Below is a DNA sequence of part of a hypothetical wild-type bacterial genome. Th

Question

Below is a DNA sequence of part of a hypothetical wild-type bacterial genome. The nucleotides are numbered 1 to 100. Transcription begins with and includes the T/A (top strand/bottom strand) base pair at position 7 (indicated by the arrow), and proceeds from left to right along the DNA. Translation of the transcript yiclds the peptide of interest. 9. 20 20 30 e 0 50 51 60 70 80 90 100 What are the first 12 nucleotides of the resulting messenger RNA (mRNA)? Indicate the 5' and 3' ends of the mRNA. (a) (2 marks) Second Letter UUU Phe UC uCC Ser uUAC Leu UCA cCU CCA UAA Stop UGA StopA UAG Stop lucG Trp CAU His CGU 1stCUA letter AUU ACU AA Asn AGU Ser letter CCG CGG AGA Arg AUG Met ACG GUU GUA GAU Asp GGUU GAG Glu GGA Gly GAA cu Gcu GCG

Explanation / Answer

9. a. Transcription begins with and includes T/A (top strand and bottom strand) base pair at 7 and proceeds from left to right.

so the template strand on which transcription takes place is the strand with 3' .......AGTCTAATGCAT........5'

and the 1st 12 nucleotides of resulting mRNA are 5'UCAGAUUACGUA 3'

b. i) Based on the DNA sequence shown in the diagram the number of nitrogen bases from 7- 100 are 93 . so the number of codons are 93/3 as each codon is a triplet codon. so the number of codons are 31. Each codon codes for one amino acid and hence there should be 31 amino acids in the peptide. but the 71,72, 73 nitrogen bases are TAA in the 5'-3' strand and ATT in template strand. So the mRNA has the codon UAA which is a stop codon and hence stops the translation at 70th nucleotide.

so the translation takes place only till 70th nucleotide hence from 7-70 has 63 nucleotides or 63/3 =21 codons which code for 21 amino acids. the the peptide has 21 amino acids.

ii) The first four codons are UCA GAU UAC GUA

So the first four amino acids are: ser asp tyr val

iii) In a bacteriial strain A, there is an insertion of an extra C/G base pair immediately after 60. so the reading frame changes, and it is called frameshift mutation or point mutation. The sequence of nitrogen bases in the codon changes the sequnce of the amino acids. so there is change in the amino acid in the peptide produced.

iv) In a bacterial strain B there is a substitution of A/T base pair at position 47 with a G/C base pair. Hence A is substituted by G and T is substituted by C. That means a purine is substituted by another purine and a pyrimidine is substituted by another pyrimidine. This kind of substitution mutation is called Transition mutation. It changes the sequence of nucletides of a codon and hence change the amino acid at that place of the peptide.

v) In a bacterial starin C, there is a substitution of A/T base pair at position 47 with a C/G base pair. Hence A is substituted by C and T is substituted by G. This kind of substitution mutation where a purine is substituted by a pyrimidine and the pyrimidine is substituted by a purine is called Transversion mutation. There will be a change in the sequnce of the nucleotide bases at that position and there will be a change in the amino acid at that position.

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