NEET Biology: Genetics and Evolution · Chapter 2

Molecular Basis of Inheritance

What to remember

DNA is a double helix of two antiparallel polynucleotide chains (A pairs with T by 2 hydrogen bonds, G pairs with C by 3); it is the genetic material, as shown by Griffith, Avery-MacLeod-McCarty and Hershey-Chase.
DNA replication is semi-conservative (Meselson and Stahl), works only in the 5′ → 3′ direction, and gives a continuous leading strand and a discontinuous lagging strand made of Okazaki fragments.
Flow of information: DNA → (transcription) → RNA → (translation) → protein. The code is a triplet, degenerate, universal and non-overlapping, with AUG as start and UAA, UAG, UGA as stop.
The lac operon is an inducible, negatively regulated system in which lactose (inducer) inactivates the repressor, so the genes z, y and a are transcribed.

Structure of DNA

A nucleotide has three parts: a nitrogenous base, a pentose sugar and a phosphate group. A base joined to a sugar is a nucleoside. Purines are adenine (A) and guanine (G). Pyrimidines are cytosine (C), thymine (T, in DNA) and uracil (U, in RNA). The sugar is deoxyribose in DNA and ribose in RNA.

A base is joined to the 1′ carbon of the sugar by an N-glycosidic bond.
A phosphate is joined to the 5′ carbon of the sugar by a phosphoester bond.
Two nucleotides are joined by a 3′-5′ phosphodiester bond, so the chain has a 5′ end and a 3′ end.
In RNA the sugar has an extra 2′-OH group.

Friedrich Meischer first isolated DNA (he called it nuclein) in 1869. Watson and Crick proposed the double helix model in 1953. They used the X-ray diffraction data of Maurice Wilkins and Rosalind Franklin.

Features of the double helix:

Two polynucleotide chains run in opposite directions (antiparallel). One runs 5′ → 3′ and the other 3′ → 5′.
The sugar-phosphate backbone is on the outside. The bases point inward.
A pairs with T by two hydrogen bonds. G pairs with C by three hydrogen bonds. A purine always pairs with a pyrimidine, so the width is the same throughout.
The helix is right-handed. The pitch is 3.4 nm. There are 10 base pairs in each turn. The distance between two adjacent base pairs is 0.34 nm.
Chargaff's rules: the amount of A equals T, and G equals C. So purines equal pyrimidines (A + G = T + C). The ratio (A + T)/(G + C) differs between species but is constant within a species.

For a human diploid cell with about 6.6 × 10⁹ bp, the total length of DNA is about 2.2 m (6.6 × 10⁹ × 0.34 × 10⁻⁹ m).

Packaging of DNA

In bacteria, the DNA is not enclosed by a membrane. It is held in a region called the nucleoid, in loops by positively charged proteins.

In eukaryotes, histones are positively charged, basic proteins rich in lysine and arginine. Negatively charged DNA wraps around them.

Level	Description
Histone octamer	Eight histone molecules: two each of H2A, H2B, H3 and H4
Nucleosome	About 200 bp of DNA wrapped around one histone octamer
Chromatin	Nucleosomes in a "beads on string" form
Higher packing	Chromatin fibres fold and coil further to form metaphase chromosomes, and this needs non-histone chromosomal (NHC) proteins

Euchromatin: loosely packed, stains light, transcriptionally active.
Heterochromatin: densely packed, stains dark, transcriptionally inactive.

The genetic material

Griffith (1928) worked with Streptococcus pneumoniae. The S (smooth, capsule) strain is virulent. The R (rough, no capsule) strain is not. Live S killed mice. Live R and heat-killed S did not. But heat-killed S plus live R killed the mice, and live S cells were recovered. Some "transforming principle" passed from dead S to R.

Avery, MacLeod and McCarty purified biochemicals from heat-killed S cells. Protease and RNase did not stop transformation. DNase stopped it. So DNA is the transforming principle.

Hershey and Chase (1952) used bacteriophages grown in two media. One medium had radioactive phosphorus (³²P), which labels DNA. The other had radioactive sulphur (³⁵S), which labels protein. After infection, they used a blender to separate the viral coats from the bacteria. Centrifugation then separated light viral coats from heavier bacterial cells. The ³²P was in the bacterial pellet, and the ³⁵S was in the supernatant. So DNA enters the bacterium and is the genetic material.

