3.5/7.4 Translation – part ii

10/11/2012 § Leave a comment

For the detailed rundown of how translation works, click here.

Otherwise, the rest of this blogpost is only a list of answers to the chapter questions of this unit. These have been checked and are all correct answers now.

Note: the actual assigned DBQ can be found at the very end of this post; no. 15 of the Chapter 6 questions on page 75.


Chapter 5 Questions

1. With reference to Figure 17, answer the following questions.

  • a) What part of the nucleotide is labelled A? The phosphate group.
  • b) What kind of bond forms between the structures labelled B? Hydrogen bonds.
  • c) What kind of bond is indicated by label C? A phosphodiester bond, which is a covalent bond.
  • d) What sub-unit is indicated by label D? That’s the sugar unit, either a deoxyribose or a ribose, but the book says it’s the entire nucleotide. It could be both, though!
  • e) What sub-unit is indicated by label E? Nitrogenous base, but the book says it’s the 3′ end, which is not correct.

2. Figure 18 shows a short section of DNA before and after replication. If the nucleotides used to replicate the DNA were radioactive, which strands in the replicated molecules would be radioactive?

Strands II and III.

3. Complete the table by filling in the missing base percentages.

30 20 20 30
22 28 28 22
24 26 26 24
24 26 26 24

4. Give a synonym for the term “nucleoside monophosphate”.

nucleotide. BAM.

5. State the name of the enzyme with each of these functions.

  • a) Creates an RNA primer during replication. RNA primase
  • b) Removes the primer at the end of replication. DNA polymerase I
  • c) Seals gaps between fragments on the lagging strand. DNA ligase
  • d) Unwinds DNA prior to replication. helicase

6. Refer to Figure 19 when answering the following questions.

  • a) State what molecule is represented. deoxyribose – a sugar molecule
  • b) State whether the molecule would be found in DNA or RNA. DNA
  • c) State the part of the molecule to which phosphates bind. part V and III
  • d) Identify the part of the molecule that refers to the 3′ end. part III


a) Outline the steps involved in DNA replication.

At the origin of replication (there are multiple on a strand of DNA), the enzyme helicase binds to the DNA strand and unwinds the helix structure of the DNA by breaking the hydrogen bonds between the nucleotides. Following this, DNA polymerase III attaches nucleotides (that are present everywhere) to a growing strand of new DNA based on complementary base pairing (A-T, G-C), connecting them with new phosphodiester bonds by hydrolyzing nucleoside triphosphates. Continuous replication happens on the leading strand in a 5′ to 3′ direction. Lagging or fragmented replication happens on the lagging strand in a 3′ to 5′ direction with the help of RNA primase (which makes the primer that starts of the lagging replication), DNA polymerase III, DNA polymerase I that gets rid of the primer later, and ligase (which adds the final phosphodiester bond). 

b) Describe the structure and function of nucleosomes.

Nucleosomes are bodies of eight histones, a type of protein. Nucleosomes are a group of four pairs of histones, to be exact, quartets stacked on top of each other. These help to coil the structure of DNA because the long strands of DNA wrap around the nucleosomes twice (with the help of a linker protein) to create its structure. This is what coils DNA and what is uncoiled to make chromatin during replication. It also helps regulate gene expression and stabilizes DNA structure.

c) Compare RNA and DNA.

characteristic DNA RNA
type of sugar deoxyribose (lacks extra oxygen molecule) ribose (contains extra oxygen molecule)
nitrogenous bases A, T, G, C A, U, G, C
number of strands double-stranded single-stranded

8. State two functions of nucleosomes.

  1. they supercoil DNA to prepare for mitosis or meiosis
  2. they regulate transcription/gene expression within a cell

9. Figure 20 shows the structure of part of a nucleic acid molecule.

a) State, with a reason, whether the molecule is part of RNA or DNA.

This is a DNA molecule because it doesn’t have an extra oxygen molecule in its sugar molecule (H instead of OH).

b) State, with a reason, whether I is the 3′ or 5′ terminal.

I is the 3′ terminal because it is the free 3-prime end where a DNA strand may end, and it is where the new nucleotide will connect with the previous nucleotide, via the phosphate group. 

c) II is guanine. State whether it is a purine or a pyrimidine base.

It’s purine, ohhh, yeah.

Chapter 6 Questions

1. In which process is transfer RNA (tRNA) involved? D: translation

2. Which two processes involve the unwinding (uncoiling) of the DNA double helix and its separation into two strands of nucleotides? D: transcription and replication

3. Where is an intron found? B: RNA

4. A certain gene codes for a polypeptide that is 120 amino acids long. Approximately how many nucleotides long is the mRNA that codes for this polypeptide likely to be?

c) 360 nucleotides/bases

5. A concept map involves drawing lines between associated concepts, and labeling the lines to show the relationships between the connected concepts. Create a concept map that links the following terms: DNA, nucleotide, mRNA, tRNA, rRNA, protein, amino acid, ribosome.

Creds to my Pa’s phone. // Click to enlarge.

6. Explain the terms universal and degenerate as they relate to the genetic code.

The genetic code is degenerate because there are 64 possible codons and different codons can code for the same amino acid, i.e. CUU and CUC can both code for LEU, and there are three codons that are the code for stopping translation. The genetic code is universal because this code can be found in all forms of life, from viruses, to plants, to the most complex of animals. This code and amino acids are what form proteins, and proteins are what constructs all organisms – whatever they may be. There are some exceptions but nearly all life in Earth uses this code.

