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MCB 250 Spring 2024
Discussion Week 5 – Feb. 22-26
1.
A promoter for an E. coli
gene that is transcribed by a -70 RNA polymerase has the following
sequence:
-30 -20 -10 +1 | | | | 5’GGCTTTACACTTTATGCTTCCGGCTCGTATGTTGTGTGGA
3’CCGAAATGTGAAATACGAAGGCCGAGCATACAACACACCT The transcription start site (+1) is identified.
a.
Identify the –10 and –35 sequences. How close are they to the consensus –10 and –35
sequences?
The -10 sequence 5' TATGTT 3' and the -35 sequence 5' TTTACA 3' are regions within a
promoter sequence. The consensus -35 sequence, which is the most common sequence found
at the -35 region, is 5' TTGACA 3'. When comparing the given -35 sequence 5' TTTACA 3' to the
consensus, we can see they are very similar, with only a single base difference. This close
similarity to the consensus sequence suggests the given promoter sequence is likely a strong
promoter that can effectively initiate transcription.
b.
What is the spacing between the –10 and the –35 sequences? How does this compare with
the consensus spacing? They are separated by 17 bases, i.e consensus spacing
.
c.
The sequence of bases in a transcribed RNA is identical (except for U's instead of T's) to the
non-template strand. Explain.
- The template strand of DNA is the strand that is used as the template for mRNA synthesis.
The mRNA sequence is complementary to the template strand.
- The non-template strand has a sequence that matches the mRNA transcript, except that in
RNA, thymine (T) bases are replaced by uracil (U) bases. So the non-template DNA strand and
the mRNA transcript are essentially identical in sequence, with U substituted for T. - The template and non-template DNA strands are complementary to each other with A-T and
G-C base pairing.
1
So in essence, the mRNA transcript matches the non-template DNA strand except for
substituting U where there are T bases. This allows the mRNA to convey the same genetic
code information as the non-template DNA strand.
d.
Define promoter strength. The promoter's strength is defined by its ability to bind with RNA polymerase for the
execution of transcription. The degree of binding strength directly influences the extent of
gene transcription.
e.
Predict the effect of the following mutations on the strength of this promoter (stronger,
weaker, no change). The changes indicated show the sequence as top strand/complementary
strand.
1)
Replacement of CTT/GAA at –23 to –20 with AAG/TTC.
No change bc promotor ends bind to -10 and -35 sites only.
2)
Replacement of G/C at –9 with A/T.
This will give sequence 5' TATATT 3'. This sequence is pretty much similar to the original one - 5'
TATAAT 3'. That's why it will be a strong promoter.
3)
Replacement of T/A at –12 with C/G.
This will give the sequence as 5' CATGTT 3'. This sequence is less similar to the original JAta sequence.
Hence, the promoter will be weak.
4)
Replacement of G/C at –38 with C/G.
No change. Promotor wont bind
.
5)
Insertion of AT/TA between –18 and –17.
No change. Promotor wont bind.
2.
Alternative splicing is a key feature of the eukaryotic transcription-translation process.
a)
In eukaryotes, alternative splicing of pre-mRNAs can result in the production of many
structurally distinct proteins from information encoded in a single gene. Explain.
Alternative splicing is a mechanism wherein segments (exons) of pre-mRNA are rearranged in
different combinations, resulting in the generation of multiple mRNA molecules from a single gene.
These mRNA variants give rise to distinct protein isoforms. The splicing machinery identifies splice
sites, allowing for the selective inclusion or exclusion of specific exons. This dynamic process
contributes to the diversification of protein functions by introducing variations in structure, function,
and interaction partners.
2
b)
Do lower eukaryotes (such as yeast, Saccharomyces cerevisiae
) have the same opportunity to
use alternative splicing to generate a vast array of proteins from a single gene? Explain.
