Name____________________________
ALE #6 DNA replication, transcription, translation
DNA Structure
1. a. What is a nucleotide? Why are nucleotides important, that is, what do your cells use them for?
A subunit of a nucleic acid. Contains a 5-carbon sugar, a phosphate, and a nitrogenous base. Our cells use them for
DNA & RNA, which store our genetic information
b.
Label the three main components (phosphate, sugar, nitrogen base) of the simplified representation of a nucleotide
below.
Phosphate
Nitrogenous base
Sugar
c.
Indicate the nitrogen bases found in DNA and RNA by completing the table below.
Nitrogen Bases Found in DNA
Nitrogen Bases found in RNA
Adenine
Adenine
Thymine
Uracil
Guanine
Guanine
Cytosine
Cytosine
DNA Replication
2. a. Why is it DNA replication necessary for all organisms on earth today?
DNA replication is necessary so that cells can divide for growth and repair. Replication also allows genetic
information to be passed from parent to offspring during the division.
b. When during the cell cycle does DNA replication occur? S phase
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3.
DNA Replication. Use the hypothetical representation of a double stranded DNA molecule, below, to complete the
following tasks.
a. Complete the base sequence of the complementary strand of the hypothetical DNA molecule diagrammed below.
b. Use dashed lines to indicate hydrogen bonding between paired bases.
c. Show how this molecule would be replicated:
o Draw the molecule partially “unzipped” while undergoing replication, followed by the resulting daughter
molecules with their correct nucleotide sequences and base pairing.
o Use two colors, one for the template (or parent) strands, and another for the newly synthesized daughter
strands.
You can check your answer to the question above by comparing your figure to Figure 10.6 in text
4. a. Name the enzyme responsible for lengthening each strand during DNA replication by repeatedly adding
nucleotides to the end of each strand.
DNA polymerase
b.
How does this enzyme “know” which nucleotide to use during DNA replication—that is, what “rules” does the
enzyme follow?
Adenine pairs with Thymine, Guanine pairs with Cytosine
c.
Name the enzyme that proofreads and corrects any errors it finds in the DNA strands after the completion of DNA
replication.
DNA polymerase
d.
What is a mutation? What are the four causes?
Any change in the nucleotide sequence of DNA. The first cause is random errors during the replication process, as
when DNA polymerase miss-matches a base pair. Replication occurs very fast and not all mistakes are corrected by proofreading. The other three casues are viruses, toxins, and radiation.
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Relating DNA Replication to the Cell Cycle
5. a. As in mitosis, the chromosomes are duplicated prior to the start of meiosis I. How many duplicated chromosomes
will there be in a human “pregamete” cell just before the start of meiosis I? 46 duplicated chromosomes.
(23 homologous pairs, and each chromosome is replicated, so it looks like an “X”)
b. How many DNA molecules are in a human “pre-gamete” cell just before the start of meiosis I? 92 DNA molecules
During DNA replication, we start with a single molecule of double-stranded DNA. We end with two molecules of
double-stranded DNA. the daughter molecules are exact copies of one another and they are attached at a centromere,
which mean they are still considered 1 chromosome. Thus for this question we have 46 duplicated chromosomes,
which means we have 92 DNA molecules. This would be the same as asking how many sister chromatids to we have
(92).
6.
c.
Just after meiosis II and cytokinesis (the division of the cytoplasm) how many chromosomes are there in each
gamete? 23. Meiosis results in ½ the number of chromosomes as the parent cell. Remember – Any time you see
the word “gamete,” think about and egg or a sperm. They must have ½ the number of chromosomes because they
get together to make a regular diploid body cell (the zygote).
d.
Just after meiosis II and cytokinesis how many DNA molecules are there in each gamete? 23. Remember that in
Anaphase II, sister chromatids split, and so each of the resulting 4 daughter cells has 23 unreplicated
chromosomes.
Diploid “pregamete” cells from a hypothetical mammal were examined under a microscope and found to contain 4
chromosomes (2 homologous pairs).
a.
Draw one of these cells in metaphase I of
meiosis. Consider the size and shape of your
chromosomes.
b.
A drug is given to this animal so that the S phase
(DNA replication) does not occur. Draw one of
these cells in metaphase I of meiosis. Draw your
chromosomes so that the size and shape is
consistent with the cell you drew in part a.
Plasma
Membrane
Plasma
Membrane
XX
XX
7. Explain why AZT prevents another nucleotide from being added to a growing DNA strand and then explain how AZT
will affect DNA replication.
The N3 portion of AZT disrupts the sugar-phosphate backbone & so another nucleotide cannot bond properly to the
growing strand. The replication of the DNA molecule will be discontinuous – it will have “nicks” that it cannot repair,
resulting in a damaged chromosome.
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8. Most nerve cells do not replicate their DNA upon reaching maturity. Suppose that a cell biologist measured the amount
of DNA in several different types of human cells:
