Fundamentholfundamenthol

Gene Regulation and Genetic Code

BiologyMolecular Basis of InheritanceFor NEET aspirants

Gene regulation and the genetic code explain how a cell reads its genes and when it switches them on. This page covers how the triplet code was proposed and deciphered by Gamow, Khorana, Nirenberg and Ochoa, the codon table and its six salient features, point and frameshift mutations, and tRNA as the adapter molecule. It then covers the levels of gene regulation and the lac operon of Jacob and Monod, with its repressor and inducer. NEET often asks codon features, stop codons and the lac genes under gene regulation and the genetic code.

On this page1Genetic code2Codon table3Mutations4tRNA adapter5Regulation6lac operon7Exam essentials8Quick revision9Solved examples10Practice
Key Points at a Glance
  1. ★ Must learn The code is a triplet: = 64 codons; 61 code for amino acids and 3 are stop codons.
  2. Gamow proposed the triplet; Khorana made RNAs of defined bases; Nirenberg's cell-free system deciphered the code; Ochoa's enzyme made RNA without a template.
  3. ★ Must learn The code is degenerate, read contiguously (no punctuation) and nearly universal.
  4. ★ Must learn AUG codes for methionine and is the initiator codon; UAA, UAG, UGA are stop codons.
  5. Point mutation in the beta globin gene: glutamate valine, causing sickle cell anaemia.
  6. Inserting or deleting 1 or 2 bases shifts the reading frame (frameshift); 3 bases keep the frame.
  7. tRNA (earlier sRNA): anticodon loop + amino acid acceptor end; clover-leaf on paper, inverted L in reality; no tRNA for stop codons.
  8. Eukaryotes regulate at four levels: transcriptional, processing, transport of mRNA, translational.
  9. ★ Must learn In prokaryotes, control of the rate of transcriptional initiation is the main control point; a repressor binds the operator.
  10. ★ Must learn lac operon: i (repressor), z (-galactosidase), y (permease), a (transacetylase); lactose or allolactose is the inducer; negative regulation.

1. The Genetic Code

  • In replication and transcription, one nucleic acid is copied into another, so complementarity makes them easy to picture.
  • Translation transfers information from a polymer of nucleotides to a polymer of amino acids.
  • ★ Exam imp There is no complementarity between nucleotides and amino acids, and none can be drawn even in theory.
  • Yet there was ample evidence that changes in nucleic acids (the genetic material) caused changes in the amino acids of proteins.
  • This led to the proposal of a genetic code that directs the sequence of amino acids during protein synthesis.

1.1 Proposing and deciphering the code

  • Proposing and deciphering the code was the most challenging task. It needed physicists, organic chemists, biochemists and geneticists.
  • ★ Exam imp George Gamow, a physicist, argued that 4 bases must code for 20 amino acids, so each code word must be a combination of bases.
  • He suggested that each code word, or codon, should be made of three nucleotides.
  • This was a bold proposal, because the permutations give = 64 codons, many more than needed.
  • Proving that the codon is a triplet was an even more daunting task.
  • Har Gobind Khorana developed a chemical method to synthesise RNA molecules with defined combinations of bases (homopolymers and copolymers).
  • Marshall Nirenberg's cell-free system for protein synthesis finally helped decipher the code.
  • Severo Ochoa's enzyme (polynucleotide phosphorylase) also helped, by polymerising RNA with defined sequences without a template (enzymatic synthesis of RNA).
  • Finally, a checker-board of the genetic code was prepared.

