University Biology: Questions and Answers Assignment

This assignment provides answers to a series of short questions covering core concepts in molecular biology. The questions address topics such as the function of restriction enzymes and the products of their action, the essential features of yeast artificial chromosomes (YACs) and their construction, the principle of quantitative PCR (qPCR) and its applications, and a supplementary methodology for protein separation using 2D gel electrophoresis. The assignment also explores methods for detecting genetically modified organisms (GMOs) in food products, the critical considerations in primer design and optimization of annealing temperature for PCR assays, the nature and consequences of splice site mutations, the differences between forward and reverse genetic screening approaches, and the process of DNA replication. The answers provided demonstrate an understanding of the key concepts and techniques in the field of molecular biology.

Running head: QUESTIONS AND ANSWERS
Questions and answers
Name of the Student
Name of the University
Author Note

1QUESTIONS AND ANSWERS
SHORT ANSWERS:
Answer 1:
a) BamHI creates 5’ overhang after cleaving at these restriction sites after first G on the
sequence
5’ GGATCC 3’
3’ CCTAGG 5’
After cleavage it produces the following overhang sequence
5’ G 3’ 5’ GATCC 3’
3’ CCTAGG 5’ 3’ G 5’
b) BcII produces a 5’ overhang after cleaving at these restriction sites after first T on the
sequence
5’ TGATCA 3’
3’ ACTAGT 5’
After cleavge it produces the following overhang sequence
5’ T 3’ 5’ GATCA 3’
3’ ACTAG 5’ 3’ T 5’
c) i. If the sequence is cut by BamHI, it will produce two fragments
A- 5’GTTCAG 3’ B- 5’ GATCCGTAATTCCTGATTCACGCTCCA 3’
3’ CAAGTCCTAG 5’ 3’GCATTAAGGACTAGTGCGAGGT 5’
d) ii. If the sequence is cut by BcII, it will produce two fragments
C- 5’ GTTCAGGATCCGTAATTCCT 3’ D- 5’ GATCACGCTCCA 3’
3’ CAAGTCCTAGGCATTAAGGACTAC 5’ 3’TGCGAGGT 5’
e) If we join A + D, it will give the ligated sequence as

2QUESTIONS AND ANSWERS
5’GTTCAGGATCACGCTCCA 3’
3’ CAAGTCCTAGTGCGAGGT 5’
f) No, the ligated fragment can neither be cut with BamHI or BcII because the
recognition and cut site has been destroyed on the newly ligated sequence for both the
restriction enzymes. This is because the restriction enzymes cut at specific sites on the
DNA sequence by recognizing the sequence.
Answer 2:
YAC or yeast artificial chromosome is a modified or recombinant from of bacterial
plasmids with essential features of yeast chromosome making it capable of replicating within
the host that is yeast as well as Escherichia. coli. In order to replicate inside yeast the size of
the vector should be large that is around 1000 kb where the insert can be around 500 kb. The
features from yeast chromosomes for making a vector suitable as yeast artcificial
chromosome to replicate inside yeast-
 Linear DNA produced from circular plasmid of bacteria by breaking the circulization
with restriction enzymes.
 Centromere to ensure that chromosome is partitioned to two daughter cells as CEN
sequences
 Telomer regions at each end of the linear DNA to stabilize the chromosome ends as
TEL sequences
 Autonomously replicating sequence (ARS) for replication and preserving YAC inside
yeast cells. Therefore these act as origin of replication.
 Yeast marker genes such as URA3 and TRP1 for selection and screening of uracil and
tryptophan autotrophy respectively inside yeast
 SUP4 supressor gene for red or white colour selection after DNA insert.

