Chapter 10. From DNA to Protein: Gene Expression
1. Add these labels to the following depiction of a cell
2. Describe the structure and function of each of the following types of RNA: mRNA,
rRNA, microRNA and tRNA
mRNA. It is the messenger RNA which is produced when one of two DNA genes
on a strand is transcribed to guide the synthesis of the other RNA strand. The
nucleotide sequence of the mRNA determines the order of the sequence of
amino acids on the polypeptide chain.
rRNA. These are made of multiple proteins and ribosomes. One of the functions
is to catalyze the formation of the peptide bond the amino acid with the protein
that is being synthesized.
microRNA. Is a small noncoding RNA molecule. It does exert the regulation of
the gene expression, when bind to an mrNA, the union formed cannot be
translated.
tRNA. Transfer RNA transports a specific amino acid to the ribosome. Each
tRNA molecule has triple of nucleotides that bind to specific triplet of nucleotides
on mRNA. In this manner, the sequence of codons in mRNA determines the
sequence of amino acids in proteins.
3. A dog was taken to a veterinary clinic because it has become increasingly
inactive and lethargic. A blood test determined that the dog had very low levels
of the steroid hormones produced by the adrenal glands, especially cortisol. The
veterinarian told the owner that the dog has “a mutation in its cortisol gene”.
Explain why this characterization is inaccurate, given that steroid hormones are
lipids. Speculate on a way that a genetic mutation could result in low levels of
steroid hormones.
It is inaccurate because genes provide codes for the synthesis of proteins
including enzymes, then low levels of cortisol could be because of a mutation in a
gene that codes for an enzyme needed for the synthesis of cortisol.
4. Describe the genetic condition of a person who has sickle-cell anemia. Use
specific details about which gene is mutated, the different effects of whether the
person is heterozygous or homozygous for the mutation, and include information
about the polypeptide that has been altered by the mutation.
This is a heritable blood disorder in people that carry two copies of the sickle
allele blood protein in humans that carries oxygen. The sickle allele differs from
the normal because it has polypeptide that differs by one amino acid from the
normal protein which causes sickle-shaped red blood cells, particularly when
oxygen levels are reduced in the blood. This cn block narrow capillaries, which
result in pain because of tissue damage.
5. Describe two possible consequences for a gene’s expression resulting from a
mutation in:
a) The promoter region of that gene.
Inhibit the ability of DNA to bind to RNA polymerase. Inability of the enzyme
to bind to the DNA, as a result the gene would not be expressed.
b) The stop codon of that gene.
A mutation of the stop codon of a gene could prevent the release of the
completed polypeptide from the translation complex. This mutation could also
include longer than normal RNA and protein.
6. Describe what happens to the “intron” and “exon” sequences of pre-mRNA when
it undergoes processing.
Pre-mRNA processing involves cutting introns out of the transcript and splicing
together the exon transcripts.
7. Describe the functions of the GTP cap and the poly-A tail on mRNA.
The cap is a chemically modified GTP molecule that helps bind the mRNA to the
ribosome. The poly-A tail aids in export of mRNA from the nucleus and
enhances stability.
8. For this fragment of an mRNA molecule
5’AUG UUU CAG CGA GGA UGA 3’
Use the codon table to characterize the peptide product from translating this
sequence.
Start with codon methionine-phenylalanine-glutamine-arginine-glycine-ending
with a stop codon
9. Describe how the result would change if the above sequence were altered so
that 6th base from the 5’ end was switched to G.
UUG is leucine, the second codon replaces phenylalanine with leucine.
10. Describe how the result would change if the above sequence were altered so
that 19th base from the 5’ end was switched to A.
The fragment shown has 18 bases, so 19th would be an addition. Because a
codon is composed of 3 nucleotides, a single “A” would not provide enough
information for translation process to continue.
11. Even though the amino acid serine is specified by UCU, UCC, UCA, UCG, AGU
and AGC, is it claimed that “the code” is not ambiguous. Explain.
Serine can be produce in response to any of these 6 codons, but each of these
encodes one amino acid. An ambiguous codon would be one with 2 or more
different amino acids.
12. Explain the comment, “the genetic code is nearly universal”, in terms of
evolutionary ancestry.
It is called universal because it is used by all species in Earth, which provides
enough evidence that it has been maintained through the evolution of living
organisms.
13. Explain why the human gene for insulin can be inserted into an E. coli bacterium
and the bacterium can then produce human insulin.
The E. coli bacterium can be manipulated to produce human insulin by
introducing the human gene for insulin into the bacteria. The universal genetic
code is the reason why this is possible.