The codons for the amino acids

First positionSecond: USecond: CSecond: ASecond: GThird position
UUUU Phe
UUC Phe
UUA Leu
UUG Leu
UCU Ser
UCC Ser
UCA Ser
UCG Ser
UAU Tyr
UAC Tyr
UAA Stop
UAG Stop
UGU Cys
UGC Cys
UGA Stop
UGG Trp
U
C
A
G
CCUU Leu
CUC Leu
CUA Leu
CUG Leu
CCU Pro
CCC Pro
CCA Pro
CCG Pro
CAU His
CAC His
CAA Gln
CAG Gln
CGU Arg
CGC Arg
CGA Arg
CGG Arg
U
C
A
G
AAUU Ile
AUC Ile
AUA Ile
AUG Met
ACU Thr
ACC Thr
ACA Thr
ACG Thr
AAU Asn
AAC Asn
AAA Lys
AAG Lys
AGU Ser
AGC Ser
AGA Arg
AGG Arg
U
C
A
G
GGUU Val
GUC Val
GUA Val
GUG Val
GCU Ala
GCC Ala
GCA Ala
GCG Ala
GAU Asp
GAC Asp
GAA Glu
GAG Glu
GGU Gly
GGC Gly
GGA Gly
GGG Gly
U
C
A
G

1.2 Salient features of the genetic code

  1. The codon is a triplet. 61 codons code for amino acids and 3 do not code for any amino acid, so they work as stop codons.
  2. Some amino acids are coded by more than one codon, so the code is degenerate.
  3. The codon is read in mRNA in a contiguous fashion. There are no punctuations.
  4. The code is nearly universal: from bacteria to humans, UUU codes for phenylalanine (Phe). Some exceptions are found in mitochondrial codons and in some protozoans.
  5. AUG has dual functions: it codes for methionine (Met) and also acts as the initiator codon.
  6. UAA, UAG and UGA are the stop (terminator) codons.
★ Very important AUG = methionine + start; UAA, UAG, UGA = stop. 64 codons in all: 61 sense codons and 3 stop codons.
Memory Trick Stop codons: U Are Away, U Are Gone, U Go Away = UAA, UAG, UGA. All three begin with U.
  • The mRNA -AUG UUU UUC UUC UUU UUU UUC- codes for Met-Phe-Phe-Phe-Phe-Phe-Phe, since both UUU and UUC code for phenylalanine.
  • Going the other way is harder. For Met-Phe-Phe-Phe-Phe-Phe-Phe, each Phe could be UUU or UUC, so many different mRNAs are possible.
  • Reading forward always gives one answer, because each codon codes for only one amino acid; reading backward does not, because the code is degenerate.
NEET Focus Count traps appear in statement sets: 64 codons, 61 for amino acids, 3 stop. The code is nearly universal, not fully: the exceptions are mitochondrial codons and some protozoans. AUG is the only codon in this list with two functions.
Key idea
A contiguous, degenerate, nearly universal triplet code: 61 codons for amino acids, AUG to start, three to stop.

2. Mutations and the Genetic Code

  • The relationship between genes and DNA is best understood by studying mutations.
  • Large deletions and rearrangements in a DNA segment are easy to understand: they may cause loss or gain of a gene, and so of a function.
  • A point mutation is a change of a single base pair.
  • ★ Exam imp A classic example is a single base pair change in the gene for the beta globin chain, which changes the amino acid glutamate to valine. This causes sickle cell anaemia.

2.1 Insertions and deletions: RAM HAS RED CAP

  • Think of a sentence made of three-letter words, just like codons: RAM HAS RED CAP.
Change made after HASSentence read in three-letter words
NoneRAM HAS RED CAP
Insert BRAM HAS BRE DCA P
Insert BIRAM HAS BIR EDC AP
Insert BIGRAM HAS BIG RED CAP
Delete RRAM HAS EDC AP
Delete R and ERAM HAS DCA P
Delete R, E and DRAM HAS CAP
  • ★ Exam imp Inserting or deleting one or two bases changes the reading frame from the point of change: these are frameshift insertion or deletion mutations.
  • Inserting or deleting three bases, or a multiple of three, adds or removes one or more codons, and so one or more amino acids.
  • In that case the reading frame stays unaltered from that point onwards.
1 or 2 bases added or lostReading frame shifts.
Every codon after the change is misread.
Frameshift mutation.
3 bases (or a multiple) added or lostReading frame is kept.
One or more amino acids gained or lost.
The rest of the protein reads normally.
Memory Trick Three keeps the frame: any change that is not a multiple of three scrambles everything that follows.
Key idea
A single base change can swap one amino acid; adding or losing 1 or 2 bases shifts the whole frame after it.