1) Nerve cells
2) Sperm cells
3) Bone cells just starting interphase
4) Skin cells in the S phase
5) Intestinal cells just beginning mitosis
She found x amount of DNA in the nerve cells. Use this fact and the information in the table below to identify
Cells A - D.
Cell Type
Nerve Cell
Cell A
Amount of DNA in Cell
x
2x
Cell B
1.6x
Cell C
0.5x
Cell D
x
Identity of Cell
Nerve Cell
intestinal cells beginning mitosis (they have replicated
already, so they should have 2x the number of DNA
molecules of a regular cell)
Skin cells in S phase (they are undergoing the process
of replication but have not completed it yet – thus 1.6x
instead of 2x)
Sperm cells (any cell with half the amount of DNA
would be a gamete –egg or sperm)
Bone cells at interphase (this part of the cell cycle is
before replication)
9. Fill in the following spaces concerning the “central dogma” of biology. Use the following terms only once:
(phenotype, nucleotides, DNA, amino acids, protein, shape ) A gene is a section of ___DNA____________ that
contains the code for the synthesis of one particular ______protein_____________. The order of
______________nucleotides______________ in a gene determines the order of nucelotides in mRNA, which
determines the order of ______amino acids____________________ in the protein the gene codes for. This order
controls the ____________shape__________ of the protein, which in turn determines the function of that protein. An
organism’s genotype determines the kind of proteins an organism can make. These proteins determine an organism’s
_____________________phenotype____________.
10. Compare the differences in RNA and DNA by completing the table below.
DNA
Number of Strands
Name of sugar in
nucleotides
RNA
2
1
Deoxyribose
ribose
ATCG
AUCG
Bases present
Where produced in cell
nucleus
nucleus
Where found in cell
nucleus
mRNA: nucleus, the cytoplasm (at
ribosomes on rough ER
tRNA: cytoplasm
rRNA: ribosomes
Name of process that
makes it
replication
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transcription
11. Here is a hypothetical gene showing the sequence of DNA nucleotides for the coding strand (i.e. the coding strand is
the strand that is transcribed). IMPORTANT!! This sequence includes the regions that code for start and stop in
translation—that is, locate both the start codon and stop codon in the mRNA before you translate it into protein!!
Coding strand of DNA:
3’ T A G G T A C T G G G G C A T T A A 5’
a.
What would be the sequence of bases in the resulting mRNA if this strand of DNA was transcribed?
mRNA: AUCC AUG ACC CCG UAA UU
b.
What is the amino acid sequence of the protein that this gene codes for? (You will need the table of codons in your
textbook to answer this question.)
The amino acid sequence is: Met – Thr – Pro
AUG was the start codon, and it codes for Met. UAA was
the stop codon, so it codes for nothing. Thus there should only be 3 amino acids here.
12. Outline the basic steps from DNA to the formation of proteins: DNA  RNA  Protein. For each step indicate where
it takes place in the cell, the name of each process involved, what is needed for each process to occur, the names of the
major enzymes involved, etc. Be sure to include the major events of each process. Refer to the steps as I outlined in
lecture, and take a look at the figures for transcription and translation in your text book.
Transcription – takes
place in the nucleus, requires: DNA, the subunits for RNA (ribose, phosphate, and AUGC), RNA polymerase.
Translation – takes place in the cytoplasm, at the ribosomes on the rough ER. Requires: Ribosomes, mRNA, tRNA,
amino acids. We did not discuss the enzymes involved in translation.
Transcription:
1.
Initiation - RNA polymerase enzyme binds to the promotor (section of DNA indicating “start of a gene”)
2.
Elongation – RNA polymerase catalyzes base pairing on the template strand (U-A, G-C)
3.
Termination – RNA polymerase reaches the “stop” sequence and the new mRNA is released.
4.
mRNA processing – non-coding regions of the mRNA are removed and the mRNA leaves the nucleus.
Translation –
1. Initiation – mRNA start codon binds to tRNA anticodon; Ribosome binds to both
2. Elongation – tRNA brings specific AAs to the ribosome as mRNA passes through the ribosome (codon –
anticodon recognition) Polypeptide chain forms
3. Termination – Ribosome reads an mRNA stop codon (no tRNA with anticodon). mRNA and protein detach
from the ribosome
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13. Hemoglobin is a protein in your red blood cells that is responsible for carrying oxygen. A mutation in the gene that
codes for hemoglobin leads to a disease called sickle cell anemia. Sickle cell hemoglobin is unable to carry oxygen
effectively, resulting in weakness in individuals who inherit one copy of this gene, death in results if the faulty gene is
inherited by from both parents. There are over 300 known mutations in the hemoglobin genes. One of these mutations
causes a condition called Hemoglobin C disease, which is not as serious as sickle cell anemia. You will need the table
of codons in your book to answer some of these questions.)
a.
Below is the part of the sequence from the coding strand of DNA for three variants of the hemoglobin gene. Circle
the mutation in the sickle cell sequence and the hemoglobin C sequence. I have underlined the mutations in red
instead of circling them

