AP Inter 2nd Year Botany Study Material Chapter 10 Molecular Basis of Inheritance

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 10th Lesson Molecular Basis of Inheritance Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 10th Lesson Molecular Basis of Inheritance

Very Short Answer Questions

Question 1.
Distinguish between Heterochromatin and Euchromatin. Which of the two is transcriptionally active?
Answer:
The chromatin that is more densely packed and stains dark is called heterochromatin. The chromatin that is loosely packed and stain light is called Euchromatin. Euchromatin is transcriptionally active chromatin.

Question 2.
Who proved that DNA is Genetic Material? What is the organism they worked on?
Answer:
Alfred Hershey and Martha chase (1952). They worked with viruses that infect Bacteria, bacteriophages.

Question 3.
What is the function of DNA polymerase?
Answer:
DNA polymerise is a highly efficient enzyme which catalyse polymerisation of a large number of nucleotides in a very short time. It also catalyse the reaction with a high degree of accuracy.

Question 4.
What are the components of a nucleotide?
Answer:
A nucleotide has three components – a nitrogenous base, a pentose sugar and a phosphate group’.

Question 5.
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) 5’TTACGTCGATAATCC-3′
b) 5’ AAUGCAGCUAUUAGG – 3’

Question 6.
Name any three viruses which have RNA as the Genetic Material.
Answer:
Tobacco mosaic virus, QB bacteriophage, HIV.

Question 7.
What are the components of a transcription unit?
Answer:
a) A promotor b) The structural Gene c) A terminator.

Question 8.
What is the difference between exons and Introns?
Answer:

Exons	Introns
1) Coding or expressed sequences.	1) Intervening sequences.
2) They appear in nature or processed RNA.	2) They do not appear in mature or processed RNA.

Question 9.
What is meant by capping and tailing?
Answer:
Adding of an unusual nucleotide (methyl guanosine triphosphate) to the 5′ -end of heterogenous nuclear RNA [hnRNA) is called capping.

Adding of adenylate residues to the 3’ – end in a template independent manner is called tailing.

Question 10.
What is meant by point mutation? Give an example.
Answer:
Change of single base pair in the gene for betaglobin chain that results in the change of aminoacid residue glutamate to valine. It results in a diseased condition called sickle cell anaemia.

Question 11.
What is meant by charging of tRNA?
Answer:
Activation of aminoacids in the presence of ATP and linked to their cognate tRNA is known as charging of tRNA or amino acylation of tRNA.

Question 12.
What is the function of the codon AUG?
Answer:
It has dual functions. It codes for methionine and also act as the initiator codon.

Question 13.
Define stop codon. Write the codons.
Answer:
Codons which do not code for any aminoacids are called stop codons. They are UAA, UAG, UGA.

Question 14.
What is the difference between the template strand and a coding strand in a DNA molecule?
Answer:
The two strands have opposite polarity and the DNA-dependant RNA polymerise also catalyses the polymerisation in only one direction that is, 5′ → 3′, the strand that has the polarity 3′ → 5′ acts as a template and is also called template strand. The other strand which has polarity 5′ → 3′ and does not code for anything is called coding strand.

Question 15.
Write any two differences between DNA and RNA.
Answer:

DNA	RNA
1) Nitrogen bases are Adenine, Guanine Thymine and cytosine.	1) Nitrogen bases ate Adenine, Guanine, Uracil and Cytosine.
2) Deoxyribose sugar is present.	2) Ribose sugar is present.

Question 16.
In a typical DNA molecule the proportion of thymine is 30% of the N bases. Find out the percentages of other N bases.
Answer:
Adenine = 30%
Guanine = 20%
Cytosine = 20%

Question 17.
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.
Answer:
RNA. Only one strand is present.

Short Answer Questions

Question 1.
Define transformation in Griffith’s Experiment. Discuss how it helps in the identifi¬cation of DNA as genetic material.
Answer:
Frederick Griffith (1928) conducted experiments on streptococcus pneumoniae and observed a transformation in bacteria. When streptococcus were grown on a culture plate, some produced smooth shiny colonies (s) while others produced rough colonies (R). Mice injected with ‘s’ shain (mucous coat) die from pneumonia infection but mice injected with R strain do not develop pneumonia.

He injected heat killed ‘s’ strain bacteria to mice, It is healthy. Finally he injected heat killed S and R strains, the mice died. He concluded that the R strain bacteria had been transformed by heat killed ‘s’ strain bacteria. Some transforming principle transferred from heat killed strain to R strain to synthesize a mucous coat and become virulent. This is due to the transfer of genetic material.

