DNA Sequences, Gene Expression, And Mutations: Transcription And Translation

Complete scenario. Read the information provided below. Determine and record sequences for the DNA Template, mRNA sequence and Protein (amino acid sequence).

Plication. 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?

1) Which of the above mutation(s) is/are the least dangerous i.e. would not be harmful to the cell? Explain.

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?

3) Which amino acids have greater redundancy (i.e. greatest number of codons to code for an amino acid)?

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)

Part II

         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.

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

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:

However, interestingly Tryptophan and Methionine are not redundant and are only coded by the codon UGG and AUG respectively.

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.