3. tRNA: the Adapter Molecule

  • From the start, Francis Crick saw that something must read the code and link it to amino acids, because amino acids have no structural features to read the code themselves.
  • He postulated an adapter molecule that would read the code on one side and bind a specific amino acid on the other.
  • ★ Exam imp tRNA, then called sRNA (soluble RNA), was known before the genetic code was proposed. Its role as the adapter was assigned much later.
  • tRNA has an anticodon loop, with bases complementary to the code.
  • It also has an amino acid acceptor end, to which it binds an amino acid.
  • tRNAs are specific for each amino acid.
  • For initiation there is another specific tRNA, the initiator tRNA.
  • ★ Exam imp There are no tRNAs for stop codons.
  • Drawn on paper, the secondary structure of tRNA looks like a clover-leaf. Its actual structure is a compact molecule shaped like an inverted L.
tRNA, the adapter molecule A clover-leaf tRNA molecule standing on a wavy blue mRNA strand that runs from its 5 prime end on the left to its 3 prime end on the right. The stem at the top of the tRNA carries its 3 prime end on the left and its 5 prime end on the right; the tRNA has a loop on each side and a loop at the bottom. Three bases at the bottom of this loop, the anticodon, pair with three bases of the mRNA, the codon, running in the opposite direction. An amino acid, drawn as an orange ball at the top left, is joined by a dotted bond to the 3 prime end. Labels: tRNA, amino acid, amino acid acceptor end, anticodon loop, anticodon, codon, mRNA, 5 prime and 3 prime ends. 3′ 5′ 5′ 3′ Amino acid Amino acidacceptor end tRNA Anticodon loop Anticodon Codon mRNA
Figure 1: tRNA, the adapter molecule. Its anticodon pairs with a codon on mRNA, and its amino acid acceptor end carries the matching amino acid.
★ Very important tRNA reads the codon with its anticodon loop and carries the amino acid on its acceptor end. There is an initiator tRNA for AUG but no tRNA for UAA, UAG or UGA.
Tips and Tricks Clover-leaf versus inverted L is a favourite statement trap: the clover-leaf is the secondary structure drawn on paper; the inverted L is the real three-dimensional shape.
Quick Recall: tap to check
What was tRNA called before its adapter role was known?
sRNA, soluble RNA.
Why is there no tRNA for UGA?
UGA is a stop codon; stop codons code for no amino acid, so no tRNA reads them.
How many mRNA sequences can code for Met-Phe-Phe?
Four: AUG followed by UUU or UUC for each Phe (2 2).
Key idea
tRNA is Crick's adapter: anticodon on one end, amino acid on the other, one kind for each amino acid.

4. Regulation of Gene Expression

  • Regulation of gene expression is a broad term; it can occur at various levels.
  • Gene expression results in a polypeptide, so it can be regulated at several steps.
  • In eukaryotes, regulation can be exerted at four levels:
  1. Transcriptional level: formation of the primary transcript.
  2. Processing level: regulation of splicing.
  3. Transport of mRNA from the nucleus to the cytoplasm.
  4. Translational level.
Memory Trick Teachers Prepare Their Tests: Transcriptional, Processing, Transport, Translational.
  • Genes in a cell are expressed to perform a particular function or a set of functions.
  • Example: E. coli makes the enzyme beta-galactosidase to hydrolyse the disaccharide lactose into galactose and glucose, which it uses as a source of energy.
  • If there is no lactose around, the bacteria no longer need to make beta-galactosidase.
  • ★ Exam imp So, in simple terms, metabolic, physiological or environmental conditions regulate the expression of genes.
  • The development and differentiation of an embryo into an adult also result from the coordinated regulation of several sets of genes.

4.1 Regulation in prokaryotes

  • ★ Exam imp In prokaryotes, control of the rate of transcriptional initiation is the main site for controlling gene expression.
  • The activity of RNA polymerase at a promoter is regulated by accessory proteins, which affect its ability to recognise start sites.
  • These regulatory proteins can act positively (activators) or negatively (repressors).
  • Access to promoter regions is often controlled by proteins that interact with sequences called operators.
  • In most operons, the operator lies adjacent to the promoter, and in most cases it binds a repressor protein.
  • Each operon has its own operator and repressor. The lac operator is present only in the lac operon and interacts only with the lac repressor.
Key idea
Cells express genes only when needed; prokaryotes control mainly the start of transcription through operators and repressors.