Question 2.
Discuss the significance of heavy isotope of Nitrogen in Meselson and Stahl’s experiment.
Answer:
Matthew Meselson and Franklin Stahl, grow E.Coli in a medium containing ¹⁵NH₄Cl and observed 15_N was incorporated in newly synthesized DNA. This heavy DNA molecule could be distinguished from themormal DNA by centrifugation in a cesium chloride density gradient.

They transformed the cells into a medium with ¹⁵NH₄Cl (normal) and took samples at various time intervals and extracted the DNA that remained as double stranded helices. The various samples were separated independently on cesium chloride (cscl) gradients to measure the densities of DNA. Thus the DNA that was extracted from the culture, one generation after the transfer from ¹⁵N to ¹⁴N medium had a hybrid density. DNA extracted from the culture after another generation (40 mts) was composed of equal amounts of this hybrid DNA and of ‘Light’ DNA.

Question 3.
A single base mutation in a gene may not always result in loss or gain of function. Do you think the statement is correct? Defined your answer.
Answer:
A single base mutation in a gene may result in loss or gain of a gene and so a function.
E.g.: A change of single base pair in the gene for betaglobin chain that results in the change of aminoacid residue glutamate to valine. It results in a diseased condition called sickle cell anaemia.

E.g.: 2) Consider a statement that is made like a genetic code is – RAM HAS RED CAP.
If we remove a letter ‘S’ in HAS, it will be RAM HAR EDC AP.

Question 4.
How many types of RNA polymerases exist in cells ? Write their names and functions.
Answer:
Three RNA polymerases exist in Cells. They are :

1. RNA polymerase I – It transcribes r RNAs (28S, 18S, and 5.8S)
2. RNA polymerase II – It transcribes the precursor of RNA, the heterogeneous nuclear RNA (hn RNA)
3. RNA polymerase III – It is responsible for transcription of tRNA 5 Sr RNA and Sn RNAs (small nuclear RNAs)

Question 5.
What are the contributions of George Gamow, H.G. Khorana, Marshall Nirenberg in deciphering the genetic code?
Answer:
George Ganow, suggested that, in order to code for all the 20 Amino’ acids, the code should be made up of three nucleotides. This was a very bold proposition, because a permutation and combination of 4³ would generate 64 codons, generating more codons than required.

H.G. Khorana developed chemical method in synthesising RNA molecules with defined combinations of bases.

Marshall Nirenberg’s cell-free system for protein synthesis finally helped the code to be deciphered. The enzyme polynucleotide phosphorylase was also helpful in polymerising RNA with defined sequences in a template independent manner (enzymatic synthesis of RNA)

Question 6.
On the diagram of the secondary structure of tRNA shown below label the location
of the following parts.
Answer:

a) Anticodon
b) Acceptor stem
c) Anticodon stem
d) 5′ end
e) 3′ end

Question 7.
Draw the schematic / diagrammatic presentation of the lac operon.
Answer:

Question 8.
What are the differences between DNA and RNA.
Answer:

DNA	RNA
1. It contsists of two strands of nucleotides.	1. It consists on only strand of nucleotides.
2. It is present more in nucleus and very little in chloroplasts and mitochondria.	2. It is present more incytoplasm and little in nucleus.
3. Deoxyribose sugar (C5H10O4) is present.	3. Ribose sugar (C5H10O5) is present.
4. Thymine and cytosine are pyrimidines.	4. Uracil and cytosine are pyrimidines.
5. DNA is made up of 4 millions nucleotides.	5. RNA is made up of 75 – 2000 nucleotides.
6. It undergoes self-replication.	6. It does not undergo self replication except in RNA viruses.
7. DNA is the genetic material.	7. It is non-genetic material (except in RNA) viruses.
8. It does not participate directly in protein synthesis.	8. RNA participate directly in protein synthesis.
9. Metabolically DNA is of one type.	9. Metabolically RNA is of three types.
10. The base puring is A = T and G ≡ C.	10. The base puring is A = U and G = C.

Question 9.
Write the important features of Genetic code?
Answer:

1. The codon is triplet. 61 codons code for aminoacids and 3 codons donot code for any aminoacids called stop codons.
2. One codon codes for only one aminoacid, hence it is unambiguous and specific.
3. Some aminoacids are coded by more than one codon, hence the code is degenerated.
4. The codon is read in mRNA in a contiguous fashion. There are no punctuations.
5. The code is nearly universal: For Ex: UUU code for phenylalanine (Phe) in Bacteria or Humans.
6. AUG has dual functions. It codes for methionine and also acts as initiator codon.