Properties needed in genetic material: it must replicate itself, be chemically and structurally stable, allow slow mutation (needed for evolution) and express itself as Mendelian characters.

RNA world: RNA was probably the first genetic material. It can act as genetic material and also as a catalyst (ribozyme). Many essential life processes (metabolism, translation, splicing) work with RNA. But RNA is unstable and mutates faster. DNA is more stable because its sugar has no 2′-OH group and it has thymine in place of uracil. So DNA evolved from RNA as a better, more stable genetic material.

Replication of DNA

Replication is semi-conservative: each new DNA has one old strand and one new strand.

Experiment	Organism	Method	Result
Meselson and Stahl (1958)	E. coli	Grown in ¹⁵N (heavy nitrogen, from ¹⁵NH₄Cl), then moved to ¹⁴N. DNA was separated by CsCl density gradient centrifugation	After one generation (about 20 min), all DNA was hybrid (intermediate density). After two generations (40 min), equal amounts of hybrid and light DNA
Taylor	Vicia faba (faba bean)	Radioactive thymidine, root tips	At the first metaphase after replication in labelled medium, both chromatids were labelled. After one more replication in non-radioactive medium, only one chromatid of each chromosome was labelled

Steps and enzymes:

Replication begins at the origin of replication.
Helicase unwinds the helix. This forms a Y-shaped replication fork.
DNA-dependent DNA polymerase adds nucleotides only in the 5′ → 3′ direction. It uses deoxyribonucleoside triphosphates, which also supply the energy.
On the template with 3′ → 5′ polarity, the new strand is made continuously. This is the leading strand.
On the template with 5′ → 3′ polarity, the new strand is made in short pieces, the Okazaki fragments. This is the lagging strand.
DNA ligase joins the Okazaki fragments.
In bacteria such as E. coli the speed is about 2000 bp per second.
In eukaryotes replication takes place in the S phase. If cell division fails after replication, the result is polyploidy.

Transcription

Transcription is the copying of one strand of DNA into RNA. The transcription unit has three regions: promoter, structural gene and terminator.

The template strand has 3′ → 5′ polarity and is copied. The coding strand has 5′ → 3′ polarity. Its sequence is the same as the RNA (with U in place of T).
The promoter is at the 5′ end (upstream) of the structural gene. RNA polymerase binds to it.
The terminator is at the 3′ end (downstream) of the coding strand.
A cistron is a segment of DNA that codes for a polypeptide.

Prokaryotes: one RNA polymerase makes mRNA, tRNA and rRNA. Sigma (σ) factor helps initiation. Rho (ρ) factor helps termination. Genes are usually polycistronic and translation can begin before transcription ends.

Eukaryotes: the genes are split. Exons are the coding sequences. Introns are the intervening non-coding sequences. The primary transcript (hnRNA) contains both.

RNA polymerase	Product
RNA polymerase I	rRNAs (28S, 18S, 5.8S)
RNA polymerase II	hnRNA, the precursor of mRNA
RNA polymerase III	tRNA, 5S rRNA and snRNAs

Processing of hnRNA: (1) Splicing removes introns and joins exons. (2) Capping adds methyl guanosine triphosphate at the 5′ end. (3) Tailing adds about 200 to 300 adenylate residues (poly-A) at the 3′ end. The result is mature mRNA, which leaves the nucleus.

Genetic code

The code was worked out by Marshall Nirenberg, Har Gobind Khorana and others. George Gamow suggested that a triplet code is needed, because 4³ = 64 combinations are enough for 20 amino acids. Khorana made synthetic RNA with known repeating sequences. Nirenberg used a cell-free system for protein synthesis.

The code is a triplet: three bases make one codon.
There are 64 codons. 61 code for amino acids and 3 are stop codons (UAA, UAG, UGA).
AUG codes for methionine and is also the start (initiation) codon. UGG is the only codon for tryptophan.
The code is degenerate (many amino acids have more than one codon), unambiguous (a codon codes for only one amino acid), universal (the same in nearly all organisms), non-overlapping and comma-less (read continuously with no punctuation).
Mutations and the code: a point mutation (as in sickle-cell) changes one amino acid. Deletion or insertion of one or two bases causes a frameshift. Insertion or deletion of three bases (or a multiple of three) adds or removes amino acids but keeps the reading frame.

tRNA and translation

tRNA is the adapter that reads the code (proposed by Francis Crick). Its two-dimensional shape is a clover leaf. Its three-dimensional shape is an inverted L. It has an anticodon loop, which pairs with the codon, and an amino acid acceptor end at the 3′ end. Each amino acid has its own tRNA.

