Download SMU-DDE-Assignments-Scheme of Evaluation PROGRAM msc

SMU-DDE-Assignments-Scheme of Evaluation
PROGRAM
SEMESTER
SUBJECT CODE &
NAME
BK ID
DRIVE
MARKS
Q.
No
1
A
MSC BIOINFORMATICS
2
MBI201
MOLECULAR AND DEVELOPMENTAL BIOLOGY
B1943
WINTER 2015
60
Criteria
Describe the functioning of lac operon.
Functioning of the lac operon:
The lac operon is an operon required for the transport and metabolism
of lactose in Escherichia coli and other enteric bacteria. It consists of
three adjacent structural genes, lac Z, lac Y and lac A.
• lacZ encodes β galactosidase (LacZ), lacY encodes β galactoside
permease (LacY) and • lacA encodes β galactoside transacetylase
(LacA)
• There are two ways by which regulation of lac operon takes place.
They are (1) Induction and (2) catabolite repression.
(1) Induction:’
The lac operon is a negatively controlled inducible operon; the lac Z,
lac Y, and lac A genes are expressed only in the presence of lactose.
• The lac regulator gene, designated as ‘I ‘gene, encodes a repressor
that is 360 amino acids long. However, the active form of lac repressor
is a tetramer containing four copies of I gene product.
• In the absence of inducer, the repressor binds to the lac operator,
which in turn prevents RNA polymerase from catalyzing the
transcription of the structural genes.
• A few molecules of lac Z, lac Y, and lac A gene products are
synthesized in the un induced state, providing a low background level
of enzyme activity.
• This background level of enzyme activity is essential for induction of
the lac operon because the inducer of the operon, allolactose, is derived
from lactose in a reaction catalyzed by β galactosidase.
• Once formed, allolactose is bound by the repressor causing the release
of repressor from the operator. In this way, allolactose induces the
transcription of lac Z, lac Y, and lac A structural genes.
Total
Marks
(Unit 5 ; Section 5.3)
10
10
Marks
SMU-DDE-Assignments-Scheme of Evaluation
(2) Catabolite repression:
• When glucose enters an E. coli cell, it is utilized directly without the
induction of any new enzymes because, the enzymes required for the
glucose catabolism are always expressed in the E. coli cells.
• Thus, the cell is primed to use this sugar, a central molecule in
carbohydrate metabolism, above all the others.
• As E. coli becomes starved for glucose, it begins synthesizing an
unusual nucleotide named cyclic adenosine 3′, 5′ monophosphate,
generally referred to as cyclic AMP, or cAMP.
• An enzyme called adenylyl cyclase is continually active in converting
ATP molecules into cAMPs.
• When the cells are starved for glucose, the cell produces a pool of
cAMP molecules. These cAMP molecules bind to proteins called
Catabolite activator proteins (CAP) also called as cAMP receptor
proteins or CRPs.
• This CAP cAMP complex then binds to a site near the lac operon’s
promoter called CAP site.
• By themselves, both RNA polymerase and cAMP CAP complex have
relatively low affinity for their respective binding sites in lac promoter
DNA. Interaction between residues in CAP and α subunit of RNA
polymerase forms a protein protein complex that binds much more
stably to promoter DNA than either protein alone.
• Thus, binding of the CAP cAMP complex to the promoter site is
required for the transcription of lac operon.
• This is an example of positive control, in which the binding of a
protein(CAP) activates transcription.
• When glucose is present in the environment and transported into the
cell, the function of adenylyl cyclase is inhibited.
• The result is that ATP is no longer converted into cAMP, the cAMP
pool drops.
• This decrease in the complex inactivates the promoter, and the lac
operon is turned off.
SMU-DDE-Assignments-Scheme of Evaluation
• When cAMP is low, because of the lower level of glucose, cAMP
CAP does not bind to the CAP site.
• As a result, RNA polymerase does not bind efficiently to the lac
promoter and only a little lac mRNA is synthesized.
• When lactose is present and the glucose is absent, maximal
transcription of the lac operon occurs. In this situation, the lac repressor
does not bind to the lac operator.
• The concentration of cAMP increases, and the cAMP CAP complex
formed binds at the CAP site, stimulating binding of RNA polymerase
strongly. This process leads to transcription of lac structural genes.
• However, in the absence of lactose, no lac mRNA is formed because
repressor bound to the lac operator prevents transcription.
2
A
Discuss the steps involved in the processing of pre m-RNA. (Unit 3, Section 3.5, Pg. No. 81)
2.5X4
10
Following are the steps involved in the processing of pre-mRNA.
1. The addition of the 5’ cap:
 Almost all eukaryotic pre-mRNAs are modified at their 5’ends
by the addition of a structure called a 5’cap.
 This capping consists of the addition of an extra nucleotide at
the 5’end of the mRNA and methylation by the addition of a
methyl group (CH3) to the base in the newly added
nucleotide/s and to the 2’ –OH group of the sugar of one or
more nucleotides at the 5’ end.
 Capping takes place rapidly after the initiation of
transcription, the 5’ cap functions in the initiation of
translation.
2. The addition of the poly (A) tail:
 Most mature eukaryotic mRNAs have from 50 to 250 adenine
nucleotides at the 3’ end (a poly (A) tail).
 These nucleotides are not encoded in the DNA, but are added
after transcription) in a process termed Polyadenylation.
 Many eukaryotic genes transcribed by RNA polymerase II are
transcribed well beyond the end of the coding sequence; the
extra material at the 3’ end is then cleaved and the poly (A)
tail is added.
3. RNA splicing:
 The other major type of modification that takes place in
eukaryotic pre-mRNA is the removal of introns by RNA
splicing.
SMU-DDE-Assignments-Scheme of Evaluation