Question 10.
Write briefly on nucleosomes.
Answer:
Nucleosome is a bead like structure of chromosomes. It consists of eight histone molecules and a DNA segment of about 150 base pairs. Each Nucleosome is separated from one another by a linker DNA sequence of about 50 base pairs. Nucleosome helps to fold DNA into a compact form in the interphase nucleus. Otherwise the length of a chromosome, when linear is many orders of magnitude greater than the diameter of the nucleus.

Intext Questions

Question 1.
Group the following as nitrogenous bases and nucleosides : Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:
Nitrogenous bases : Adenine, Thymine, Uracil, Cytosine,
Nucleosides : Cytidine, Guanosine.

Question 2.
If a double stranded DNA has 20% of cytosine, calculate the percent of adenine in the DNA.
Answer:
Cytosine = 20%
Guamine = 20%
Adenine = 30%
Thymine = 30%

Question 3.
If the sequence of one strand of DNA is written as follows : Write down the sequence of complementary strand in 3′ → 5′ direction.
5′ – ATGCATGCATGCATGCATGCATGCATGC – 3‘
Answer:
3′ – TACGTACGTACGTACGTACGTACGTACG – 5′

Question 4.
If the sequence of the coding strand in a transcription unit is written as follows.
5 – ATGCATGCATGCATGCATGCATGCATGC – 3 write down the sequence of m RNA
Answer:
3′ AUGC AUGC AUGC AUGC AUGC AUGC AUGC – 5

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise semi conservative mode of DNA replication? Explain.
Answer:
The two strands would separate and acts as a template for the synthesis of new complementary strands. After completion of replication, each DNA would have one parental and one newly synthesised strand.

Question 6.
Depending upon the chemical 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:
DNA polymerases
RNA polymerases.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer:
Hershey and chase purified biochemicals (proteins, DNA, RNA etc) from the heat killed ‘S’ cells. They discovered the DNA alone from S bacteria caused R bacteria to become transformed. They also discovered thal protein digesting enzymes and RNA digesting enzymes did not affect transformation. So the transforming principle was not a protein or RNA. Digestron with DNAse did inhibit transformation suggesting that the DNA caused the transformation.

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

a)

RNA	tRNA
1) It contains more nucleotides.	1) It contains lesser nucleotides.
2) It moves important information from the DNA to ribosome.	2) It transports aminoacids into a growing protein chain.

b)

Template strand	Coding strand
1) It is complementary strand it serves as the template for making the coding strand.	1) It contains coding genes. It is to be transcribed that is the side make ‘sense’.
2) It runs from 3’to 5′.	2) It runs from 5′ to 3′.

Question 9.
List two essential roles of ribosomes during translation.
Answer:

1. Ribosome acts as the site where protein synthesis takes place from individual aminoacids.
2. Ribosome acts a catalyst for forming peptide bond.
   E.g. : 23 S r-RNA in bacteria acts as ribozyme.

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:
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. It works in a coordinated manner to metabolise lactose into glucose and galactose. In Lac operon lactose acts as an inducer. It binds to the repressor and inactivates it. Once the lactose binds to the repressor, RNA polymerase binds to the promotor region.

Hence, three structural genes express their product and respective enzymes are produced. These enzymes act on lactose or metabolise it into glucose and galactose. After sometime, when the level of the inducer decreases, it causes the synthesis of the repressor from regulator gene. The repressor binds to the operator gene and prevents RNA polymerase from transcribing the operon. Hence the transcription is stopped.

Question 11.
Explain (in one or two lines) the functions of the followings :
a) Promoter b) tRNA c) Exons.
Answer:
a) Promoter:
It is a region of DNA that helps in initiating the process of transcription. It serves as the binding site for RNA polymerase. :

b) tRNA :
It is a small RNA that reads the genetic code present on mRNA it carries specific aminoacids to mRNA on ribosome during translation of proteins.

c) Exons:
Exons are coding sequences of DNA in Eukaryotes that transcribe for proteins.