5. The lac Operon

  • ★ Exam imp The lac operon was worked out by the geneticist François Jacob and the biochemist Jacques Monod, the first to explain a transcriptionally regulated system.
  • Here lac refers to lactose.
  • In the lac operon, a polycistronic structural gene is regulated by a common promoter and regulatory genes.
  • Such an arrangement is very common in bacteria and is called an operon. Examples: lac, trp, ara, his and val operons.
  • The lac operon has one regulatory gene (the i gene) and three structural genes (z, y and a).
  • The letter i does not stand for inducer; it comes from the word inhibitor.
GeneCodes forRole
i (regulatory)The repressor of the lac operonSwitches the operon off
zBeta-galactosidase (-gal)Hydrolyses the disaccharide lactose into galactose and glucose
yPermeaseIncreases the permeability of the cell to -galactosides
aTransacetylaseThird enzyme of the operon
  • All three gene products of the lac operon are needed for the metabolism of lactose.
  • In most other operons too, the genes are needed together in the same or a related metabolic pathway.
Memory Trick Gene order z, y, a gives the enzymes in order: Good People Talk = Galactosidase, Permease, Transacetylase. And i is for Inhibitor.

5.1 How lactose switches the operon on

  1. Lactose is the substrate of beta-galactosidase, and it regulates the switching on and off of the operon. So it is called the inducer.
  2. In the absence of a preferred carbon source such as glucose, lactose added to the medium is transported into the cells by permease.
  3. A very low level of lac operon expression must be present all the time; otherwise lactose cannot enter the cells.
  4. The repressor is synthesised all the time (constitutively) from the i gene.
  5. Without inducer, the repressor binds the operator and prevents RNA polymerase from transcribing the operon.
  6. With an inducer, such as lactose or allolactose, the repressor is inactivated by interacting with the inducer.
  7. RNA polymerase then gets access to the promoter, and transcription of z, y and a proceeds.
The lac operon A row of DNA regions: p, i, p, o, z, y, a. Top, in the absence of inducer: the i gene makes repressor mRNA and the repressor, which binds the operator (o) and stops RNA polymerase from transcribing the operon. Bottom, in the presence of inducer: the inducer binds the repressor and makes it inactive, so z, y and a are transcribed into one lac mRNA, which is translated into beta-galactosidase, permease and transacetylase. p i p o z y a In absence of inducer Repressor mRNA Repressor Repressor binds to the operator region (o) and prevents RNA polymerase from transcribing the operon p i p o z y a In presence of inducer Repressor mRNA Inducer (Inactive repressor) lac mRNA Transcription Translation β-galactosidase permease transacetylase
Figure 2: The lac operon. Without an inducer the repressor sits on the operator; the inducer inactivates the repressor, so the z, y and a genes are transcribed and translated.
  • ★ Exam imp Regulation of the lac operon can be seen as regulation of enzyme synthesis by its substrate.
  • Glucose or galactose cannot act as inducers for the lac operon.
  • The operon stays on only as long as lactose is present. Once lactose is used up, the repressor is no longer inactivated, binds the operator again and switches the operon off.
  • Regulation of the lac operon by the repressor is called negative regulation.
  • The lac operon is also under positive regulation, but that is beyond the scope of this level.
  • Transcription and translation are energetically very expensive, so they are tightly regulated, and regulation of transcription is the primary step.
  • The lac operon is the prototype operon in bacteria: it codes for the genes of lactose metabolism and is regulated by the amount of lactose in the medium.
No inducerRepressor made by i binds the operator.
RNA polymerase is blocked.
No lac mRNA; operon off.
Inducer presentLactose or allolactose binds and inactivates the repressor.
RNA polymerase reaches the promoter.
lac mRNA made; -galactosidase, permease, transacetylase formed.
★ Very important The inducer works by inactivating the repressor; the repressor itself is made all the time. Lactose (or allolactose) induces; glucose and galactose do not.
NEET Focus Statement sets mix up the genes: z codes for beta-galactosidase, y for permease and a for transacetylase, while i codes for the repressor. Other traps: i means inhibitor, not inducer; the repressor is constitutive; repressor control is negative regulation.
Quick Recall: tap to check
In Figure 2, which region of the operon does the repressor bind?
The operator (o), next to the promoter.
What does the letter i of the i gene stand for?
Inhibitor, not inducer; the i gene codes for the repressor.
Name the two inducers of the lac operon.
Lactose and allolactose. Glucose and galactose cannot induce it.
Key idea
lac operon: the constitutive repressor blocks the operator until lactose (or allolactose) inactivates it, so substrate switches on its own enzymes.