3
A
This occurs in the nucleus following transcription, but before
the RNA moves to the cytoplasm.
 Consensus sequences and the spliceosome splicing require the
presence of three sequences in the intron.
 A cut is made at the 3’ splice site and, simultaneously, the
3’end of exon 1 becomes covalently attached (spliced) to the
5’ end of exon 2.
4. RNA editing:
 A long-standing principle of molecular genetics is that genetic
information ultimately resides in the nucleotide sequence of
DNA, except in RNA viruses.
 This information is transcribed into mRNA, and mRNA is
then translated into a protein.
 The assumption that all information about the amino acid
sequence of a protein resides in DNA is violated by a process
called RNA editing.
 In RNA editing, the coding sequence of an mRNA molecule
is altered after transcription, and so the protein has an amino
acid sequence that differs from that encoded by the gene.
Explain any five features of genetic code.
Explanation for any five of the following features of genetic code:
1. The genetic code is non-overlapping:

Each nucleotide is a part of only one triplet codon.
Successive codons are read in order.

The translation starts at initiation codon and continues till the
termination codon in a single translational frame.

The nucleotide sequence is read without any punctuation
because, if the reading frame is displaced by a single base, it
remains shifted throughout the sequence and so the message
will code for a different amino acid.
2. The codes are highly degenerate:

The amino acids methionine and tryptophan have single
codons, whereas, three amino acids (Leucine, Serine,
Arginine) have six codons, five amino acids have four,
isoleucine has three, and nine amino acids have two.

Codons representing the same or related amino acids tend to
be similar in sequence.

Often the codons that differ in the last base will code for a
single amino acid.

The reduced specificity of the third base in a codon is called
as third base degeneracy. Codons that specify the same
amino acid are termed ‘synonyms’.
5X2
10
SMU-DDE-Assignments-Scheme of Evaluation
3. The arrangement of code table is nonrandom:

Most synonyms differ only in their third nucleotide.

XYU (where X and Y are two different nucleotides) and
XYC always specify the same amino acid; XYA and XYG
do so in all except in two cases (i.e., (case 1) UGA codes for
stop codon, UGG codes for tryptophan and (case 2) AUA
codes for Isoleucine whereas AUG codes for methionine).

These observations indicate that the code evolved so as to
minimize the deleterious effects of sudden change in
nucleotide bases (i.e., mutation).
4. UAG, UAA and UGA are stop codons:

These three codons (are also called nonsense codons) do not
specify any amino acids, but give signal to ribosome (protein
synthetic machinery) for polypeptide chain elongation.

The UAG, UAA and UGA are often referred to as ‘amber’,
ochre, and ‘opal’ codons respectively.
5. AUG and GUG are chain initiation codons:

The codons AUG and less frequently, GUG specify the
starting point for polypeptide chain synthesis.

