Biology Essay Example
BIO3309 Molecular Biology Problem Set #2
BIO3207 CELL & MOLECULAR BIOLOGY 2
COURSE ASSESSMENT WEIGHTING = 20%
: Thursday 10th October 2013 DUE
(Failure to submit by due date may incur a late penalty as per course specifications)
STUDENT NAME: ___________________________________
100 = / 5= / MARK STUDENT NUMBER: ____________________
[10 marks total] / 10
A dideoxynucleotide sequencing gel obtained from the beginning of a certain DNA segment is shown below.
The nucleotide sequence of this DNA section is 5’GAACATCAGGATCGTACCAGGGTAA3’
2 INTRONS & EXONS
[5 marks total] / 5
single loop lying under molecule shown)note:(
.eight(a) According to the figure, an estimate of introns in the gene is
.intron ends and a base which is conserved located in the intronsEukaryotic pre-mRNA splice sites are indicated by two sequences that are conserved at the
(b) AG/GU is the typically designating sequence at the mRNA 5′ end.
.(c) At the 3′ end of a mRNA, NCAG/ is the typical designating sequence
(d) The base A is very important in splicing and is present within the introns.
3. RESTRICTION ENZYMES
[15 marks total] / 15
(a key to amino acids is given on page two of this report).MWVYME? cleave a segment of cDNA that encodes the peptide definitelyIndicate (by circling the appropriate number(s)) which, if any, of the restriction enzymes listed below will
I GTACRsa (ii)
I AATTY) Apo(iii
4. [10 marks] / 10
Albumin and a-fetoprotein are present-day members of the same protein family. Shown below is the exon-intron map of the genomic DNA coding for each of these proteins.
Using the diagram above (and arrows ↓ to indicate any similarities) indicate the likely relationships between the present-day albumin and a-fetoprotein genes.
NB: exons have expanded slower than introns.
remember exons are likely to be conserved but may be subject to processes like duplication resulting in novel new proteins)hint: that coded for the related primordial protein from which the present day albumin and a-fetoprotein genes are likely to have arisen (primordial gene(ii) Considering that evolution works from simple to complex, draw a diagram of the likely first
5. [10 marks total] / 10
The sequence of a cDNA encoding a 10 amino-acid protein is shown below:
(N.B. This sequence is intentionally presented as 3’->5’ because it is the sequence of the DNA replicate arising from the reverse transcription of a mRNA. The standard convention is to present nucleotide sequences 5’->3’)
3’ CTCTTACGGTACTCTACGCCATTGCTTTTATACTATCTAATCCTGA 5’
Using different colours, on the sequence above:
codons in this cDNA START the ALLBox the
codons in this cDNASTOP the ALL ircle C
5’ CTCTTACGGTACTCTACGCCATTGCTTTTATACTATCTAATCCTGA 3’
that this sequence encodes.10 amino acid protein the coding sequence for the Underline
The three translation start codons (3′)TAC(5′) are contained in this cDNA. Actually this is in correspondence with the mRNA sequence in the polarity that is opposite called AUG. An open reading frame must be used for following the actual translation initiation site simply because (3′)ATC(5′) , (3′)ATT(5′) and (3′)ACT(5′) codons are not in it corresponding to UAG, UAA and UGA mRNA sequences in the opposite polarity. The sole open reading frame coding for a protein 40-aa starts with the second triplet set in in row tow and continues till two ATT stop codons are attained. The coding sequence is as shown; the first stop codon next to the three start codons is enclosed inside the box while the start codons are circled.
6. [10 marks] / 10
ESTIMATING GENE NUMBERS
proteins theoretically could be expressed by the haploid human genome than by the haploid E.coli genome?more, how many E.coli DNA encodes protein and that the regular size of proteins is same in humans and E.coli cells comprise an average four genomes totalling 0.017pg of DNA. Supposing that every the sequence in E.coli The best-educated guess today is that not more than about 1 percent of the human genome constitutes protein-coding sequences (exons). A diploid somatic human cell comprises 5 picogram (pg) of DNA and fast developing
2.5 pg of DNA/haploid genome are present in each human cell. There are 0.025 pg of protein-coding DNA/haploid human genome if 1 percent of this codes for protein. 0.00425 pg DNA per haploid genome, or 0.017 pg DNA/4 gemotes are there for an E.coli cell that is rapidly growing. If the ration is these two values is taken, 5.9 is the result. Therefore compared to E.coli cells somatic human cells have sixfold more capacity in coding of proteins. However as indicated by the calculation of actual ratio of protein-coding is probably more, this is because of the prevalence that is greater of the genes that are tandemly repeated in human cells than in bacterial cells and also the presence of genes that are duplicated.
7. [10 marks total]
EFFECT OF DELETIONS ON GENE TRANSCRIPTION
Analysis of the influence of deletions following gene transcription is a foremost method for distinguishing DNA control elements. The following diagram summarises the effects of different deletions after the transcription of the eukaryotic 5S-rRNA gene. In figure (A), transcribed mutants are indicated with a “+” and those that are not transcribed are indicated with a “-“. Figure (B) represents an archetypal autoradiographic electrophoretogram used in the examination of the deletions.
Based on this data:
Is the transcription of the 5S-rRNA gene dependent on upstream sequences?
As indicated by the data, an upstream of 100bp from the RNA commencing site can be erased without any obvious effect on the molecule size or the 5S rRNA amount. Regarding this, it can be reported that upstream sequences seems to have very little importance for 5S RNA production.