3QUESTIONS AND ANSWERS
 HIS3 for histidine slection
The features of YACs are developed from yeast centromere shuttle-plasmids.
To replicate inside bacteria, the YAC contains features from bacterial plasmid-
 Beta-lactamase gene bla
 Bacterial pMB1 as bacterial origin of replication
 Multiple cloning site for restriction enzymes
 Bacterial slectable marker such as ampicillin resistant gene AmpR from pBR322
plasmid (Bajpai, 2014).
Fig 1: a) linear form of YAC and b) circular form of YAC (Image retrieved from Bajpai,
2014)
Answer 3
The principle of quantitative PCR is based on the measurement of amount of
amplification that is not possible with conventional PCR. This type of PCR does not require

4QUESTIONS AND ANSWERS
agarose gel electrophoresis to visualize the amplicon and its quantity because analysis of
melting curve will determine the amount. It uses flurescent dyes in the polymerase chaun
reaction to measure the amount of amplication by flashes of fluorescence. The PCR mixture
contains DNA molecules or target DNA, primers, DNA polymerase and buffer in
conventional PCR but in q-PCR, the addition of flurophor or a probe that will bind to the
target DNA during amplification inside the thermal cycler will measure the fluorescence of
the the fluorophor when it reaches an excited wavelength. This will allow the measurement of
one or more products of amplification at each PCR cycle and the data generated will be
calculated as relative gene expression of samples on a computer software. The process goes
as follows- denaturation at 95 °C, primer binding at 50-60 °C and elongation of
polymerization at 68-72 °C (Furda et al., 2014).
Answer 4:
Since the crude protein extract after resolving it with SDS-PAGE contained many
proteins of similar molecular weight therefore these molecules having similar mases should
be separated based on their electrical charge. The technique applied is 2-D gel electrophoresis
in which the purified proteins at first are separated on the basis of their electrical charge and
then separated on the basis of their molecular weight in SDS PAGE. This will allow speration
of similar molecular weight proteins because two unrelated proteins will always differ in their
net charges on the sequences. The process begins when the isoelctric focusing is used as the
first dimension followed by SDS-PAGE. The process followed-
 First the 8M urea is used to denature the cell extract and layered on a glass tube
containing ampholytes.
 Then electric field is passed through the ampholyte which seperates the solution based
on the net charge. The polyanionic ampholytes will remain at one end and polycaionic
at other end of the tube thereby establishing a pH gradient.

5QUESTIONS AND ANSWERS
 The proteins keep on migrating at each end till it reaches isoelectric pH thereby
resolving the protein by one uinit of charge.
 Then a second dimension is used where polyacrylamide gel is run under electric field
for resolving proteins by molecular weight and it makes use of SDS (Lodish et al.,
2000).
Figure 2: 2-D gel electrophoresis (Image retrieved from Lodish et al., 2000)
Answer 5:
Presence of gentically modified organisms can be detected in food vua various
amplication and sequencing methods. One common strategy to identify and quantify GMO is
by using qPCR. This type of detection technique is benficial for both processed and
unprocessed food because amplicons of size 100 are also amplified. Firstly the presence of
GMO in food is assessed using a screening method where the common transgenic elements
that are likely to be present in GMO are targeted and it includes p35S that is 35S promoter

6QUESTIONS AND ANSWERS
originated in cauliflower mosaic virus and tNOS that are terminators for nopaline syntghesis
in Agrobacaterium tumefacians. Other than this there are marlers distinct for the specific
GMO such as t35S PpCAMBIA and Cry3Bb can be used for detection. Primers are
developed specifc for the target sites such as the markers and it amplies the target sites which
are further detected as fluroesecence in qPCR. The second step is more specific detection via
targeting construct specific markers that is the markers present in the cassette of the
transgenic gene in the GMO or in the junction of plant genome and the transgenic cassette.
Finally the quantification of the detected transgene or its markers is performed from the
tested food or sample (Fraiture et al., 2015).
Answer 6:
a) Primer design for PCR is a critical area where specific features of the primers
determine the success rate of PCR reaction. The optimal length of the primers that
include both forward and reverse primers should be around 18-22 base pairs in length.
This is because the length is adequate for specific and easy primer binding to the
template during annealing process. Primers should have melting temperature (Tm)
between 52-58 o C. The GC content of the primer should be high that is around 40-
60% of the total primer bases. The last five bases at 3’ end should be either C or G
which will help specific and strong binding to the template.
b) Optimal primer annealing temperature is maintained to ensure stability between the
DNA-DNA hybrid. The annealing temperature should not be too high so that primers
are not sufficiently bound to the template and product yield is low. On the other hand
if the annealing temperature is too low it may lead to non-specific binding on the
template resulting in undesired products due to increased mismatches. The optimal
annealing temperature should be calculated based on the melting temperature of the
primer as well the product. The optimum annealing temperature should be between

