Answers to HW Questions
Text Ques. Pg.241
• The DNA would be subjected to
degradation, damage or distortion. The
DNA is the original blueprint and must
remain in the nucleus to be protected.
• The central dogma of molecular genetics
states that genes found within a DNA
sequence are transcribed into m-RNA
which then exits the nucleus & is
translated into protein by ribosomes in the
cytoplasm.
• Ribosomes are the site of protein
synthesis
• M-RNA is formed by transcription and
carries the message from the DNA to
direct the protein being formed.
• T-RNA contains the anti- codon and
delivers the amino acids to the ribosome
• R-RNA – helps to form & maintain
ribosomes.
Transcription
Translation
Location
In the nucleus
Purpose
Copies a DNA
sequence (a
gene) into a
template that
can be used by
the ribosome to
build a protein.
In the cytoplasm
(ribosomes)
Uses the mRNA
template in order to
build the specified
protein.
• A start codon (AUG) tells the ribosome
where to begin the production of a
polypeptide chain.
• A stop codon tells the ribosome when a
polypeptide chain ends. (UAA,UGA,UAG)
• CCU AGU UCC AGG UCC GUU AAA
UCG UAC GGG GUU
11. Using the genetic code, decipher the following mRNA
sequence:
5 - GGCAUGGGACAUUAUUUUGCCCGUUGUGGUGGGGCGUGA - 3
• DNA sequence divided into codons
• 5'–GGC-AUG-GGA-CAU-UAU-UUU-GCC-CGUUGU-GGU-GGG-GCG-UGA–3'
• The start codon is the second codon, AUG;
therefore, translation into protein commences
with the amino acid methionine.
• Translation results in the following protein:
• Met-Gly-His-Tyr-Phe-Ala-Arg-Cys-Gly-Gly-Ala
• The last codon, UGA, is a stop codon.
13. The amino acid sequence for a certain peptide is
Leu–Tyr–Arg–Trp–Ser. How many nucleotides are necessary
in the DNA to code for this peptide?
• Since there are five amino acids found in
this polypeptide, there must be at least five
codons. Since each codon consists of
three nucleotides, 15 (5×3) nucleotides
would be required to code for this peptide
sequence.
Ques. Pg.249
1. Explain the role of the following in transcription: RNA
polymerase,
.
poly-A polymerase, and spliceosomes
• RNA polymerase plays many roles in the process of
transcription. It recognizes the promoter region upstream
of a gene to
• be transcribed and binds to this site. Binding to the
promoter region results in DNA’s double helix opening
up. Once
• bound to the DNA template strand, RNA polymerase
starts to build the mRNA complementary strand using
• ribonucleotides. Finally, RNA polymerase recognizes the
termination sequence that signals the end of a gene and
ceases
• transcription.
#1 cont’d
• Poly-A polymerase plays a role in the posttranslational modification of mRNA. It adds
approximately 200 to 300 adenine nucleotides to
the 3' end of the primary mRNA transcript
• Splicesomes found in eukaryotic organisms cut
introns (noncoding regions) from the primary
mRNA transcript and anneal the remaining
exons (coding regions) together.
2.A short fragment of a particular gene includes the
following sequence of nucleotides:
3- TACTACGGTAGGTATA - 5
Write out the mRNA sequence.
• DNA 3'–TACTACGGTAGGTATA–5'
• RNA 3'–AUGAUGCCAUCCAUAU–5'
3. A short fragment of a particular gene includes the
following sequence of nucleotides:
3- GGCATGCACCATAATATCGACCTTCGGCACGG - 5
(a) Identify the promoter region. Justify your answer.
(b) Explain the purpose of the promoter in transcription.
• (a) 3' –
GGCATGCACCATAATATCGAACCTTTCGGCA
CGG – 5'
• The red area represents the promoter region.
The promoter region contains a high
concentration of A and T bases. Since A and T
only share two double bonds between them,
RNA polymerase will expend less energy in
opening up the double helix at this point. (
Known as the tata region)
# 3 Cont’d
• (b) The purpose of the promoter is twofold
in the process of transcription. First, the
promoter lies upstream of a DNA
sequence that represents a gene. Hence,
it acts as a signal for RNA polymerase to
bind and transcribe the gene found
downstream. Second, it also is an area
where the DNA double strand is able to be
unwound more easily because of its high
concentration of adenine and thymine.
4. Differentiate between introns and
exons.
