Question

1. even a gene mutation that is a point mutation, meaning it affects one nucleotide base, can still make a major change

consider the below information for normal hemoglobin:

portion of
hemoglobin dna gga ctc ctc
mrna ccu gag gag
amino acids proline - glutamic acid - glutamic acid

sickle cell anemia is caused by a point mutation known as a substitution. show what would occur if the first t (\thymine\) dna base in the portion shown above experienced a mutation with a substitution of a (\adenine\).

sickle cell hemoglobin:
portion of mutated hemoglobin dna: ____
mrna: ____
amino acids: ____

Explanation:

Step1: Identify the original DNA sequence

The original portion of hemoglobin DNA is \( \text{GGA CTC CTC} \). We need to substitute the first \( \text{T} \) (thymine) with \( \text{A} \) (adenine). Let's break down the DNA sequence: the second triplet is \( \text{CTC} \), so the first \( \text{T} \) is in the second triplet. So the original DNA triplets are \( \text{GGA} \), \( \text{CTC} \), \( \text{CTC} \). Substituting the first \( \text{T} \) in \( \text{CTC} \) (so \( \text{CTC} \) becomes \( \text{CAC} \)?) Wait, no: wait, DNA bases are A, T, C, G. The original DNA is GGA (first triplet), CTC (second), CTC (third). The first T is in the second triplet: C T C. So substituting T with A: so the second triplet becomes C A C? Wait, no, wait: the original DNA is GGA CTC CTC. Let's write each base:

First triplet: G, G, A

Second triplet: C, T, C

Third triplet: C, T, C

We need to substitute the first T (in the second triplet, position 2 of the second triplet) with A. So the second triplet becomes C, A, C? Wait, no, wait: the problem says "the first T (thymine) DNA base in the portion shown above". So the portion is GGA CTC CTC. Let's list all bases:

1: G (first triplet, first base)

2: G (first triplet, second base)

3: A (first triplet, third base)

4: C (second triplet, first base)

5: T (second triplet, second base) → this is the first T

6: C (second triplet, third base)

7: C (third triplet, first base)

8: T (third triplet, second base)

9: C (third triplet, third base)

So substituting base 5 (T) with A. So the new DNA sequence is:

First triplet: GGA

Second triplet: CAC (since C (base4), A (base5, substituted), C (base6))

Third triplet: CTC (unchanged, since we only substitute the first T, which is in the second triplet)

Wait, no: wait, the original DNA is GGA CTC CTC. So the triplets are GGA, CTC, CTC. The first T is in the second triplet (CTC). So substituting that T with A: CTC → CAC. So the mutated DNA sequence is GGA CAC CTC.

Step2: Transcribe DNA to mRNA

DNA to mRNA: A pairs with U, T pairs with A, C pairs with G, G pairs with C.

First triplet: GGA → mRNA: CCU (since G→C, G→C, A→U)

Second triplet: CAC → mRNA: GUG (C→G, A→U, C→G)

Third triplet: CTC → mRNA: GAG (C→G, T→A, C→G)

So mRNA sequence: CCU GUG GAG

Step3: Translate mRNA to amino acids

Use the genetic code:

- CCU: Proline (same as original first amino acid)

- GUG: Valine (original second was GAG→Glutamic Acid; GUG codes for Valine)

- GAG: Glutamic Acid (same as original third? Wait, no: original third was GAG→Glutamic Acid. Wait, original mRNA was CCU GAG GAG, amino acids Proline-Glutamic Acid-Glutamic Acid.

After mutation, mRNA is CCU GUG GAG. So:

- CCU: Proline

- GUG: Valine

- GAG: Glutamic Acid

So amino acids: Proline-Valine-Glutamic Acid? Wait, no, wait: wait the third triplet in DNA is CTC, which transcribes to GAG (since T→A, so CTC→GAG). So mRNA is CCU (from GGA), GUG (from CAC), GAG (from CTC). So amino acids:

CCU: Proline

GUG: Valine

GAG: Glutamic Acid

Wait, but in sickle cell anemia, the mutation is in the beta-globin gene, where a GAG (glutamic acid) is changed to GTG (valine) in DNA (so mRNA GUG). Wait, maybe I made a mistake in the triplet. Wait original DNA: GGA CTC CTC. Let's check the original mRNA: CCU GAG GAG. So DNA CTC → mRNA GAG (since T→A, so CTC (DNA) → GAG (mRNA: C→G, T→A, C→G). So the second DNA triplet is CTC, mRNA GAG. If we substitute the T in CTC (DNA) with A, so DNA becomes CAC (since C-T-C → C-A-C). Then mRNA from CAC is GUG (C→G, A→U, C→G). So mRNA is CCU (from GGA), GUG (from CAC), GAG (from CTC). So amino acids:

CCU: Proline

GUG: Valine

GAG: Glutamic Acid

Wait, but in sickle cell, the mutation is usually in the sixth codon, where GAG (glutamic acid) becomes GTG (valine) in DNA (so mRNA GUG). So maybe the original DNA is different? Wait maybe the original DNA is GGA CTC CTC, but maybe the first T is in the third triplet? Wait no, the problem says "the first T (thymine) DNA base in the portion shown above". Let's count the Ts: in GGA (no Ts), CTC (one T), CTC (one T). So first T is in the second triplet (CTC), position 2 of the second triplet. So substituting that T with A: CTC → CAC (DNA). Then mRNA: CAC → GUG. So amino acid for GUG is Valine. So the amino acid sequence becomes Proline-Valine-Glutamic Acid? Wait, but in sickle cell, the change is from Glutamic Acid to Valine in the sixth position, but maybe this is a different portion. Anyway, let's proceed with the steps.

Answer:

• Portion of mutated hemoglobin DNA: \( \text{GGA CAC CTC} \)
• mRNA: \( \text{CCU GUG GAG} \)
• Amino Acids: \( \text{Proline-Valine-Glutamic Acid} \)

(Note: In typical sickle cell mutation, the DNA change is \( \text{CTC} \to \text{CAC} \) (or \( \text{CTT} \to \text{CAT} \)) in the beta-globin gene, leading to \( \text{GAG} \to \text{GUG} \) in mRNA, and Glutamic Acid \( \to \) Valine. The third triplet here remains \( \text{CTC} \) (mRNA \( \text{GAG} \), Glutamic Acid) as only the first T is substituted.)