7QUESTIONS AND ANSWERS
the range 50-60 °C depending on the composition of primer (Bustin & Huggett,
2017).
Answer 7:
Pre-mRNA formed after transcription undergoes splicing mechanism to form mRNA
trancripts required for proper translation of protein. Splicing depends on the recognition
of cis-sequences of exon-intron and regulatory regions. Any point mutation occurred in
these consensus sequences then it may cause improper splicing recognition of exon and
intron thereby forming aberrant transcripts from the muated genes. This type of mutation
is known as splice-site mutation. Mutation can occur in any region of the intron or the
exon causing disruption of the intronic or exonic splice sites or regulatory region splice
sites. This results in errorneous splicing process and undesired intron removal leading to
changed open reading frames (Anna & Monika, 2018).
Fig 3: splice site mutation: 1: shows that a point mutation is present in the junction of intron
and exon 2 which leads to improper recognition by the splicing mechanism and
misincorporation of intron in the altered mRNA whereas in 2, point mutation has is present in
exon 2 which does not allow the splicing enzymes to recognize exon 2 and thus the altered
2

8QUESTIONS AND ANSWERS
mRNA has excluded exon 2 during splicing (Image retrieved from National Cancer Institute,
2020).
Answer 8:
Genetic screnning is a process used for identifying the function of a gene and
understand gene interaction. Screening is mainly done by introducing mutation in genes and
assessing the desired phenotype in the affected organism. It can be classified as forward
genetic approach and reverse genetic approach. In forward genetic approach random
mutations are in the gene are screened to understand which genes are responsisble for a
particular phenotype. On the other hand reverse genetics is applied where specific genes are
targeted for mutation to find out the phenotype produced. Forward genetic approach is also
known as “pehenotype to gene” approach whereas reverse genetic approach is also known as
“gene to phenotype”. These approaches are applied in animal model experiment where
diseases are identified by experimenting in animal models such as mouse. Forward genetics is
useful is finding out the causes of many human diseases and its characteristics. Gene
targeting or knockout in mouse is an example of reverse genetic approach to find out the
function of the gene. Disruption of a gene will associate with the disease responsible for its
absence (Gondo et al., 2017).
Answer 9:
Replication is the first process of central dogma where the double stranded DNA
(dsDNA) is copied into new set of double stranded DNA before it is divided into the daughter
cells making two pairs of DNA strands from one of paternal DNA strand. The process begins
with recognition of the ori region by an enzyme known as DNA A, which opens the duplex at
the ori site. The ori site is rich in A-T base pairs. This facilitates the binding of another
enzyme called helicase that that unwinds the dsDNA forming two single strands. The structre
formed is known as replication fork. To stabilise these single strands, single stranded binding

9QUESTIONS AND ANSWERS
proteins (SSBs) binds to the strands and removes any tortional strain and renaturation of
DNA. Another enzyme known as topoisomerase II or DNA gyrase that requires ATP to
introduce negative supercoiling by cleaving both strands, passing a segment through the
break to the other site and then seals the cut. In the presence of gyrase and SSBs, primase
comes into action that adds primer (a short stretch of RNA nucleotides) to the
complementary regions of the single strands. This is the initiation phase of replication. Primer
addition will be followed by addition of complementary nucelotides to the single strands right
after the primer. There two strands of DNA known as lagging and leading starnd based on the
direction of nucleotide addition. Addition of nucleotides to the leading strand occurs from 5’
to 3’ in a continuous process and requires only one primer but the synthesis of lagging strand
is a discontinuous process away form the replication fork requiring many primers bound to its
complementary strand. Finally the elongation process ceases and the primers are removed by
DNA polymerase I. DNA pol I also fills in the gap of primer site and does proof reading of
the correct incorporation of nucleotides. The enzyme ligase joins the phosphodiester bonds
between the nucleotides and seals any gap. The replication process is terminated when the
replication fork reaches Ter sites or termination sites (Duderstadt et al., 2014).
Fig 4: replication process (Image retrieved from A Level Biology, 2020)
Answer 10:
Sanger sequencing is a first generation sequencing which is absed on chain
termination by incorporation of dideoxynucleotides to find out an unknown sequence. The