• Exons are segments of DNA that code for
part of a specific protein, while introns are
the noncoding regions of a gene.
5. In eukaryotes, mRNA must be modified before it is
allowed to exit the nucleus.
(a) Describe the modifications that are made to eukaryotic
primary transcript.
• The modifications that are made to the primary
mRNA transcript include capping and tailing and
the excision of introns. Capping involves the
addition of a 7-methyl guanosine to the 5' end of
the primary mRNA transcript. Tailing consists of
the addition of 200 to 300 adenine nucleotides to
the 3' end of the primary mRNA transcript by
poly-A polymerase. The excision of introns is
carried out by splicesomes, which then join the
remaining exons together.
5.(b) Explain how these modifications ensure that
mRNA survives in the cytoplasm and is translated
into a functioning protein.
• The capping and tailing of the primary mRNA
transcript ensures that when the transcript exits
the nucleus, it is not degraded by nucleases and
phosphatases found in the cytoplasm. Capping
also plays a role in the initiation of the process of
translation. Introns are excised to ensure that
when the mRNA transcript is translated into
protein, it does not contain amino acids that are
extraneous. These extraneous amino acids
would interfere with the proper folding and
functioning of the protein.
8. Explain the ramifications to the process of transcription if
the following occurs:
(a) The termination sequence of a gene is removed.
(b) Poly-A polymerase is inactivated.
• 8. (a) If the termination sequence of a gene is
removed, RNA polymerase will continue to
transcribe and build a primary mRNA transcript
beyond the gene.
• (b) If poly-A polymerase is inactivated, the 200 to
300 protective adenine nucleotides that are
added at the end of a primary mRNA transcript
will no longer be added. This will leave the
mRNA susceptible to degradation on exiting the
nucleus.
8. (c) The enzyme that adds the 5 cap is nonfunctional.
(d) Spliceosomes excise exons and join the remaining
introns together
• (c) If the enzyme that adds the 5' cap is
dysfunctional, the mRNA is susceptible to
degradation on exiting the nucleus.
• (d) If splicesomes excise exons and join
introns together, then the mRNA transcript
will consist of noncoding sequences.When
the mRNA is translated, it will produce a
nonfunctional protein.
8.(e) The RNA polymerase fails to recognize the
promoter sequence.
• (e) The RNA polymerase fails to recognize
the promoter sequence.
Ques. Pg. 254
1. Differentiate between the following:
(a) P site and A site
(b) codon and anticodon
• (a) The A and P sites are found in a ribosome
that is translating an mRNA sequence into
protein. The A (acceptor) site is where tRNA
molecules bring in the appropriate amino acid.
The P (peptide) site is where peptide bonds are
formed between adjoining amino acids on a
growing polypeptide chain.
• (b) A codon is a triplet of bases on mRNA that
encodes a single amino acid. An anticodon is a
triplet of bases on tRNA that recognizes and
pairs with a codon on the mRNA.
3. List the possible anticodons for
threonine, alanine, and proline.
Amino
Possible
Acid
codon
ACU
Threonine ACC
Possible anti-codon
ACA
ACG
UGA
UGG
UGU
UGC
Alanine
GCU
GCC
GCA
GCG
CGA
CGG
CGU
CGC
Proline
CCU
CCC
CCA
CCG
GGA
GGG
GGU
GGC
4. Errors in the process of transcription are occasionally
made. Explain why an error in
the third base of a triplet may not necessarily result in a
mutation. Provide an example to support your answer using
the genetic code.
• An error in the third base of a codon in mRNA
may not necessarily result in an error during the
process of translation because more than one
codon encodes a particular amino acid. The
codons differ by the third nucleotide. For
example,proline can be encoded by the codons
CCU, CCC, CCA, and CCG. If a mistake is
made in the third nucleotide of the codon, it is
negligible. It does not matter what the third
nucleotide is—the two first nucleotides, CC, will
always code for proline.
6. The following sequence was isolated from a fragment of mRNA:
5 - GGC CCA UAG AUG CCA CCG GGA AAA GAC UGA GCC CCG - 3
Translate the sequence into protein starting with the start codon.
• The genetic code (Figure 7, Section 5.2, p.
240 of the Student Text) must be used. AUG
is always the start codon and encodes for
the amino acid methionine:
• 5'–GGC-CCA-UAG-AUG-CCA-CCG-GGAAAA-GAC-UGA-GCC-CCG–3'
• Met–Pro–Pro–Gly–Lys–Asp–Stop