7. Given a certain amino acid sequence, to what extent can the following be predicted?

a) The sequence of the mRNA that coded for it. You can trace only three of the nucleotides that formed the amino acid. You would be able to narrow it down to the few that code for that amino acid but not the precise sequence.

b) The sequence of the gene that coded for it. You can trace only the three nucleotides that formed the amino acid from the mRNA strand, and from that, you find the complementary base pairings (on a DNA strand) back to the original DNA code, but that would only give you 3 bases still and there are still a few options so you wouldn’t be able to find the exact sequence. It would also be harder because of the introns present in eukaryotes. 

8. A nucleic acid is composed of nucleotides. What is the composition of each of the following?

  • a) a gene – composed of 1000 bases/nucleotides
  • b) a codon – three bases/ribosome nucleotides
  • c) a nucleotide – a sugar molecule (ribose/deoxyribose), a phosphate group, and a base
  • d) a polypeptide – amino acids in a chain

9. Compare transcription and translation.

Both transcription and translation are part of protein synthesis. Both move in a 5′ to 3′ direction. Transcription and translation both involve enzymes and nucleotides that originally come from the strands of DNA molecules found in the cell. Both processes are done within the cell but transcription is only done inside the nucleus while translation is in the cytoplasm or on the endoplasmic reticulum. Transcription is the replication and rewriting of the original DNA sequence into the complementary RNA sequences while translation is the process that takes those sequences and turns them into proteins, or polypeptides (chains of amino acids). Translation involves ribosomes while transcription does not. The enzymes used for transcription and translation are different – in fact, translation doesn’t need any enzymes – so, the units required for the two processes are different (i.e. transcription requires DNA polymerase I and III, ligase, and RNA primase while translation requires ribosomes, tRNA, and an abundance of amino acids). 

10. Describe the genetic code.

The genetic code is the language of DNA where the nucleotide bases are the letters that make up the words – codons. Codons are three-lettered words (three bases) that code for specific amino acids. Since there are four different types of bases (A, G, C, T/U) and three spots, there are a total of 64 possible codons. There are only 20 amino acids that make up proteins, so there are more than enough codons to code for each protein. The genetic code is used by tRNA molecules that pick up specific amino acids for translation and the production of proteins. The genetic code is degenerate and universal (#6). 

11. Determine two possible DNA sequences that could code for: Phe-Ile-Val-Leu.


~ Imagine an HL ribbon here. ~

12. Outline how the physical structure of tRNA is related to its function.

tRNA is structured so that it can play its part in translation. Every tRNA molecule has a CCA sequence on the 3′ end so that an amino acid can attach to the tRNA by the tRNA activating enzyme (which recognizes the tRNA by its shape and chemical properties). tRNA has some sections that are double stranded because of complementary base pairings and hydrogen bonds that loops the tRNA’s structure. The three-dimensional or clover-leaf shape ensures correct binding. tRNA also has a loop of seven bases on one end which is the anticodon loop. The three bases at the very end of the anticodon loop make up the actual anticodon of the tRNA and is what differentiates tRNA amongst each other. This anticodon determines what the amino acid of the tRNA is and is what binds to the mRNA strand during translation to provide the amino acid it carries for the growing polypeptide chain.

13. Compare the role of enzymes in the processes of transcription and replication.

All of the enzymes in the processes of transcription and replication are different. In replication, the helicase is what unzips the DNA strands but in transcription, RNA polymerase does that. DNA polymerase III is what adds new nucleotides to the growing daughter strand but RNA polymerase also does that in transcription. DNA replication also needs other enzymes like ligase, RNA primase, and DNA polymerase I because it can be done continuously (therefore it’ll only need the DNA polymerase III) in a 5′ to 3′ direction or in fragments – in a 3′ to 5′ direction, in which case it will need all of those enzymes. Transcription is done entirely by RNA polymerase, which is a very versatile enzyme. 

14. Place the following events of translation in the correct sequence:

  • a) joining of the small and large ribosome subunits
  • b) binding of the met-tRNA to the start (AUG) codon
  • c) covalent bonding between two amino acids
  • d) formation of polysomes
  • e) binding of a second tRNA to the A-site

b, a, e, c, d

15) α-neo-endorphin is a natural painkiller that was discovered in the hypothalamus. It is an oligopeptide, consisting of ten amino acids linked together by peptide bonds. In the early 1980s, Shoji Tanaka and a team of molecular biologists designed a gene with a base sequence that would give the desired amino acid sequence when translated. In one of the earliest examples of genetic modification, they synthesized this gene and inserted it into the bacterium Escherichia coli. The bacterium successfully synthesized α-neo-endorphin. The base sequence of the sense strand of the gene is: ATGTATGGCGGTTTCCTTCGTAAGTATCCGAAGTAA

a) The base sequence of the mRNA produced by E. coli from this gene was not identical to that of the sense strand. Predict the differences in base sequence.

G, and C, are still complimentary and T from the original gene will match will still be matched with an A in the E. coli gene, but A will have partnered up with U on the mRNA strand produced by the E. coli. 

b) Deduce the amino acid sequence of α-neo-endorphin from the base sequence of the gene.



c) Explain why the base sequence that was inserted into E. coli is not the only one that codes for α-neo-endorphin.

It isn’t the only one that codes for α-neo-endorphin because between the twelve amino acids coded for in the sequence above, all of them have other codons available that can code for the same exact codon. Knowing that, there are many other amino acid sequences similar but not exactly like the first sequence that will still create the same chain of amino acids. This is because of the degenerateness of the genetic code.

d) Predict how the amino acid sequence is modified, after translation, to produce the final form of α-neo-endorphin.

The amino acid sequence is modified when one amino acid is removed in between transcription and translation (post-translational modification).

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