Transcription & Translation; the Effect of Mutations in Biological Sciences

This article explores the effect of mutations on transcription and translation in Biological Sciences. It covers missense, frame-shift, silent, and insertion mutations. The article also provides solved assignments, essays, and dissertation on Desklib. The subject is Biological Sciences with course code BIOL1010. The article is relevant to college and university students.

Biological Sciences (BIOL1010) NAME: _______________________________
ASSIGNMENT #1: Transcription & Translation; the Effect of Mutations
TEXTBOOK PAGES: Chapter 25, section 25.3 pages 130-140
DUE: Monday October 1st, 2018
INTRODUCTION
Gene expression is the process whereby the DNA sequence of a gene is used to synthesize a protein that
has a function in the cell. Through this activity you will follow a wild-type (normal) DNA sequence
through the steps of transcription to produce messenger RNA (mRNA) and translation to produce a
chain of amino acids (called a protein). The normal or wild-type protein product of translation
contributes to normal cell function. Sometimes however, mutations or errors in DNA and/or RNA
sequences can occur resulting in the production of a defective protein with abnormal function in the
cell; the consequence of this being unhealthy cells, disease and even death of the individual. Mutations
may be the result of errors from copying the DNA (i.e. in DNA replication) which can be transferred to
mRNA, damage from ultraviolet light or hazardous chemicals which may damage DNA or RNA directly.
Although DNA/RNA mutations may lead to disease it is important to also note that not all mutations do;
this is particularly true when there is no change to the actual amino acid sequence of the protein
produced. The effect of different types of DNA mutations i.e. missense, frame-shift, silent, and insertion,
will be examined.
PART I
Scenario #1 – Normal (wild-type DNA / RNA sequences) ( /5)
1) The DNA template sequence for a portion of the haemoglobin gene is shown below. This is the wild-
type (normal) DNA sequence that codes for a protein product with normal function. Take this sequence
of DNA through transcription and translation to complete the messenger RNA (mRNA) product and the
amino acid (protein) product. Remember that we follow the rules of complementary base pairing to
create the mRNA and U replaces T in RNA. Also each mRNA codon (group of 3) is replaced by an amino
acid according to the “The Genetic Code” – see during translation.
DNA Template: TAC CAA GTA GAC CTC CTT CTC GTG CAT CTT GTG ATC
mRNA: AUG GUU CAU CUG GAG GAA GAG CAC GUA GAA CAC UAG
Protein (amino acid: Met Val His Leu Glu Glu Glu His Val Glu His stop
sequence)
2) Determine the mRNA sequence – recording your results above. Remember COMPLEMENTARY BASE
PAIRING. AUG GUU CAU CUG GAG GAA GAG CAC GUA GAA CAC UAG
3) Using the genetic code determine the amino acid (protein) product for the mRNA template. Record
your results above.
The sequence of amino acid would be as follows:

Methionine, Valine, Histidine, Leucine, Glutamine, Glutamine, Glutamine, Histidine, Valine, Glutamine,
Histidine
4) The mRNA and protein product (amino acid sequence) represent the wild-type haemoglobin RNA and
protein. These sequences will be used for comparison to the sequences produced in the following
scenarios (PART II and PART III below).
PART II (chosen scenario:4)
1) CHOOSE ONE of the following scenarios #2 (Missense point mutation), OR #3 (Frame-shift mutation)
OR #4 (Insertion) – read the information provided below. Determine and record sequences for the DNA
Template, mRNA sequence and Protein (amino acid sequence) for the chosen scenario. ( /5)
DNA Template: TAC CAA GTA GAC CTC CTT CTC GTG CAT*C CTT GTG ATC
TAC CAA GTA GAC CTC CTT CTC GTG CAT CCT TGT GAT C
mRNA: AUG GUU CAU CUG GAG GAA GAG CAC GUA GGA ACA CUA G
Protein (amino acid : Met Val His Leu Glu Glu Glu His Val Gly Thr Leu ….
sequence)
Scenario #2 – Missense Point Mutation
A mutation to the normal (wild-type) DNA sequence from Scenario #1 (PART I) has occurred at position
#14; an adenine (A) has replaced a thymine (T). Modify your original DNA template sequence. Take this
sequence through transcription (to produce mRNA) and translation (to produce the amino acid – protein
product). What happens? Does the mutation alter the amino acid sequence? Fill in the information in
TABLE 1 below.
Scenario #3 – Frame-shift Mutation
A mutation to the normal (wild-type) DNA sequence from Scenario #1 has occurred. A base is deleted
accidentally during DNA replication. Remove the cytosine (C) from position #25 of the original DNA
sequence. Modify your original DNA template sequence shifting nucleotides over to maintain groupings
of three. Take this sequence through transcription (to produce mRNA) and translation (to produce the
amino acid – protein product). What happens? Does the mutation alter the amino acid sequence? Fill in
the information in TABLE 1 below.
Scenario #4 – Insertion
Another mutation has occurred. An extra nucleotide was inserted during DNA replication. Add a cytosine
(C) to the wild-type DNA sequence after the T at position #27 i.e. the new C will be at position #28 and
everything else will shift over to maintain groupings of three nucleotides. Take this sequence through
transcription (to produce mRNA) and translation (to produce the amino acid – protein product). What
happens? Does the mutation alter the amino acid sequence? Fill in the information in TABLE 1 below.