In lower eukaryotes such as yeast, alternative splicing is a phenomenon that occurs, but it is less
frequent and complex compared to higher eukaryotes. This is primarily attributed to the simpler gene
structures and fewer introns present in yeast genomes. Instead of relying heavily on alternative
splicing for protein diversity, yeast employs alternative mechanisms. These mechanisms include gene
duplication, diverse gene expressions, and post-translational modifications like phosphorylation or
acetylation. These alternative strategies contribute to the generation of protein diversity in yeast,
compensating for the relatively limited role of alternative splicing in comparison to higher eukaryotes.
c)
What other events can lead to additional diversity in the repertoire of proteins that could be
encoded by a single gene? Hint: think of the N and C terminal ends. What about a post-
translational mechanism. Name one. Indeed, various splicing-related mechanisms play crucial roles in enhancing protein diversity.
Alternative start codon usage allows for the translation of different N-terminal sequences, affecting
protein localization and function. Alternative polyadenylation generates mRNA transcripts with
varying 3' UTR lengths, impacting mRNA stability, localization, and translation efficiency. Additionally,
alternative exon usage at the C-terminal end can lead to variations in protein domains or interaction
sites.
Furthermore, post-translational modifications, including phosphorylation, glycosylation, acetylation,
and ubiquitination, modify protein properties, stability, and cellular localization. These modifications
contribute significantly to the diversification of functional proteins, adding an additional layer of
complexity to the regulation of cellular processes. The combined effects of these splicing-related
mechanisms and post-translational modifications contribute to the intricate and diverse landscape of
functional proteins in cells.
3
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Related Questions
15
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Urgently needed
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Lesson 2
Focus Questions
1. What chemicals and molecules are needed for PCR, and what is the function of each
component?
2. Examine the 150 base promoter sequence below.
Kaylee Kauff
5'TAGAAAAGGA AGGTGGCTCC TACAAATGCC ATCATTGCGA TAAAGGAAAG
GTATCATTC AAGATGCCTC TGCCGACAGT GGTCCCAAAG ATGGACCCCC
ACCCACGAGG AGC ATCGTGG AAAAAGAAGA CGTTCCAACC ACGTCTTCAA3'
Write in the sequence of the complementary strand and mark the 3' and 5' ends of the
complementary strand.
43
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G-Unit 5: DNA anc X
https://docs.goog x
y google classroom x
A Classes
QLSeyPC6Kmoa0k5JJd1DWGzqqRwaQQobHNOOdqFX_aDbV_6-bKw/formResponse
of Distance Le...
Launch Meeting - Z..
4 2020 HORNETS 4N...
StudentVUE
Clear s
How is a protein made in the cell? *
One strand of DNA in the ribosome combines with amino acids.
Two strands of DNA in the nucleus combine with amino acids.
One strand of RNA in the ribosome is the template (model) for an amino acid
sequence.
Two strands of RNA in the nucleus are the template (model) for an amino acid
sequence.
Use the chart to determine the correct amino acid that this DNA would
code for - ATA
EGFL
Y.
UCAGUCAG
Gutame
Aspart
ac
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1_30*_SP23 - General Biology I (for majors)/1
of
us page
The anticodon sequence created from the following DNA: TACGGGGCTGAGATT
F1
Select one:
O a. Tyr-Gly-Ala-Glu-lle
O b. AUGCCCCGACUCUAA
c. UACGGGGCUGAGAUU
O d. Met-Pro-Arg-Leu-STOP
F2
#
80
F3
$
000
000
F4
%
F5
MacBook Air
F6
&
r
F7
DII
F8
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Final Assessment for Exploring M X
forms/d/e/1FAIpQLSe8RrfvHEr59pISGjbEneL041GxKTvyw9-Xc_EA7Um_8xqReA/viewform?hr_submission-Chg1990yzQ8SEAjwvY285...
First mRNA base (5) end of codon)
A
Sequence C
Sequence D
5. A DNA sequence of "ACG" will code for the amino acid
UUG
CUC
CUA
CUG
wh
GUC
GJA
GUG
Cys
100
AUU
AUC lle ACC
AUA
ACA
AUG MACG]
Val
M
Second mRNA base
UCA
UCG
The
GCU
GCC..