10QUESTIONS AND ANSWERS
reaction mixture requires single stranded DNA template, specific primer sets, enzyme that is
DNA poylmerization for addition of nucelotides, four types of dNTPs or deoxynucleotides
and ddNTPs or dideoxynucleotides that terminates the chain elongation due to the absence of
hydroxyl group at the 3’ of the ribose thereby not forming phosphodiester bond with the
incoming dNTP. The unknown DNA sequence samples are separated equally into four tubes,
where each of the test tubes contain all components of the reaction reaction but each test tube
differs in the content of ddNTPs. Every test tube will contain only type of dNTP. The
reaction follows the PCR amplification where the final product is visualized under gel
electrophoresis that reveals the sequence by indication of the positions of four lanes relative
to each test tube reaction. The sequence is read form the lower band to the top. On the other
hand pyrosequencing which is a second generation sequencing where the nucleotide sequence
of the DNA are determined by release of pyrophosphate molecules. It follows similar reaction
components such as sanger sequencing that include DNA template, dDNTPS, DNA
polymerase and primers but additional compnents are required to detect the release of light
based on chain reaction. The additional components are ATP sulfurylase, luciferase and
apyrase. The reaction begins as addition of dDNTP next to the primer by DNA polymerase
where apyrase removes any dNTP not incorporated or mismatched. Sequential addition of a
nucleotide bases A, T, G and C releases PPi for every correct base pairing. This
pyrophosphate (PPi) is converted to ATP by ATP sulfurylase, further ATP acts as a substrate
for luciferase mediated reaction where luciferin is converted to oxyluciferin and light is
generated for each nucleotide addition (Ihle et al., 2014).
Answer 11:
Allele dropout is the loss of one allele during amplification in polymerase chain
reaction of DNA. The reason for allele dropout is the failure of amplification of one allele. It
is difficult to recognize the mistake in the PCR reaction because PCR reaction is successful

11QUESTIONS AND ANSWERS
but the gentic information is not available. Allele dropout is the reason for causing
homozygosous or heterozygosous mutations.
To determine the mechanism of allele dropout, a study was conducted for genotyping
of a human gene that is MEST. PCR analysis of the short region in the promoter sequence of
MEST revealed that led to the genotype patterns of non-mendelian type with regard to three
single nucleotide polymorphisms or SNPs. This problem could not be resolved by primer
redesign or standard optimization of PCR. It was established that presence of methylation on
cytosine and certain DNA structures such as G-quadruplexes in the promoter region was
responsible for allele dropout leading to misinterpreted genotyping. These guanine bases
forms a connection between each other through hoogsteen bonds. G4 structures comprise of
four guanine nucelotides that stack upon one another and this structure formation is
facilitated by the cations present in the buffer. These structures inhibit the Taq polymerase
activity and the affect is intensified when these G4 regions are methylated (Stevens et al.,
2017).
LONG ANSWERS:
Answer 1:
a) Gene expression profiling of carcinogen X should be performed with the help of RT
PCR or reverse transcriptase PCR which will quantify the gene expression or amount
of transcript from the gene expression of X.
b) The procedure to be followed for RT PCR of carcinogen X is as follows
 Purification of whole genome from the cells through cell disruption
 Separation of total cellular RNA for conversion into cDNA