sickle cell anemia:
3’...T G A G G A T T C C T C...5’
ACU CCU AAG GAG

hemoglobin C disease: 3’...T G A G G A C A C C T C...5’
ACU CCU GUG GAG

b.
normal hemoglobin:
3’...T G A G G A C T C C T C...5’
What is the mRNA sequence of the normal hemoglobin gene?
mRNA of normal hemoglobin: ACU CCU GAG GAG
c.
What is the amino acid sequence of the normal hemoglobin gene?
The amino acid of normal hemoglobin: Thr – Pro – Glu - Glu
d.
e.
Exactly what is the difference between the normal hemoglobin protein and the hemoglobin protein in a person
with sickle cell anemia and a person with hemoglobin C disease?

normal hemoglobin vs. sickle cell: Underneath the coding strands above I wrote in the corresponding mRNA
codes. Based on this, the AA sequence for the sickle cell disease is: Thr – Pro – Lys – Glu. The normal
hemoglobin AA sequence is Thr – Pro – Glu – Glu. Thus we have a difference of one AA for the sickle –cell
disease

normal hemoglobin vs. hemoglobin C: The AA sequence for hemoglobin C disease is Thr – Pro – Val – Glu,
based on the mRNA code above. Again, this means there is a one AA difference for the hemoglobin – C
disease
As mentioned above, there are over 300 known mutations in the hemoglobin genes. While many of these
mutations lead to diseases, some of these mutations do not change the ability of the hemoglobin protein to do its
job. Describe how it is possible that a mutation in the hemoglobin gene doesn't affect how the hemoglobin protein
works. Substitution of a third base pair in a triplet code may not change the AA it codes for, since there is
redundancy in the genetic code. For example, GUA, GUG, GUC, and GUU all code for Valine.
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14. Cystic fibrosis is due to the inheritance of two faulty CFTR alleles. The normal CFTR allele codes for a membrane
protein found in cells involved with secretion, for example: mucous secreting cells in the respiratory tract, sweat
glands in the skin, digestive enzyme secreting cells of the pancreas, etc. The CFTR protein is actually a
glycoprotein, that is, it is a protein that has been modified to by the addition of several monosaccharides.
a. Trace the pathway of the normal CFTR protein from the organelle where CFTR is made to its final location in the
cell membrane.
ribosome
 ER vescicle
Golgi
 Golgi vescicle
 Plasma membrane
b.
The function CFTR protein is to pump chloride ions (i.e. salt) into the cells lining the lungs and cells lining various
ducts in the body. In cystic fibrosis, the protein never makes it to the plasma membrane of these cells. Use this
information to explain why people with CF experience the following symptoms:
 Salty sweat – no chloride ion channel (the CFTR proteins) to regulate the salt balance of the sweat glands

Mucous collects in the lungs – no chloride ion channel (the CFTR proteins) , so salt builds up in the lungs,
water enters the lungs to dilute the salt, provide a warm, wet medium in which bacteria can grow. White
blood cells attack the bacteria. All of the these are components of mucus.

Chronic respiratory infections – mucous build up in the lungs traps bacteria and viruses – they are not able to
be removed fast enough to prevent infections



Problems digesting food - no chloride ion channel (the CFTR proteins) at the pancreatic duct, and so this duct
becomes clogged with salts/mucous
Reduced fertility in women – We did not discuss this in class. Women need to have a certain quality
(viscosity) of cervical fluid in order for sperm to be able to live for several days around ovulation and to travel
to the egg. Without the chloride ion channels, the cervical fluid may become too thick and block the sperm’s
ability to travel to the egg
Sterility in males - no chloride ion channel (the CFTR proteins) at the vas deferens, which becomes clogged
with salt, blocking the movement of sperm.
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