GCA
GIGA
CAA
CAG
Tyr
UAC
UGC
UAA Stop UGA Stop A
UAG Stop UGG Tro
AAU
AAC
MT
AAC
GAC
GAG
2
Lys
Asp
CGA
CGG
AGU
AGC
AGA
AGG
Lys
GGU
GGC
GGA
GGG
Arg
DUAU
Gly
C
Third mRNA base (3' end of codon)
Lys (K)
DELL
C
GU
A
C
DEOTOCCAGUCAGUCAGUCHOSOTOS
M
G
(F)
A
(LS1-1) *
GU
A C
CUGA
OPCUGACU GACUSACCO
AD
A
G
OCO
Trp (W)
\u«
9799%3Fo
33
4
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4:39 PM Thu Jul 25
<
Cames
A
Help Center?
Spokane
Today 10:10 AM
You study cells in which DNA synthesis rates are influenced by a ligand that binds to a receptor tyrosine kinase (RTK: must polymerize
and cross-phosphorylate to transmit its signal). To determine which protein(s)-A, B, C, or D- are responsible for the activation of DNA
synthesis, you construct mutant versions of the RTK that retains one or more of the tyrosines that are the protein docking sites (+
means the tyrosine is present and gets phosphorylated," means the tyrosine is not present). You express these mutant versions in cells
and measure the level of DNA synthesis.
(A)
(B)
740
751
BP
771
B
1009
1021
100-
50-
山
protein
P-site
1
2 3 4
5 6 7
8 9
A
740, 751
B
771
C
1009
D
1021
First: Maintenance of which docking protein(s) binding site is required maximize DNA synthesis? [Select]
Second: Which protein has no detectable role in DNA synthesis? | Select]
MacBook Pro
Edit
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Question e
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Drag the labels onto the diagram to identify the anatomical terminology describing body orientation and direction (human a
Posterior
(caudal)
Distal
Superior
(dorsal)
Proximal
Inferior
(caudal)
Anterior
(ventral)
Posterior
(dorsal)
Pearson
arch
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Questions and situational problems
I. A particular gene codes for a mature mRNA containing 900 bases, which is
translated into a 40 kDa protein. A mutant form of the gene created by a single
point mutation yields an 830-base mature mRNA yielding a 37 kDa protein with
modified enzymatic activity. The analysis shows that the mutation has resulted
in a 22 amino acid deletion within the protein. What is the most likely effect of
the mutation? Explain.
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Problem #2
DNA
TACT TAA A AAT G
A
(TS)
DNA
(CS)
MRNA
Amino Acids:
Please continue to the next page.
Activity 2- Reinforcing Protein Synthesis
Use your codon chart to determine the amino acid sequence. Read through the strand and ONLY start on AUG
and STOP when it tells you to stop. Follow example below:
Example:
DNA > AGA CGG TAC CTC CGG TGG GTG CTT GTC TGT ATC CTT CTC AGT ATC
MRNA > UCU GCC AUG GAG GC ACC CAC GAA CAG ACA UAG GAA GAG UCA UAG
protein >
start - glu - ala -thre - hist – asp -glu – threo - stop
1. DNA → CCT CTT TAC ACA CGG AGG GTA CGC TAT TCT ATG ATT ACA CGG TTG CGA TCCATA ATC
MRNA
protein >
2 DNA → AGA ACA TAA TAC CTC TTA ACA CTC TAA AGA CCA GCA CTC CGA TGA ACT GGA GCA
mRNA >
pratein >
3. DNA > TAC CTT GGG GAA TAT ACA CGC TGG CTT CGA TGA ATC CGT ACG GTA CTCGCC ATC
MRNA >
protein →
4. Fill in the diagram below.
DNA
MRNA
ERNA
Amino
Acids
Please continue to the next page.
Universal Genete Code Chart
Mensenger RNA Codans and the Amina Asd for Which They Code
SECONDBASE…
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Question In Image Thank you!
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Questions related to SARS-CoV-2 Inhibitor Science Article (Zhang et al.)
1) What is the function of the viral main protease?
2) What is the amino acid recognition sequence that is cleave by this main protease? Indicate where
this protease cuts?
3) Would interfering with this protease effect other proteases in human cells?
4) How closely related are the amino acid sequences between SARS-CoV-2 and SARS-COV? What does
this suggest about focusing research on inhibiting this enzyme?