PART III
1) Complete scenario #5. Read the information provided below. Determine and record sequences for
the DNA Template, mRNA sequence and Protein (amino acid sequence). ( /5)
DNA Template: TAC CAA GTA GAC CTC CTT CTC GTG CAT CTT GTG ATC
TAC CAG GTA GAC CTC CTT CTC GTG CAT CTT GTG ATC
mRNA: AUG GUC CAU CUG GAG GAA GAG CAC GUA GAA CAC UAG
Protein (amino acid : Met Val His Leu Glu Glu Glu His Val Glu His Stop
sequence)
Scenario #5 – Silent Point Mutation
Another mutation has occurred! One nucleotide has been accidentally substituted for another
nucleotide during DNA replication. Replace the adenine (A) for a guanine (G) at position #6 of your
original DNA template. Make the new mRNA sequence and the new amino acid (protein) sequence.
What happens? Does the mutation alter the amino acid sequence? Fill in the information in TABLE 1
below.
PART IV – RESULTS ( /10)
TABLE 1
MUTATION
Protein Product Changes from Original
YES OR NO
** If YES then describe change to protein i.e. amino acid changes – BE SPECIFIC**
Mutation likely harmful to cell/individual
YES OR NO
** NOTE: as long as there is a change to the protein’s amino acid sequence – there is potential to be
harmful**
Missense Point Mutation

Frame-shift Mutation
Insertion
Original Template Mutation
Protein product changes from original? Yes
Description of the change Insertion at the 27th position of the DNA template
resulting in a complete frame shift
Is the mutation harmful? Yes, the mutation is harmful because on account
of the frame shift the amino acids in the
sequence has changed resulting in the formation
of a new/faulty protein
Silent Point Mutation
Original Template Mutation
Protein product changes from original? No
Description of the change Replacement of Adenine to Guanine at the 6th
position of the DNA template resulting in the
formation of the same amino acid/ no change in
the protein expression
Is the mutation harmful? No

QUESTIONS: DUE Monday October 1st, 2018
1) Which of the above mutation(s) is/are the least dangerous i.e. would not be harmful to the cell?
Explain. ( /2)
Silent mutation is the least dangerous mutation and it would not be harmful to the cell. Silent mutations
are not harmful because there is no change whatsoever in the amino acid sequence and the same
protein is produced which is necessary to maintain the normal physiological functions. These mutations
do not elicit an observable response on the phenotypic expression. These mutations have been hence
classified as a type of neutral mutation. Hence, it can be stated that silent mutations alters the sequence
of the nucleotide bases present in the DNA template. However, the altered nitrogenous bases do not
alter the protein expression and hence the mutations are not harmful to the cell.
2) Examine the Genetic Code -it is said to be degenerate, meaning more than one triplet codon codes for
the same amino acid. Why is this characteristic of the genetic code advantageous to organisms? ( /2)
The statement ‘genetic code is degenerate’ refers to the fact that one triplet codon has the ability to
code for more than one protein. On a broader perspective it can be explained as the ability of different
genetic codons to code for a single protein (Tee & Wong, 2013). For instance, ACU, ACC, ACA and ACG
are different genetic codons that code for the same protein that is Threonine. Since more than one
codon has the ability to code for the same amino acid. Hence, it helps in masking the effect of some
mutations (silent and nonsense mutations).
3) Which amino acids have greater redundancy (i.e. greatest number of codons to code for an amino
acid)? ( /2)
Genetic redundancy can be defined as the process where two or more genes are responsible for the
same function and the inactivation of one of the gene would not elicit an effect on the phenotype of the
organism (Mauro & Chappell,2014). For instance the codon UUU or UUC both code for the same protein
Phenylalanine. Therefore it should be noted that genetic redundancy of the codes often lead to the
cause of synonymous and non synonymous mutations for the protein coding regions of a particular
gene.
The greatest rate of redundancy is shown by the following amino acids:
Leucine: 6
Serine: 6
Arginine: 6
However, interestingly Tryptophan and Methionine are not redundant and are only coded by the codon
UGG and AUG respectively.
TOTAL POINTS: /31
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References:
Mauro, V. P., & Chappell, S. A. (2014). A critical analysis of codon optimization in human
therapeutics. Trends in molecular medicine, 20(11), 604-613.
Tee, K. L., & Wong, T. S. (2013). Polishing the craft of genetic diversity creation in directed
evolution. Biotechnology advances, 31(8), 1707-1721.