However, they also specify amino acids methionine and
valine respectively at internal positions in a polypeptide
chain.
6. The code is unambiguous:

Each triplet codon specifies a particular amino acid and only
one amino acid.
7. The standard genetic code is not universal:
4
A

DNA studies in 1981 revealed that the genetic codes of certain
mitochondria are variants of standard genetic codes i.e., in
mammalian mitochondria, AUA, as well as standard AUG, are
a methionine (initiation codon); UGA specifies tryptophan
rather than stop; AGA and AGG are ‘Stop’ rather than
Arginine.

Thus, the standard genetic code, although very widely utilized,
is not universal.
Describe the phases of fertilization. Add a note on the structure of ovum.
(Unit 10, Section 10.4, Page No. 268)
6
10
Explanation for the following phases of fertilization:
a) Passage of sperm through corona radiata
b) Penetration of zona pellucida
c) Fusion of plasma membranes of the oocyte and the sperm.
SMU-DDE-Assignments-Scheme of Evaluation
d) Completion of second meiotic division of oocyte and formation of
female pronucleus.
e) Formation of male pronucleus
f) Fusion of pronuclei
5
A
4
Structure of ovum:
 The ovum that is shed from the ovary is not fully mature.
 It is still a secondary oocyte, which is undergoing division to
shed off the second polar body, and is surrounded by the zona
pellucida.
 Some cells of corona radiata remain adherent to the zona
pellucida. The ovum has no nucleus as it is undergoing second
meiotic division, and instead has a spindle.
 The first polar body lies in the perivitelline space between the
cell membrane and the zona pellucida.
 The ovum measures about 0.1 mm in size.
Describe the physical properties of DNA.
(Unit 1, Section 1.3, Pg. No. 16)
Description for the following five physical properties of DNA:
5X2
10
1. Antiparallel orientation:
 The DNA strands have got directionality.
 The orientation of the strands is anti-parallel, with one being
5’ to 3’ and the other 3’ to 5’.
 The direction of the naming starts from 5’ through 3’ region.
 Both the chains have the same geometry but the base pairing
holds them together with opposite polarity.
2. Grooves:
 Twin helical strands form the DNA backbone.
 The strand backbones are closer together on one side of the
helix than on the other.
 The major groove occurs where the backbones are far apart,
the minor groove occurs where they are close together.
3. Sense and antisense:
 A DNA strand is called "sense" if its sequence is the same as
that of a messenger RNA copy that is translated into protein.
 The sequence on the opposite strand is called the "antisense"
sequence.
4. Supercoiling:
 DNA can be twisted like a rope in a process called
supercoiling
 With DNA in its "relaxed" state, a strand usually circles the
axis of the double helix once every 10.4 base pairs, but if the
DNA is twisted, the strands become more tightly or more
loosely wound.
SMU-DDE-Assignments-Scheme of Evaluation

If the DNA is twisted in the direction of the helix, this is
positive supercoiling, and the bases are held more tightly
together.
 If they are twisted in the opposite direction, this is negative
supercoiling, and the bases come apart more easily.
5. Quadruplex structures:
 In humans, the telomeres are the lengths of single stranded
sequences, enriched with GT bases.
 In order to stabilize the ends, the telomeres form structures of
stacked sets of four-base units, called “G-quadruplex”.
 In addition to these quadruplexes, the DNA also form loops
called “Telomere loops” or simply “T-loops”, where in the
single-stranded DNA curls around, forming a long circle. The
T-loops are stabilized by telomere binding proteins.
6
A
Define mutation. Discuss the three types of DNA repair mechanisms.
(Unit 8, Section 8.2, Page No. 195)
1
10
Definition:
Sudden heritable changes in the genetic material are called mutations.
9
Explanation for the following three types of DNA repair
mechanisms:
(1) Mismatch repair:
Occurs immediately after DNA synthesis, uses the parental strand
as a template to correct an incorrect nucleotide incorporated into the
newly synthesized strand.
(2) Excision repair:
It entails removal of a damaged region by specialized nuclease
systems and then DNA synthesis to fill the gap.
(3) SOS response:
A bacterial deoxyribonucleic acid (DNA) repair system in which
cell division and DNA replication are blocked, and DNA repair,
recombination, and mutation genes are induced.
*A-Answer
Note –Please provide keywords, short answer, specific terms, specific examples (wherever necessary)
**********