Translation is the making of a polypeptide using the sequence in mRNA.

1. Aminoacylation (charging): an amino acid is activated by ATP and joined to its tRNA by aminoacyl tRNA synthetase.
2. Initiation: the small ribosomal subunit binds to mRNA at the start codon AUG. The initiator tRNA binds there. Then the large subunit joins.
3. Elongation: the ribosome has two sites for tRNAs. Amino acids are joined by peptide bonds. The enzyme activity (peptidyl transferase) is a ribozyme, the 23S rRNA in bacteria. The ribosome moves along mRNA in the 5′ → 3′ direction.
4. Termination: a release factor binds to a stop codon, and the polypeptide is released.

Ribosomes are 70S (50S + 30S) in prokaryotes and 80S (60S + 40S) in eukaryotes. Many ribosomes on one mRNA form a polysome. The mRNA also has untranslated regions (UTRs) at both the 5′ end (before the start codon) and the 3′ end (after the stop codon). They are needed for efficient translation but are not translated.

Regulation of gene expression: the lac operon

Gene expression can be regulated at transcription, processing, mRNA transport and translation. In prokaryotes the main control is at transcription. François Jacob and Jacques Monod worked out the lac operon in E. coli.

Part	Role
i gene (regulatory)	Codes for the repressor
Promoter (p)	Binding site for RNA polymerase
Operator (o)	Binding site for the repressor
z gene	β-galactosidase, which breaks lactose into galactose and glucose
y gene	Permease, which brings lactose into the cell
a gene	Transacetylase

No lactose: the repressor binds the operator. RNA polymerase is blocked, so z, y and a are not transcribed.
Lactose present: lactose is the inducer. It binds the repressor and inactivates it. The repressor leaves the operator and RNA polymerase transcribes z, y and a. The operon is switched on. This is negative regulation, because the repressor shuts the genes off.

Human Genome Project (HGP)

It ran from 1990 and was completed in 2003. It was a very large project of an international group.

Goals: find all the genes in human DNA, find the sequence of the 3 × 10⁹ base pairs, store the data in databases, improve tools for data analysis, transfer the technologies to industry, and study the ethical, legal and social issues.

Methodologies: (1) Expressed Sequence Tags (ESTs) identify the genes expressed as RNA. (2) Sequence annotation sequences the whole genome (coding and non-coding) and then assigns functions. DNA is cut into fragments and cloned in vectors such as BAC (bacterial artificial chromosomes) and YAC (yeast artificial chromosomes). Frederick Sanger's method was used for sequencing.

Salient features:

The human genome has about 3 × 10⁹ base pairs. An average gene has about 3000 bases. Dystrophin is the largest known gene, with 2.4 million bases.
The number of genes is about 30,000.
About 99.9% of the nucleotide bases are the same in all people.
The functions of over 50% of the discovered genes are unknown.
Less than 2% of the genome codes for proteins.
A very large part of the genome is repetitive sequences.
Chromosome 1 has the most genes (2968) and the Y chromosome has the fewest (231).
About 1.4 million single nucleotide polymorphisms (SNPs) were found.

DNA fingerprinting

Alec Jeffreys developed it. It compares the repetitive DNA that differs between people. Satellite DNA is repetitive, non-coding DNA. It forms a large part of the genome and separates as small peaks during density gradient centrifugation. A kind of satellite DNA with a variable number of tandem repeats (VNTR) differs in length from person to person. The number of repeats varies, but is inherited from parents.

Steps: isolate DNA → digest with restriction endonucleases → separate by gel electrophoresis → transfer to a nylon or nitrocellulose membrane (Southern blotting) → hybridise with a labelled VNTR probe → detect by autoradiography.

Applications: forensic tests, paternity disputes and studying genetic diversity in populations. The pattern is unique to each person except identical twins.

