DNA Replication and Protein Synthesis Review
1. Describe what each of the following scientists contributed to our understanding of the structure of
DNA.
a. Chargaff - determined that nitrogen bases were found in equal quantities. Adenine is paired with
thymine and cytosine is paired with guanine.
b. Franklin - used x-rays to discovered that DNA was in a helix
c. Avery, MacLeod and McCarthy - Did experiments with rough and smooth bacteria to determine
that DNA was the ‘transforming principle’ not protein
d. Hershey and Chase - Did experiments using radioactivity and viruses to determine that DNA was
the source of hereditary information.
e. Watson and Crick tied everything together – A = T, C = G and that DNA is in a double helix
2. Describe the structure of a nucleotide. A deoxyribose sugar, a phosphate and a nitrogen base.
3. The replication of DNA is called ‘semiconservative’. What does this mean? How does the cell benefit
from replicating in this way?
Semiconservative replication means that one strand of the original is used as a temple for copying the
second strand. This saves energy and materials (because only one strand is made) and reduces the
possibility of error because there are bases to match during replication.
4. Compare or contrast:
a. chromatin to chromosome - a strand of uncoiled DNA, that can be replicated or used to create
proteins is a chromatin. A coiled chromatin, which is inactive, is a
chromosome.
b. sister chromatid to homologue - a sister chromatid is one half of a chromosome (either the original
strand or its copy). A homologue (or homologous chromosome) is
one version of a chromosome pair.
5. What are two benefits of organizing genetic information into chromosomes? - increased genetic
variation, improved genetic stability, increased complexity, allows the organism to be diploid.
6. Compare the structure of prokaryotic DNA to eukaryotic DNA - Prokaryote structure: circular shape
with no individual chromosomes, one replication site with most of the genome expressed.
Eukaryote structure: double helix shape separated into individual chromosomes. Multiple
replication sites with only a portion of the genome expressed.
7. Describe the function of each of the following enzymes, then identify when (initiation, elongation or
termination) this enzyme carries out its function.
a. helicase - opens and untwists the DNA strands. Initiation.
b. topoisomerase - keeps the opened strands of DNA from twisting back up. Initiation.
c. binding proteins – keeps the opened strands of DNA from closing back up. Inititation.
d. DNA polymerase - gathers up and glues the individual nucleotides (3 at a time) together, creating
the new DNA strand. Elongation.
e. RNA primase - glues the RNA primers to the start of the leading strand and at the beginning of each
Okazaki fragment.
f.
RNA primer - serves as a ‘hook’ for the incoming nucleotides to attach to at the start of the leading
strand and at the start of each Okazaki fragment.
g. Ligase - glues the Okazaki fragments together.
8. What is photolyase? How does it correct DNA replication mistakes? UV light splits the bond between
thymine and adenine and causes the thymines to attach to each other. Photolyase is an enzyme that
splits the thymine and reattaches them to the adenines, correcting the DNA mutation.
9. How does excision repair correct DNA replication mistakes? An enzyme cuts the incorrect base and a
base on either side out and removes it from the strand. DNA polymerase comes in and replaces these
three bases with the correct ones.
10. Compare the structure of RNA to the structure of DNA. The backbone of DNA is made up of a
deoyribose sugar, a phosphate and a nitrogen base. The backbone of RNA is the same, except the sugar
is a ribose sugar. The backbone of DNA is a double helix, in RNA it is a single helix. IN DNA, adenine is
paired with thymine. In RNA, adenine is paired with uracil.
11. Pre-messenger RNA is the ‘product’ created during transcription. It’s modified by snRNA
into mRNA. Translation follows this modification, with the final product of this process
being a protein (polypeptide).
12. For each of the following types of RNA, describe its function and when (transcription, translation or
in-between) it performs this function.
a. mRNA - translates the information provided by DNA into something that can be understood by the
other forms of RNA. Transcription.
b. tRNA - reads the mRNA strand, goes into the cytosol, gathers up the nucleotides (3 at a time) and
brings them back to the ribosome. Translation.
c. rRNA - creates a ribsome. As a ribosome, it glues the nucleotides together, creating an amino acid.
Translation.
d. snRNA - cuts the introns out of mRNA and glues the exons together, creating mRNA. In-between
transcription and translation.
13. Compare the process of:
a. ‘initiation’ in DNA replication to ‘initiation’ in transcription.
Initiation – DNA: Helicase opens and unwinds the DNA strand. Topoisomerase and binding proteins
keep the strand from returning to its original shape. Primase glues RNA primers
onto the leading strand and at the first Okazaki fragment.