Question 12.
Briefly describe the following :
a) Transcription b) Translation.
Answer:
a) Transcription:
It is the process of synthesis of RNA from DNA template. A segment of DNA gets copied into mRNA during the process. The process of transcription starts at the promotor region of the template DNA and terminates at the terminator region The segment of DNA between these two regions is known as transciption unit. The transcription requires RNA polymerase enzyme, a DNA template, four types of nucleotides, and certain cofactors such as Mg²⁺

During transcription, three events occur. They are : a)Initiations b)Elongation, c) Termination. The DNA dependant RNA polymerase and certain initiation factors bind at the double stranded DNA at the promotor region of the template strand and initiate the process of transcription, RNA polymerase moves along the DNA and leads to the unwinding of DNA duplex in two separate strands.

Then one of the strands, called sense strand acts as template for mRNA synthesis. The epzyme RNA polymerase utilises nucleoside triphosphates as raw material and polymerizes them to form m RNA according to the complimentary bases present on the template DNA. This process of opening of helix and elongation of polynucleotide chain continuous until the enzyme reaches terminator region. As RNA polymerase reaches the terminator region, the newly synthesized mRNA transcripted alpng with enzyme is released. Another factor called terminator factor (p) required for the termination of the transcription.

b) Translation :
It is the process of polymerizing amino acid to form a polypeptide chain. The triplet sequence of base pairs in mRNA defines the order and sequence of amino acids in a polypeptide chain. This process invoivs 3 steps, a) Initiation b) Elongation c) Termination. During the initiation, tRNA gets charged when the aminoacid binds by using ATR The start codon (AUG) present on mRNA is recognized only by the charged tRNA. The ribosome acts as an actual site for the process of translation and contains two separate sites in.a large subunit for the attachment of subsequent aminoacid.

The small subunit of ribosome binds to mRNA at the initiation codon (AUG) followed by the large subunit. Then, it initiates the process of translation. During the elongation process, the ribosome moves one codon dowonstream along with mRNA so as to leave the space for binding of another charged tRNA, The aminoacid brought by tRNA get linked with the previous aminoacid through a peptide bond and this process continues resulting in the formation of a polypeptide chain. When the ribosome reaches one or more STOP codon (UAA, UAG and UGA), the process of translation gets terminated. The polypeptide chain is released and ribosomes get detached from mRNA.

Question 13.
How the polymerization of nucleotides can be prevented in a DNA molecule?
Answer:
Due to Lack of Helicase enzyme, unwounding does not occurs. The single stranded Binding proteins cover the DNA strands preventing them from annealing into a double strand.

Question 14.
In an experiment, DNA is treated with a compound which trends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive base pairs increases. From 0.34 nm to 0.44 nm calculate the length of DNA double helix (which has 2 × 10⁹ bp) in the presence of saturating amount of this compound.
Answer:
2 × 10⁹ × 0.44 × 10^-9 bp.

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:
RNA is more labile and prone to degradation (owing to the presence of 2’OH group in its ribose). Hence heat killed ‘s’ strain may not have retained its ability to transform the ‘R’ strain into virulent form if RNA was its genetic material.

Question 16.
You 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:
Use of ¹⁵N will be inappropriate because method of detection of ³⁵P and ¹⁵N is different (³²P being a radioactive Isotope while ¹⁵N is not radioactive but is the heavier Isotope of Nitrogen). Even if ¹⁵N was radioactive, them its presence would have been detected both inside the cell as well as in the supernatant because ¹⁵N would also get incorporated in amino group of aminoacids in proteins. Hence the use of ¹⁵N would not give any conclusive results.

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:
Functional m RNA of structural genes need not always include all of its exons. This alternate splicing of exons in sex-specific, tissue – specific, and even developmental stage specific. By such alternate splicing of exons a single gene may encode for several Isoproteins and/or proteins of similar class’. In absence of such a kind of splicing, there should have been new genes for every protein/Isoprotein. Such an extravagency has been avoided in natural phenomena by way of alternate splicing.

Question 18.
Can you recall what centrifugal force is, and think why a molecule with higher mass/ Density would sediment faster?
Answer:
Proteins have lower density when compared to others. So a molecule with higher mass would sediment faster than molecules with light weight (density).

Question 19.
Do Retroviruses follow central Dogma? Give one example.
Answer:
Genetic material of Retroviruses in RNA. At the time of synthesis of protein RNA is reverse transcribed to its complementary DNA first/which is opposite to the central Dogma. Hence Retroviruses are not known to follow central Dogma.

Question 20.
If there are 2.9 × 10⁹ complete turns in a DNA molecule. Estimate the length of the molecule (1 angstrom = 10^-8 cm).
Answer:
1 turn of DNA = 3.4 nm.
Number of turns are = 2.9 × 10⁹
The length of the DNA molecule = 2.9 × 10⁹ × 3.4 nm.