucleotides
2) 48502 bp
3) 4.6 x 106bp
4) 3.3 x 109bp
Answer:
4) 3.3 x 109bp

Question 42.
A nucleoside is
1) Base + Sugar
2) Base + Phosphate
3) Sugar + Phosphate
4) Base + Sugar +Phosphate
Answer:
1) Base + Sugar

Question 43.
A false statement is w.r.t. properties of genetic material
1) Self replication
2) chemical and structural stability
3) Fast mutation
4) Ability of self expression
Answer:
3) Fast mutation

Question 44.
Which of the following is not a character of RNA.
1) RNA is unstable and degradable
2) RNA mutates at faster rate than DNA
3) RNA evolves slowly
4) RNA is catalytic / reactive
Answer:
3) RNA evolves slowly

Question 45.
In their side by side experiments Hershy and Chase incorporated 35 S into
1) Bacteria DNA
2) Protein capsid of phage
3) Cytosol of bacteria
4) Phage DNA
Answer:
2) Protein capsid of phage

Question 46.
A group of closely placed structural genes and regulatory genes like regulator promoter and operator which function together in a coordinated manner is called
1) Regulation
2) Genetic code
3) Operon
4) Monocistronic gene
Answer:
3) Operon

Question 47.
Heterogenous RNA is converted to mRNA by
1) Splicing
2) Capping
3) Tailing
4) All the above
Answer:
4) All the above

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 48.
The number of effective codons for the synthesis of twenty amino acids is
1) 64
2) 32
3) 60
4) 61
Answer:
4) 61

Question 49.
Identify the correct statement
1) UGU codes for tyrosine
2) ACG codes for Aspargin
3) CGA codes for serine
4) GUA codes for valine
Answer:
4) GUA codes for valine

Question 50.
Which is not applicable to genetic code?
1) It is non ambiguous
2) It is non-contiguous
3) It is nearly universal
4) It is degenerate
Answer:
3) It is nearly universal

Question 51.
The functional unit of gene is
1) Intron
2) Recon
3) Muton
4) Cistron
Answer:
4) Cistron

Question 52.
Processed m RNA contains
1) Exons and Introns
2) Introns
3) Exons
4) Capping at 3 ’ end
Answer:
3) Exons

Question 53.
The RNA that plays structural and catalyst role during translation
1) m RNA
2) tRNA
3) rRNA
4) s RNA
Answer:
3) rRNA

Question 54.
Ochoa’s enzyme is
1) DNAligase
2) Peptidyltransferase
3) DNA polymerase
4) Polynucleotide phosphorylase
Answer:
3) DNA polymerase

Question 55.
The gene which synthesizes repressor is
1) Operator gene
2) Regulator gene
3) Promoter gene
4) Structural gene
Answer:
4) Structural gene

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 56.
Which is target for the attachment of repressor protein produced by the regulator gene?
1) Promoter
2) Inducer
3) Operator
4) Structural gene
Answer:
3) Operator

Question 57.
The rule that ‘in a DNA the amount of adenine equals to that of the thymine and that guanime equals to that of cytosine’, is known as
1) Allen’s rule
2) Jordan’s rule
3) Chargaff’s rule
4) Stahl’s rule
Answer:
3) Chargaff’s rule

Question 58.
Purines are
1) Dicyclic and heterocyclic
2) Dicyclic and homocyclic
3) Monocyclic
4) Homocyclic
Answer:
1) Dicyclic and heterocyclic

Question 59.
A phosphate group is linked to 5i-0H of a nucleotide through
1) Phosphoester linkage
2) Glycosidic linkage
3) Hydrozen linkage
4) Both (1) & (2)
Answer:
1) Phosphoester linkage

Question 60.
The inducer of Lac operon binds to
1) Repressor gene
2) Repressor protein
3) Operator gene
4) Promoter gene
Answer:
2) Repressor protein

AP Inter 2nd Year Botany Important Questions Chapter 10 Molecular Basis of Inheritance

Question 61.
Hershey & Chase experiment conclusively proved that
1) DNA is genetic material of many organisms
2) RNA is genetic material of all organisms
3) DNA or RNA are genetic material
4) DNA is genetic material of Bacteriophage
Answer:
1) DNA is genetic material of many organisms

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