NEET traps

Hershey and Chase: DNA has phosphorus (³²P), not sulphur. Protein has sulphur (³⁵S), not phosphorus.
Griffith did not prove that DNA is the genetic material. Avery, MacLeod and McCarty did.
Leading strand is continuous and grows toward the fork. Lagging strand is made in fragments.
Template strand (3′ → 5′) is not the same as coding strand (5′ → 3′). RNA matches the coding strand.
Exons are expressed and introns are removed. Capping is at 5′ and tailing is at 3′.
AUG is both the start codon and the codon for methionine. Stop codons code for no amino acid.
Degenerate code means many codons for one amino acid, not one codon for many.
The repressor is made by the i gene, not by the operator. The inducer binds the repressor, not the operator.
Euchromatin is active and light. Heterochromatin is inactive and dark.
DNA fingerprinting uses VNTR, not coding genes. The blot is called Southern.

One-liners

1. A pairs with T by 2 hydrogen bonds, and G pairs with C by 3.
2. Pitch of DNA is 3.4 nm, with 10 bp per turn and 0.34 nm between base pairs.
3. A nucleosome has about 200 bp of DNA on a histone octamer.
4. Histones are rich in lysine and arginine and are positively charged.
5. Griffith's experiment used Streptococcus pneumoniae and mice.
6. DNase destroyed the transforming principle in the experiment of Avery, MacLeod and McCarty.
7. Hershey and Chase used ³²P and ³⁵S in bacteriophages.
8. Meselson and Stahl used ¹⁵N and CsCl density gradient centrifugation.
9. DNA polymerase works only in the 5′ → 3′ direction.
10. DNA ligase joins Okazaki fragments.
11. RNA polymerase II makes hnRNA.
12. Mature mRNA has a cap at the 5′ end and a poly-A tail at the 3′ end.
13. UAA, UAG and UGA are stop codons.
14. Ribosome catalysis in translation is done by rRNA, a ribozyme.
15. In the lac operon, lactose is the inducer and the i gene makes the repressor.