6. Exam Essentials

Pairs to Match

List IList II
George GamowProposed a triplet code
Har Gobind KhoranaChemical synthesis of RNA with defined bases
Marshall NirenbergCell-free system that deciphered the code
Severo OchoaPolynucleotide phosphorylase, template-independent RNA synthesis
AUGMethionine and initiator codon
UAA, UAG, UGAStop codons
UUUPhenylalanine, from bacteria to humans
Sickle cell anaemiaGlutamate to valine in beta globin
Anticodon loopPairs with the codon on mRNA
Amino acid acceptor endBinds the amino acid
Jacob and MonodElucidated the lac operon
i geneRepressor
z geneBeta-galactosidase
y genePermease
a geneTransacetylase
Exceptions
  • Only 3 of the 64 codons code for no amino acid: UAA, UAG and UGA.
  • The code is nearly, not fully, universal: exceptions occur in mitochondrial codons and some protozoans.
  • AUG is the codon with a dual role, methionine and start.
  • There are no tRNAs for stop codons.
  • Only lactose or allolactose induces the lac operon; glucose and galactose cannot.
  • The i gene is named from inhibitor, not inducer.
  • Adding or deleting three bases does not shift the reading frame.

Numbers to Remember

  • 4 bases, 20 amino acids, = 64 codons: 61 for amino acids, 3 stop.
  • 4 levels of regulation in eukaryotes.
  • lac operon: 1 regulatory gene and 3 structural genes.
  • 5 operons named: lac, trp, ara, his and val.

7. Quick Revision

  • No complementarity exists between nucleotides and amino acids, so a genetic code was proposed.
  • Gamow (physicist): triplet code, 64 codons.
  • Khorana: RNA homopolymers and copolymers; Nirenberg: cell-free system; Ochoa: polynucleotide phosphorylase.
  • 61 codons for amino acids, 3 stop codons (UAA, UAG, UGA).
  • Degenerate, contiguous with no punctuation, nearly universal (exceptions: mitochondria, some protozoans).
  • AUG: methionine and initiator codon.
  • Sickle cell anaemia: one base change, glutamate to valine in beta globin.
  • 1 or 2 bases inserted or deleted: frameshift; 3 bases: frame kept.
  • tRNA (sRNA) is Crick's adapter: anticodon loop and amino acid acceptor end.
  • Initiator tRNA for AUG; no tRNA for stop codons; clover-leaf on paper, inverted L in reality.
  • Eukaryotic regulation: transcriptional, processing, transport, translational.
  • Prokaryotes: rate of transcriptional initiation; activators and repressors; operator next to promoter.
  • lac operon (Jacob and Monod): i repressor; z -galactosidase, y permease, a transacetylase.
  • Repressor is constitutive and binds the operator; lactose or allolactose inactivates it.
  • Negative regulation; regulation of enzyme synthesis by its substrate.

8. Solved Examples

Solved Example 1
An mRNA reads 5′-AUG GGC UUU UAA-3′. What does it code for?
(A) Met-Gly-Phe
(B) Met-Gly-Phe-Stop-Met
(C) Met-Gly-Leu
(D) Gly-Phe
Solution:

Answer: (A). AUG = Met, GGC = Gly, UUU = Phe, and UAA is a stop codon, so translation ends after Phe.

