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the following dna sequence represents a portion of the non - template (…

Question

the following dna sequence represents a portion of the non - template (coding) strand from an exon of the human sonic hedgehog gene.
5 - ctg cac ttc tgc ctg gtc - 3
the gene’s reading frame is indicated by the spacing between the three - nucleotide codons. these codons can be translated using a table like the one included below, which shows the amino acid encoded by each possible mrna codon.
(the codon table is shown with first letter, second letter, and third letter axes, and various codons and their corresponding amino acids or stop signals)
given the codon table above, what would be the effect of a mutation that inserts a g at the beginning of the sonic hedgehog dna sequence?
choose 1 answer:
a the first amino acid in the sequence would change from leucine (leu) to alanine (ala), and the subsequent amino acids would also change.
b the first amino acid in the sequence would change from leucine (leu) to alanine (ala), but the subsequent amino acids would remain the same.
c the first amino acid in the sequence would change from leucine (leu) to valine (val), and the subsequent amino acids would also change.
d the first amino acid in the sequence would change from leucine (leu) to valine (val), but the subsequent amino acids would remain the same.

Explanation:

Brief Explanations
  1. First, analyze the original DNA sequence (nontemplate/coding strand): \( 5' - CTG\ CAC\ TTC\ TGC\ CTG\ GTC - 3' \). The mRNA transcribed from the coding strand (since coding strand has the same sequence as mRNA, except T→U) is \( 5' - CUG\ CAC\ UUC\ UGC\ CUG\ GUC - 3' \). The codons are CUG, CAC, UUC, UGC, CUG, GUC. Using the codon table, CUG codes for Leu.
  2. Now, consider the mutation: inserting a G at the beginning. The new DNA sequence becomes \( 5' - GCT\ GCA\ CTT\ CTG\ CTG\ TCG - 3' \) (wait, no—wait, the original is CTG CAC TTC TGC CTG GTC. Inserting G at the start: GCTG CAC TTC TGC CTG GTC? Wait, no, the reading frame is three - nucleotide codons. Wait, original DNA (coding) is \( 5' - CTG\ CAC\ TTC\ TGC\ CTG\ GTC - 3' \), so mRNA is \( 5' - CUG\ CAC\ UUC\ UGC\ CUG\ GUC - 3' \), codons: CUG (Leu), CAC (His), UUC (Phe), UGC (Cys), CUG (Leu), GUC (Val).
  3. After inserting G at the beginning, the new DNA (coding) sequence is \( 5' - GCT\ GCA\ CTT\ CTG\ CTG\ TCG - 3' \)? No, wait, the original is a string of nucleotides: CTG CAC TTC TGC CTG GTC (each codon is 3 nucleotides: CTG, CAC, TTC, TGC, CTG, GTC). Inserting G at the start: GCTG CAC TTC TGC CTG GTC? No, the correct way is that the new DNA sequence is \( 5' - GCT\ GCA\ CTT\ CTG\ CTG\ TCG - 3' \)? Wait, no, the original is 6 codons (18 nucleotides: 6×3). Inserting G at the start makes the DNA sequence length 19, but the reading frame will shift. Wait, no—wait, the original coding strand: \( 5' - CTG\ (C)\ AC\ (T)\ TC\ (T)\ GC\ (C)\ TG\ (G)\ TC - 3' \) (codons: CTG, CAC, TTC, TGC, CTG, GTC). After inserting G at the start, the new sequence is \( 5' - GCT\ GCA\ CTT\ CTG\ CTG\ TCG - 3' \)? No, I think I made a mistake. Wait, the correct approach: the original mRNA (from coding strand) is \( 5' - CUG\ CAC\ UUC\ UGC\ CUG\ GUC - 3' \) (codons: CUG, CAC, UUC, UGC, CUG, GUC). After inserting G at the beginning of the DNA (coding strand), the new DNA is \( 5' - GCT\ GCA\ CTT\ CTG\ CTG\ TCG - 3' \)? No, wait, the original DNA is \( 5' - C\ T\ G\ C\ A\ C\ T\ T\ C\ T\ G\ C\ C\ T\ G\ G\ T\ C - 3' \) (let's write each nucleotide: positions 1 - 18: C(1), T(2), G(3), C(4), A(5), C(6), T(7), T(8), C(9), T(10), G(11), C(12), C(13), T(14), G(15), G(16), T(17), C(18)). Inserting G at position 1: new positions 1 - 19: G(1), C(2), T(3), G(4), C(5), A(6), C(7), T(8), T(9), C(10), T(11), G(12), C(13), C(14), T(15), G(16), G(17), T(18), C(19)). Now, the codons are formed by groups of 3: codon 1: GCT (positions 1 - 3: G, C, T), codon 2: GCA (positions 4 - 6: G, C, A), codon 3: CTT (positions 7 - 9: C, T, T), codon 4: CTG (positions 10 - 12: C, T, G), codon 5: CTG (positions 13 - 15: C, T, G), codon 6: TCG (positions 16 - 18: T, C, G), and there's an extra nucleotide at the end, but let's focus on the first few. Wait, no—actually, the original mRNA was in frame with codons CUG (Leu), CAC (His), etc. After inserting G, the reading frame is shifted. The new mRNA (from the new coding strand, T→U) is \( 5' - GCU\ GCA\ CUU\ CUG\ CUG\ UCG - 3' \) (wait, no, the new DNA coding strand is \( 5' - GCT\ GCA\ CTT\ CTG\ CTG\ TCG - 3' \)? No, the correct mRNA from the new coding strand (with G inserted) is \( 5' - GCU\ GCA\ CUU\ CUG\ CUG\ UCG - 3' \) (since DNA coding strand has T, mRNA has U). Now, codon 1: GCU. From the codon table, GCU (first letter G, second U? Wait no, codon table: first letter (row), second (column), third (column). GCU: first letter G, second C, third U? Wait, no, the codon is G - C - U. First letter G (row G), second letter C (column C), third letter U (third letter U). Looking at the cod…

Answer:

A. The first amino acid in the sequence would change from leucine (Leu) to alanine (Ala), and the subsequent amino acids would also change.