Practice questions

Which pair of bases is joined by three hydrogen bonds in DNA?
1. Guanine and cytosine
2. Cytosine and thymine
3. Adenine and thymine
4. Adenine and guanine
Answer
A. Guanine and cytosine
G pairs with C by three hydrogen bonds. A pairs with T by two.
The distance between two adjacent base pairs in a DNA double helix is
1. 10 nm
2. 0.34 nm
3. 3.4 nm
4. 34 nm
Answer
B. 0.34 nm
The pitch is 3.4 nm and has 10 base pairs, so each step is 0.34 nm.
A histone octamer is made of
1. two molecules each of H2A, H2B, H3 and H4
2. eight molecules of H1
3. four molecules each of H2A and H2B only
4. one molecule each of H2A, H2B, H3 and H4 and four NHC proteins
Answer
A. two molecules each of H2A, H2B, H3 and H4
Eight histone molecules, two of each of four types, form the core of a nucleosome.
In Griffith's experiment, mice died when they were injected with
1. heat-killed R cells together with live R cells
2. live R cells alone
3. heat-killed S cells alone
4. heat-killed S cells together with live R cells
Answer
D. heat-killed S cells together with live R cells
Live R cells changed into virulent S cells by the transforming principle from the dead S cells.
In the work of Avery, MacLeod and McCarty, transformation was stopped by treating the extract with
1. protease
2. DNase
3. RNase
4. both protease and RNase
Answer
B. DNase
Only DNase stopped transformation, so DNA is the transforming principle.
In the Hershey and Chase experiment, the bacteriophage protein was labelled with
1. radioactive carbon (¹⁴C)
2. heavy nitrogen (¹⁵N)
3. radioactive sulphur (³⁵S)
4. radioactive phosphorus (³²P)
Answer
C. radioactive sulphur (³⁵S)
Protein has sulphur but DNA does not. DNA has phosphorus.
The technique used to separate heavy, hybrid and light DNA in the experiment of Meselson and Stahl was
1. gel electrophoresis
2. X-ray diffraction
3. paper chromatography
4. CsCl density gradient centrifugation
Answer
D. CsCl density gradient centrifugation
DNA of different densities forms bands at different positions in a caesium chloride gradient.
Which enzyme joins the Okazaki fragments of the lagging strand?
1. Helicase
2. DNA polymerase
3. DNA ligase
4. RNA polymerase
Answer
C. DNA ligase
DNA ligase seals the nicks between the fragments.
DNA-dependent DNA polymerase adds nucleotides
1. only in the 5′ → 3′ direction
2. only in the 3′ → 5′ direction
3. in either direction, depending on the strand
4. only to the 5′ end of a free nucleotide chain
Answer
A. only in the 5′ → 3′ direction
This is the reason the lagging strand has to be made in short pieces.
In a transcription unit, the promoter is located
1. inside the exons of the structural gene
2. at the 5′ end (upstream) of the structural gene
3. at the 3′ end of the structural gene
4. next to the terminator
Answer
B. at the 5′ end (upstream) of the structural gene
RNA polymerase binds the promoter before the structural gene. The terminator is at the 3′ end.
In eukaryotes, the RNA polymerase that makes hnRNA is
1. RNA polymerase I
2. RNA polymerase III
3. RNA polymerase II
4. a single polymerase that makes all RNAs
Answer
C. RNA polymerase II
Polymerase I makes rRNAs, II makes hnRNA, III makes tRNA, 5S rRNA and snRNAs.
The poly-A tail of eukaryotic mRNA is added
1. at the 5′ end of hnRNA
2. inside the introns
3. at both ends of hnRNA
4. at the 3′ end of hnRNA
Answer
D. at the 3′ end of hnRNA
Tailing adds about 200 to 300 adenylate residues at the 3′ end. Capping is at the 5′ end.
The codon that is both the start codon and the codon for methionine is
1. UGG
2. AUG
3. UAA
4. GUG
Answer
B. AUG
AUG is the only codon for methionine and it starts translation. UGG is for tryptophan.
The three-dimensional shape of a tRNA molecule is
1. a clover leaf
2. a straight rod
3. a double helix
4. an inverted L
Answer
D. an inverted L
The clover leaf is the two-dimensional drawing. Folding makes the inverted L.
In the lac operon, the z gene codes for
1. β-galactosidase
2. permease
3. the repressor
4. transacetylase
Answer
A. β-galactosidase
β-galactosidase breaks lactose into glucose and galactose. Permease is made by y, and the repressor by i.
In DNA fingerprinting, the DNA fragments are transferred from the gel to a nylon or nitrocellulose membrane by
1. autoradiography
2. Western blotting
3. Southern blotting
4. Northern blotting
Answer
C. Southern blotting
Southern blotting transfers DNA. Autoradiography is the later step that detects the labelled probe.
Given below are two statements: Statement I: In DNA of any organism, the amount of (A + T) is equal to the amount of (G + C). Statement II: In DNA the total number of purines equals the total number of pyrimidines.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Chargaff's rules give A = T and G = C, so purines equal pyrimidines (Statement II is correct). But (A + T) and (G + C) are not equal in general (Statement I is incorrect).