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Task #3: Design primers
1 agagtctcct cagacgccga gatgctggtc atggcgcccc gaaccgtcct cctgctgctc
61 tcggcggccc tggccctgac cgagacctgg gccggtgagt gcgggtcggg agggaaatgg
121 cctctgccgg gaggagcgag gggaccqcag gcgggggcgc atgacctcag gagccgcgcc
181 gggaggaggg tcgggcgggt ctcagcccct cctcaccccc aggctcccac tccatgtggt
241 atttctacac ctccgtgtcc cggcccqgcc gcggggagcc ccgcttcatc tcagtgggct
301 acgtggacga cacccagttc gtgaggttcg acagcgacgc cgcgagtccg agagaggagc
361 cgcgggcgcc gtggatagag caggaggggc cggagtattg ggaccggaac acacagatct
421 acaaggccca ggcacagact gaccgagaga gcctgcggaa cctgcgcttc tactacaacc
481 agagcgaggc cgttgcqtga ccccqgcccg gggcgcaggt cacgactccc catcccccac
541 gtacggcccg ggtcgccccg agtctccggg tccgagatcc gcctccctga ggccgcggga
a) First you'll need to design primers to PCR-amplify amino acids 45-56.
i)
Record the sequences of the forward and reverse primers, in the 5' to 3' direction. Each
primer must be 8 nucleotides long (note that normally primers are much longer than this).
Show intermediate steps…
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PLEASE HELP, LONG ANSWERS WOULD BE APPRECIATED
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Question:-
1. Transcribe and translate the given region of DNA, note that the promoter region is to the right of the sequence, and translation must start from the start codon. Include 3' and 5' for RNA transcript and N- and C- terminus in your final peptide sequence. Use the 1-letter form of amino acids when writing the sequence of polypeptide
5' CCCCAGCGTAAGTTTATGGTTACTCATGAA 3'
3' GGGGTCGCATTCAAATACCAATGAGTACTT 5'
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Short questions
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Practice question 1D)
In the boxes on the image, label the N and C termini of the csRu protein. .
Also on the schematic above, draw the csRH protein as you would find it at the plasma
membrane. Be sure to label the N and C termini of the protein you draw to indicate its
orientation.
The csRu MRNA and protein are represented in the diagram below.
mRNA 5'
AUG
3'
ribosome
protein
DNA
ligand
binding
domain
signal
kinase
transmembrane
binding
domain
sequence
domain
domain
cytoplasm
plasma membrane
extracellular fluid
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- Lesson 2 Focus Questions 1. What chemicals and molecules are needed for PCR, and what is the function of each component? 2. Examine the 150 base promoter sequence below. Kaylee Kauff 5'TAGAAAAGGA AGGTGGCTCC TACAAATGCC ATCATTGCGA TAAAGGAAAG GTATCATTC AAGATGCCTC TGCCGACAGT GGTCCCAAAG ATGGACCCCC ACCCACGAGG AGC ATCGTGG AAAAAGAAGA CGTTCCAACC ACGTCTTCAA3' Write in the sequence of the complementary strand and mark the 3' and 5' ends of the complementary strand. 43arrow_forwardG-Unit 5: DNA anc X https://docs.goog x y google classroom x A Classes QLSeyPC6Kmoa0k5JJd1DWGzqqRwaQQobHNOOdqFX_aDbV_6-bKw/formResponse of Distance Le... Launch Meeting - Z.. 4 2020 HORNETS 4N... StudentVUE Clear s How is a protein made in the cell? * One strand of DNA in the ribosome combines with amino acids. Two strands of DNA in the nucleus combine with amino acids. One strand of RNA in the ribosome is the template (model) for an amino acid sequence. Two strands of RNA in the nucleus are the template (model) for an amino acid sequence. Use the chart to determine the correct amino acid that this DNA would code for - ATA EGFL Y. UCAGUCAG Gutame Aspart acarrow_forward1_30*_SP23 - General Biology I (for majors)/1 of us page The anticodon sequence created from the following DNA: TACGGGGCTGAGATT F1 Select one: O a. Tyr-Gly-Ala-Glu-lle O b. AUGCCCCGACUCUAA c. UACGGGGCUGAGAUU O d. Met-Pro-Arg-Leu-STOP F2 # 80 F3 $ 000 000 F4 % F5 MacBook Air F6 & r F7 DII F8arrow_forward
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