Initiation – RNA: A promoter opens and unwinds a small portion of DNA. RNA polymerase is attached to
the promoter.
b. ‘elongation’ in DNA replication to ‘elongation’ in transcription.
Elongation – DNA: nucleotides are brought to the DNA strand (3 at a time) and glued in by DNA
polymerase until all of the strand has been matched.
Elongation – RNA: nucleotides are brought to the RNA strand (3 at a time, ‘codons’) and glued in by
RNA polymerase. All work is done in the ‘transcription bubble’. The DNA strand
winds up as the transcription bubble moves.
c. ‘termination’ in DNA replication to ‘termination’ in transcription.
Termination – DNA: when all nucleotides have been matched, RNA primers are replaced by
nucleotides and glued in, completing the strands.
Termination – RNA: when the AAAAAA code has been reached, no more nucleotides are
added. RNA polymerase is released and the DNA strand finishes winding up.
14. Compare:
a. polypeptide to protein - a polypeptide and a protein is the same.
b. codon to anti-codon - 3 nucleotides of mRNA make up a codon. 3 nucleotides of tRNA make up an
anticodon.
c. intron to exon – 3 nucleotides of mRNA that don’t code for any information (blanks) are introns. 3
nucleotides of mRNA that code for information (create an amino acid) are exons.
15. Describe the structure of a ribosome. Ribosomes are made up of ribosomal RNA. A ribosome is made
up of two parts, a small subunit at the bottom and a large subunit at the top.
16. How do retroviruses violate the ‘central dogma of molecular biology’? The central dogma of molecular
biology says that information pass from DNA to RNA which is made into proteins. Retroviruses violate
this because they don’t have DNA of their own. They only have RNA. They use their RNA to make an
organism’s DNA produce the proteins they need.
17. What happens during the ‘initiation’ step in translation? Initiation begins at the mRNA sequence AUG.
Initiation begins when a small ribosomal subunit attaches to the end of an mRNA strand just in front
of this codon. Once the small ribosomal subunit is in place translation begins. A tRNA carrying the
amino acid methionine (also known as the anticodon UAC) attaches to the mRNA at the “P” site.
A large ribosomal subunit then attaches to the mRNA forming a complete ribosome with the tRNA
that’s already there.
18. What happens during the ‘elongation’ step in translation? Begins when tRNA returns with the next
amino acid and attaches it to the mRNA at the “A” site of the ribosome. The first tRNA lets go of the
methionine and leaves the mRNA to go back and get another amino acid. The methionine is
attached to the amino acid that was just placed on the mRNA. The tRNA (and the mRNA it’s bonded
to) move over from the “A” site to the “P” site. A new amino acid fills the now empty “A” space – the
tRNA in the “P” space is released – the two amino acids are attached and the polypeptide chain
(protein) gets longer. The cycle continues until all of the mRNA strand has been translated.
19. What happens during the ‘termination’ step in translation? When the ribosome encounters one of
the three stop codons, termination starts. The completed polypeptide , the last tRNA and the
ribosomal subunits are released so the process can start again.
20. Compare a point mutation with a frameshift mutation. A point mutation is a change in a single base.
(Example: a ‘c’ is replaced by a ‘t’). If this isn’t corrected, a single amino acid is changed. In a
frameshift mutation, a base is added or deleted. This causes all the bases (and the amino acids) to be
incorrect.
21. Which of the four types of mutations can have the greatest affect on an organism? Why?
Translocation will have the greatest affect because it involves multiple bases on different
chromosomes.
22. Would DNA polymerase or RNA polymerase have a higher accuracy rate? Why?
DNA polymerase has a higher accuracy rate than RNA polymerase. This is because DNA
replication cannot tolerate many errors, but it is relatively easy to replace incorrect RNA
molecules with new, correct RNA molecules.
Possible essay questions:
1. A) Describe the structure and function of the parts of a eukaryotic chromosome.
B) Describe the evolutionary importance of organizing genes into chromosomes.
C) How does the structure and function of the chromosome differ in prokaryotes?
2. A) Explain the role of each of the following in protein synthesis in eukaryotic cells.
 RNA polymerase
 snRNA
 Codons
 Ribosomes
 tRNA
B) The central dogma of molecular biology does not apply to some viruses. Explain why.
3. A) Compare/contrast the process of DNA replication to the process of protein synthesis.
B) There are two types of mutation possible during protein synthesis, point and framework.
Describe:
 one of each of these mutation types
 what affect each would have on the resulting RNA strand
 which would have a greater affect on the organism and why.