Given below are two statements: Statement I: In the Hershey and Chase experiment, DNA of the virus entered the bacterial cell. Statement II: Radioactive sulphur was found mostly in the bacterial pellet after centrifugation.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Statement I is correct. Statement II is incorrect because the ³⁵S labelled protein coats stayed in the supernatant.
Given below are two statements: Statement I: The leading strand is made discontinuously. Statement II: Okazaki fragments are joined by DNA ligase.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
The leading strand is made continuously, so Statement I is incorrect. Statement II is correct.
Given below are two statements: Statement I: The template strand of a transcription unit has 5′ → 3′ polarity. Statement II: Exons are removed during splicing.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
The template strand has 3′ → 5′ polarity. Introns, not exons, are removed. Both statements are incorrect.
Which of the following statements about the genetic code are correct? 1. The code is degenerate. 2. The code is overlapping. 3. AUG is the initiation codon.
1. 1 and 3 only
2. 1 and 2 only
3. 2 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 3 only
The code is non-overlapping, so statement 2 is wrong.
Given below are two statements: Statement I: Eukaryotic mRNA has a cap at the 5′ end and a poly-A tail at the 3′ end. Statement II: Introns are removed from hnRNA by splicing.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both statements are correct. Capping, tailing and splicing together make mature mRNA from hnRNA.
Which of the following statements about the Human Genome Project are correct? 1. Less than 2% of the genome codes for proteins. 2. Chromosome 1 has the fewest genes. 3. About 99.9% of nucleotide bases are the same in all people.
1. 1 and 2 only
2. 1 and 3 only
3. 2 and 3 only
4. 1, 2 and 3
Answer
B. 1 and 3 only
Chromosome 1 has the most genes (2968), and the Y chromosome has the fewest (231).
Given below are two statements: Statement I: A nucleosome has about 200 bp of DNA wrapped around a histone octamer. Statement II: Histones are negatively charged proteins that bind DNA through ionic attraction.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Statement I is correct. Histones are positively charged (rich in lysine and arginine), so Statement II is incorrect.
Given below are two statements: Statement I: In the lac operon, the repressor binds to the promoter. Statement II: Lactose acts as an inducer by inactivating the repressor.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
The repressor binds to the operator, not the promoter, so Statement I is incorrect. Statement II is correct.
Given below are two statements: Statement I: DNA fingerprinting is based on the coding genes of an individual. Statement II: Identical twins have different DNA fingerprints.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Fingerprinting uses VNTR (repetitive DNA), and identical twins share the same pattern. Both statements are incorrect.
Assertion (A): DNA replication is semi-conservative. Reason (R): Each daughter DNA molecule has one parental strand and one newly made strand.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
This is the meaning of semi-conservative replication, which was shown by the experiments of Meselson and Stahl.
Assertion (A): The genetic code is degenerate. Reason (R): AUG is the start codon.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true, but the start codon has no relation to degeneracy. Degeneracy means that most amino acids have more than one codon.
Assertion (A): The lac operon is switched on when lactose is present. Reason (R): Lactose binds to the operator and prevents the repressor from binding.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
Lactose (the inducer) binds to the repressor and inactivates it, not to the operator. So R is false.
Assertion (A): In eukaryotes, the primary transcript is translated immediately after it is made. Reason (R): The hnRNA is spliced, capped and tailed before it becomes mRNA.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
The primary transcript must be processed first, so A is false. R is true.
Assertion (A): DNA fingerprinting can tell two unrelated people apart. Reason (R): VNTR sequences are identical in all individuals.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
The number of repeats in VNTR differs among people, which is why the method works. So R is false.
Assertion (A): After one generation in ¹⁴N medium, all the DNA of E. coli grown earlier in ¹⁵N was of heavy density. Reason (R): DNA replication is semi-conservative.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
After one generation all DNA was hybrid (intermediate density), so A is false. R is true.
Match List I with List II. List I: a. Griffith b. Avery, MacLeod and McCarty c. Hershey and Chase d. Meselson and Stahl. List II: i. Bacteriophages and a blender ii. ¹⁵N and CsCl density gradient iii. Transforming principle in mice iv. DNase stopped transformation.