14. Distinguish between the consequences of “silent” and “nonsense” mutations.
Silent mutation can occur because of the redundancy of genetic code but,
because it may also translate into the production of the same amino acid, there
would be no phenotypic difference. Nonsense mutation could result in a
premature stop codon where the polypeptide chain would be shorter than the
original protein.
15. Explain the relationship between codons and anticodons.
The codon and anticodon bind together via hydrogen bonds, same way than two
strands of DNA, this biding is associated with the delivery of the amino acid on
the tRNA molecule.
16. Describe the activity of a single ribosome, and then describe how more than one
ribosome can be active in that same process.
The ribosome is the molecular platform where the translation of mRNA occurs.
Polyribosomes allow more than one ribosome to move along a strand of mRNA
at one time.
17. For the ribosome shown to the right, state the anticodon sequence for the next
tRNA to bind (at the bottom of the figure) and then use the table from question 8
to find what will be the next amino acid incorporated into the peptide.
It will be UAC.
18. Compare the actions of RNA polymerase and ribosomes.
RNA polymerase is an enzyme that catalyzes the formation of RNA from a DNA
template. RNA carries the genetic code to the ribosomes and then the
ribosomes carry out the actual synthesis of amino acids.
19. Secreted proteins, including those that act as hormones, typically interact with
two membrane-bound organelles prior to secretion. Describe the activities of the
two organelles in terms of preparing the polypeptide for secretion out of the cell.
Secreted proteins interact with rough endoplasmic reticulum and plasma
membrane. If a newly formed polypeptide lacks of a signal sequence for
destination within the endomembrane system, it will usually be secreted from the
vesicles that fuse with the plasma membrane.
20. Human insulin has 51 amino acids and yet it is a product of the INS gene, which
has nearly 40,000 base pairs. Explain how this is possible.
A gene contains the entire nucleic acid sequence for the synthesis of any given
protein like insulin. The gene must also include all the DNA sequence necessary
for the transcription process including all non-coding instructions necessary for
insulin to be produced, packaged and delivered for suitable functioning in the
cell.
21. Add these labels to the diagram below:
RER phospholipid bilayer mRNA polypeptide ribosome
Phospholipid bilayer - inside of cell - lumen of rer - mRNA - polypeptide - ribosome
22. Researchers have determined that a short chain polypeptide signal (nuclear
localization signal, NLS) comprised of eight amino acids must be attached to a
protein if it is to enter the nucleus. When the NLS-protein complex docks with a
pore in the nuclear membrane the signal causes the pore in the nuclear
membrane to open. The sequence for this NLS peptide is: -Pro-Pro-Lys-LysLys-Arg-Lys-Vala) Write out a mRNA stand that could produce this sequence.
5’-CCU-CCU-AAA-AAA-AAA-CGU-AAA-GUU-3’
b) What is the minimum number of DNA nucleotides needed to produce this
sequence?
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c) Explain why methionine is not part of the above sequence.
Methionine has the codon AUG which is the message for a ribosome that
signals the initiation of protein translation from mRNA
d) Write out the DNA sequence that this mRNA sequence comes from.
3’-AAC-TTT-ACG-TTT-TTT-TTT-AGG-AGG-5’
e) Is this the only possible DNA sequence that could code for this NLS?
Explain your answer.
No, the codons in this peptide chain have redundant codons that would
produce the same amino acid.
23. Below are the results of an experiment used to determine the function of the
NLS: In scenario Y to the left, an NLS-protein-red-dye complex that was injected
into a cell is later found in the nucleus. In scenario Z, the protein-red-dye
complex, lacking the NLS peptide, is injected into the cell, and it is later found
only in the cytoplasm. For each of the three scenarios below A, B and C, predict
where the protein-NLS complex will be found when injected into a cell, and
explain why.
a) A cytosolic protein, normally found only in the cytoplasm, is bound to the
NLS peptide.
It will be found in the nucleus because the nuclear localization signal
involves a localization tag (a specific amino acid sequence), interacting
with the nucleus.
b) A nuclear protein is attached to a mutated form of the NLS missing the
final valine.
The correct signal sequence is necessary to bind to the receptor protein of
the nucleus. Without this sequence, a channel will not form and the protein
is unable to move into the nucleus.
c) A nucleaer protein is attached to a NLS that was produced from the DNA
sequence: TAC-GGG-GGT-TTT-TTC-TTC-GCT-TAC-CAC-stop
This sequence contains two starts codons before stop. This mutations
makes the NLS ineffective because it lacks of the proper sequence to
provide a signal for it to bind to the receptor protein on the surface of the
nucleus.