Using figure (a) only, what is the putative location of the control region that regulates
: regions which definitely do not reduce expression are given hintthe 5S-rRNA gene? (
by “+” above )
The region between +50 and +85 deletions eventually eliminates the 5S-rRNA gene transcription. Because of this the control section for this gene has to be between +50 and +85.
8. [10 marks] / 10
REAL TIME PCR
parasites previously extracted from human faeces.Entamoeba histolyticaThe image below shows Real Time PCR analysis of a series of cultures of
Reference: Roy et al. (2005) J Clinical Microbiology 43: 2168-2172.
8(a) [2 marks] / 2
27.25; What is this value?: 10,000 value?: TWhich cell number produced the lowest C
8(b) [4 marks] / 4
Baseline What does the plot with the grey circles represent?
8(c) [4 marks] / 4
Threashold bar represents the line at which ether its intersection with the application demarcate the CTWhat does the threshold bar in the image represent?
9. [10 marks] / 10
(a) You want to amplify the DNA between 2 stretches of sequence as shown below:
GACCTGTGGAAGC ————————— CATACGGGATTG5’
CTGGACACCTTCG —————————- GTATGCCCTAAC3’
You check in the fridge and you find you have a collection of eight primers:
Of the eight primers, one and seven would be the most appropriate forward and reverse primer pair for allowance of the DNA amplification by PCR.
(b) Calculate the melting AND annealing temperatures for the following primer set.
Reverse – TCCTCCGCTTATTGATATGC 58
41 (%GC) + 59.9 – (675/primer length)
59.9 + 41 (0.45) – (675/20) = 44.6
(4 x 9) + (2 x 11) = 58
Forward – CTTGGTCATTTAGAGGAAGTAA
(2 x 14) + (4 x 8) = 60
59.9 + 41 (0.363) – (675/22) = 44.13
(41 (%GC) +59.9 – (675/primer length)
57/42 = Annealing temperature
10. [10 marks] / 10
)330; MW ssDNA = 660 MW dsDNA = Note: (RI digested pBR328 to ensure the best chance of successful ligation? Eco would you add to your 50ng of insert
much for ligations, how insert : vectorRI digested pBR328. On the basis of the optimal ratio of EcoRI site of pBR328 (4.9 Kb) and detecting successful insertions using blue/white selection. You have been given 50 nanograms of EcoYou have been employed to clone a 100bp fragment into the
Convertion of molecular weight to daltons
daltons 6 x 660 = 4.9 Kb =3.234 x 103Vector (pBR328) 4.9 x 10
daltons 4100 x 660 = 6.6 x 10
Moles in vector
Moles = Mass divided by MW
moles -15 = 15.46 x 106g (50ng) / 3.234 x 10-9which is 50 x 10
moles of ends -15as each piece is made up of two ends we multiply by 2 = 30.92 x 10
The mass of insert needed
add x 3 moles of ends (insert:vector = 3:1)
moles of insert -15 = 92.76 x 10-153 x 30.92 x 10
Moles = Mass/MW,
-15 = 92.76 x 104Mass / 6.6 x 10
4 x 6.6 x 10-15hence Mass = 92.76 x 10
-11= 612.216 x 10
-9= 6.12216 x 10
There is a need to divide by 2 as each piece has two ends
this amounts to;
(or 3.06 x 10-9g)3.06 ng
owing to this calculation, 3.06 of insert can be used
AMINO ACID / mRNA CODON TABLE
U Cys UGU Tyr UAU Ser UCU Phe UUU
C Cys UGC Tyr UAC Ser UCC Phe UUC U
A StopUGA Stop UAA Ser UCA Leu UUA
Trp UGG Stop UAG Ser UCG Leu UUG
Arg CGU His CAU Pro CCU Leu| CUU
C Arg CGC His CAC Pro CCC Leu CUC C FIRST
A Arg CGA Gln CAA Pro CCA Leu CUA LETTER
|G Arg CGG Gln CAG Pro CCG Leu| CUG
Ser AGU Asn AAU Thr ACU Ile AUU
Ser AGC Asn AAC Thr ACC Ile AUC A
Arg AGA Lys AAA Thr ACA Ile AUA
Arg AGG Lys AAG Thr ACG Met AUG
U Gly GGU Asp GAU Ala GCU Val GUU
Gly GGC Asp GAC Ala GCC Val GUC G
A Gly GGA Glu GAA Ala GCA Val GUA
Gly GGG Glu GAG Ala GCG Val GUG
Three letter symbols, names (followed by one letter symbols) are:
= aspartic acid(D)Asp= asparagine(N); Asn = arganine(R); Arg = alanine(A); Ala
= glycine (G)Gly= glutamine(Q); Gln= glutamic acid (E); Glu = cysteine(C); Cys
= lysine(K)Lys= leucine(L); Leu= isoleucine(I); Ile= histidine(H); His
= serine(S)Ser = proline(P); Pro = phenylalanine(F); Phe = methionine(M); Met
= valine (V)Val = tyrosine(Y); Tyr = tryptophan(W); Trp = threonine(T); Thr
CHEMICAL PROPERTIES OF AMINO ACIDS
Ala; Val; Leu; Ile; Pro; Met; Phe; Trp : Non-polar
Gly; Ser; Thr; Cys; Tyr; Asn; Gln : Polar
Asp; Glu: Negatively charged (acidic)
Lys; Arg; His : Positively charged (basic)
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