Solved Example 2
Match List I with List II.
List I: A. i gene, B. z gene, C. y gene, D. a gene
List II: I. Permease, II. Transacetylase, III. Repressor, IV. Beta-galactosidase
Choose the correct answer:
(A) A-III, B-IV, C-I, D-II
(B) A-IV, B-III, C-I, D-II
(C) A-III, B-I, C-IV, D-II
(D) A-III, B-IV, C-II, D-I
Solution:

Answer: (A). i codes for the repressor, z for beta-galactosidase, y for permease and a for transacetylase.

Solved Example 3
Read the statements about the genetic code.
A. It is degenerate.
B. It has punctuation between codons.
C. AUG codes for methionine.
D. It is fully universal with no exceptions.
E. UGA is a stop codon.
Choose the correct answer:
(A) A, C and E only
(B) A, B and C only
(C) B, D and E only
(D) A, C, D and E only
Solution:

Answer: (A). B is wrong: the code is read contiguously, with no punctuation. D is wrong: it is nearly universal, with exceptions in mitochondria and some protozoans.

Solved Example 4
Arrange the events when lactose is added to E. coli growing without glucose.
A. The inducer inactivates the repressor
B. Lactose enters through permease made at a low basal level
C. RNA polymerase transcribes z, y and a
D. RNA polymerase gains access to the promoter
Choose the correct answer:
(A) B, A, D, C
(B) A, B, D, C
(C) B, D, A, C
(D) B, A, C, D
Solution:

Answer: (A). Lactose must first enter (B), then inactivate the repressor (A); only then can RNA polymerase reach the promoter (D) and transcribe the genes (C).

Solved Example 5
One base is deleted near the start of a structural gene. What is the most likely effect?
(A) Only one amino acid changes
(B) One amino acid is lost and the rest are normal
(C) The reading frame shifts, so the amino acids after the deletion change
(D) No effect, because the code is degenerate
Solution:

Answer: (C). Deleting one base is a frameshift mutation; every codon after it is read in a new frame.

Solved Example 6
Which statement about tRNA is NOT correct?
(A) It was earlier called sRNA
(B) Its anticodon pairs with the codon
(C) There is a tRNA for each stop codon
(D) Its actual shape is an inverted L
Solution:

Answer: (C). Stop codons code for no amino acid, and there are no tRNAs for them.

9. Practice Questions

Practice Questions
  1. How many codons code for amino acids, and how many are stop codons?Answer: 61 code for amino acids; 3 (UAA, UAG, UGA) are stop codons, out of 64.
  2. Match List I with List II.
    List I: A. Gamow, B. Khorana, C. Nirenberg, D. Ochoa
    List II: I. Polynucleotide phosphorylase, II. Cell-free protein synthesis, III. Triplet code, IV. Chemical synthesis of RNA with defined bases
    (A) A-III, B-IV, C-II, D-I (B) A-IV, B-III, C-II, D-I (C) A-III, B-II, C-IV, D-I (D) A-III, B-IV, C-I, D-IIAnswer: (A). Gamow proposed the triplet, Khorana made defined RNAs, Nirenberg used a cell-free system and Ochoa's enzyme made RNA without a template.
  3. Statement I: Regulation of the lac operon by its repressor is negative regulation.
    Statement II: The lac operon is also under positive regulation.
    (A) Both Statement I and Statement II are correct (B) Statement I is correct, Statement II is incorrect (C) Statement I is incorrect, Statement II is correct (D) Both are incorrectAnswer: (A). Repressor control is negative; a positive control also exists, though it is not studied at this level.
  4. Which of the following cannot act as an inducer of the lac operon? (A) Lactose (B) Allolactose (C) Glucose (D) Both lactose and allolactoseAnswer: (C). Glucose (and galactose) cannot induce the lac operon.
  5. Arrange the levels at which eukaryotic gene expression can be regulated, from first to last.
    A. Transport of mRNA to the cytoplasm
    B. Formation of the primary transcript
    C. Translation
    D. Splicing
    (A) B, D, A, C (B) B, A, D, C (C) D, B, A, C (D) B, D, C, AAnswer: (A). Transcriptional, processing (splicing), transport, then translational.
  6. Read the statements.
    A. The operator lies adjacent to the promoter in most operons.
    B. The lac operator interacts with every repressor in the cell.
    C. In prokaryotes, the rate of transcriptional initiation is the main control point.
    D. Activators act negatively on transcription.
    Choose the correct answer: (A) A and C only (B) A, B and C only (C) B and D only (D) A, C and D onlyAnswer: (A). B is wrong: the lac operator interacts only with the lac repressor. D is wrong: activators act positively.
  7. In the medium where E. coli was growing, lactose was added, which induced the lac operon. Why does the lac operon shut down some time after lactose is added?Answer: Permease brings lactose in and beta-galactosidase breaks it down into glucose and galactose. Once the lactose is used up, no inducer is left to inactivate the constitutively made repressor, so it binds the operator again and transcription stops.