1. a-iii, b-iv, c-i, d-ii
2. a-iii, b-iv, c-ii, d-i
3. a-iv, b-iii, c-i, d-ii
4. a-i, b-iv, c-iii, d-ii
Answer
A. a-iii, b-iv, c-i, d-ii
Griffith used mice (iii), Avery and others used enzymes (iv), Hershey and Chase used phages (i) and Meselson and Stahl used ¹⁵N (ii).
Match List I with List II. List I: a. RNA polymerase I b. RNA polymerase II c. RNA polymerase III d. Single polymerase of bacteria. List II: i. hnRNA ii. 28S, 18S and 5.8S rRNA iii. tRNA, 5S rRNA and snRNA iv. mRNA, tRNA and rRNA.
1. a-i, b-ii, c-iii, d-iv
2. a-ii, b-i, c-iii, d-iv
3. a-ii, b-i, c-iv, d-iii
4. a-iii, b-i, c-ii, d-iv
Answer
B. a-ii, b-i, c-iii, d-iv
Pol I makes the large rRNAs (ii), pol II makes hnRNA (i), pol III makes tRNA (iii) and the bacterial enzyme makes all three RNAs (iv).
Match List I with List II. List I: a. i gene b. Operator c. z gene d. y gene. List II: i. β-galactosidase ii. Permease iii. Codes for the repressor iv. Binding site of the repressor.
1. a-iv, b-iii, c-i, d-ii
2. a-iii, b-iv, c-ii, d-i
3. a-iii, b-iv, c-i, d-ii
4. a-i, b-iv, c-iii, d-ii
Answer
C. a-iii, b-iv, c-i, d-ii
i makes the repressor (iii), the operator binds it (iv), z makes β-galactosidase (i) and y makes permease (ii).
Match List I with List II. List I: a. Capping b. Tailing c. Splicing d. Aminoacylation. List II: i. Addition of poly-A at the 3′ end ii. Addition of methyl guanosine triphosphate at the 5′ end iii. Removal of introns iv. Joining of an amino acid to tRNA.
1. a-ii, b-i, c-iii, d-iv
2. a-iii, b-i, c-ii, d-iv
3. a-i, b-ii, c-iii, d-iv
4. a-ii, b-i, c-iv, d-iii
Answer
A. a-ii, b-i, c-iii, d-iv
Capping is at the 5′ end (ii), tailing is at the 3′ end (i), splicing removes introns (iii) and aminoacylation charges tRNA (iv).
Match List I with List II. List I: a. Pitch of the helix b. Base pairs per turn c. Distance between two base pairs d. DNA in one nucleosome. List II: i. 0.34 nm ii. 3.4 nm iii. About 200 bp iv. 10.
1. a-ii, b-iv, c-i, d-iii
2. a-ii, b-i, c-iv, d-iii
3. a-iii, b-iv, c-i, d-ii
4. a-i, b-iv, c-ii, d-iii
Answer
A. a-ii, b-iv, c-i, d-iii
Pitch is 3.4 nm (ii), there are 10 bp per turn (iv), spacing is 0.34 nm (i) and a nucleosome has about 200 bp (iii).
Match List I with List II. List I: a. UAA b. AUG c. UGG d. Degeneracy of the code. List II: i. Initiation codon for methionine ii. The only codon for tryptophan iii. Stop codon iv. More than one codon for an amino acid.
1. a-i, b-iii, c-ii, d-iv
2. a-iii, b-i, c-ii, d-iv
3. a-iii, b-i, c-iv, d-ii
4. a-iii, b-ii, c-i, d-iv
Answer
B. a-iii, b-i, c-ii, d-iv
UAA is a stop codon (iii), AUG starts translation (i), UGG codes only for tryptophan (ii) and degeneracy means many codons for one amino acid (iv).
A DNA sample has 30 per cent adenine. What is the percentage of guanine in it?
1. 30 per cent
2. 70 per cent
3. 20 per cent
4. 40 per cent
Answer
C. 20 per cent
A = T = 30%, so A + T = 60%. The remaining 40% is G + C, so G = 20%.
A template strand has the sequence 3′-TACGGATTC-5′. The mRNA made from it is
1. 5′-GAAUCCGUA-3′
2. 5′-ATGCCTAAG-3′
3. 5′-UACGGAUUC-3′
4. 5′-AUGCCUAAG-3′
Answer
D. 5′-AUGCCUAAG-3′
The mRNA is complementary and antiparallel to the template, with U in place of T. So it is 5′-AUGCCUAAG-3′.
E. coli grown for many generations in ¹⁵N medium is moved to ¹⁴N medium and allowed to divide three times. What fraction of the DNA molecules is of hybrid density?
1. 1/4
2. 1/2
3. 3/4
4. 1/8
Answer
A. 1/4
After 3 generations there are 8 DNA molecules. Only 2 contain one ¹⁵N strand (hybrid) and 6 are light, so 2/8 = 1/4.
The coding part of an mRNA reads AUG GCU UUA UAG. How many amino acids are in the polypeptide made from it?
1. 4
2. 2
3. 12
4. 3
Answer
D. 3
AUG, GCU and UUA code for amino acids. UAG is a stop codon, so there are 3.
A mutation in the i gene of the lac operon makes the repressor unable to bind the operator. The result is that
1. the z, y and a genes are expressed even when lactose is absent
2. the genes are never expressed, even in lactose
3. the genes are expressed only when lactose is present
4. the genes are expressed only when glucose is present
Answer
A. the z, y and a genes are expressed even when lactose is absent
Without repressor binding, RNA polymerase can always transcribe the genes, so the expression is continuous.
What is the length of a DNA piece that has 1000 base pairs?
1. 3.4 nm
2. 340 nm
3. 34 nm
4. 3400 nm
Answer
B. 340 nm
Each base pair adds 0.34 nm, and 1000 x 0.34 nm = 340 nm. This is about 100 helical turns.
Bacteriophages whose DNA was labelled with ³²P were allowed to infect bacteria. After blending and centrifugation, most of the radioactivity was found in
1. the supernatant
2. the supernatant and the pellet in equal amounts
3. the bacterial pellet
4. neither, because the label is lost on blending
Answer
C. the bacterial pellet
The labelled DNA enters the bacterial cells, which settle in the pellet. The empty viral coats stay in the supernatant.

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