Common Mistakes to Avoid

Watch out
  • Saying all 64 codons code for amino acids. Correct: 61 do; 3 are stop codons.
  • Calling AUG a stop codon. Correct: AUG codes for methionine and is the start codon; UAA, UAG and UGA stop.
  • Expanding i as inducer. Correct: i comes from inhibitor; it codes for the repressor.
  • Giving z to permease. Correct: z codes for beta-galactosidase, y for permease and a for transacetylase.
  • Naming glucose as the inducer. Correct: lactose or allolactose induces; glucose and galactose cannot.
  • Saying the repressor is made only when lactose is absent. Correct: it is made constitutively; the inducer only inactivates it.
  • Calling a three-base insertion a frameshift. Correct: three bases add one codon and keep the reading frame.
  • Calling the code fully universal. Correct: it is nearly universal, with exceptions in mitochondria and some protozoans.

Frequently Asked Questions

Why is the genetic code a triplet?

There are only four bases but twenty amino acids. Single bases give 4 codes and pairs give 16, which are too few. George Gamow argued that three bases give 64 combinations, enough for all twenty amino acids. Experiments by Khorana, Nirenberg and Ochoa later proved the triplet code.

What are the salient features of the genetic code?

The code is a triplet, with 61 codons for amino acids and 3 stop codons. It is degenerate, read contiguously without punctuation, and nearly universal, with exceptions in mitochondria and some protozoans. AUG codes for methionine and is the initiator codon; UAA, UAG and UGA are stop codons.

What is a frameshift mutation?

A frameshift mutation is the insertion or deletion of one or two bases in a gene. Because codons are read as contiguous triplets, every codon after the change is read in a new frame. Inserting or deleting three bases instead adds or removes one codon and keeps the reading frame.

Why is tRNA called an adapter molecule?

Amino acids cannot read the code themselves. Crick proposed an adapter that reads the codon on one side and carries an amino acid on the other. tRNA does exactly this: its anticodon loop pairs with the codon and its amino acid acceptor end binds the specific amino acid.

At what levels is gene expression regulated in eukaryotes?

Eukaryotes can regulate gene expression at four levels: at transcription, when the primary transcript forms; at processing, by regulating splicing; at the transport of mRNA from the nucleus to the cytoplasm; and at translation.

What is an operon?

An operon is an arrangement, common in bacteria, in which a polycistronic structural gene is controlled by a common promoter and regulatory genes. The genes in an operon usually work together in one metabolic pathway. Examples are the lac, trp, ara, his and val operons.

How does lactose switch on the lac operon?

The i gene makes the repressor all the time, and it binds the operator to block RNA polymerase. When lactose or allolactose enters the cell, it binds the repressor and inactivates it. RNA polymerase can then reach the promoter and transcribe the z, y and a genes.

Why does the lac operon shut down some time after lactose is added?

The enzymes made by the operon break down the lactose. When the lactose is used up, nothing is left to inactivate the repressor, which is still being made. The repressor binds the operator again and transcription stops, so the enzymes are made only while their substrate is present.

Previous year questions on Gene Regulation and Genetic Code

11 questions from past papers, each with a step-by-step solution.

Show all 11 questions

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