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Chapter 22: Problem 119

The base sequences in mRNA that code for certain amino acids are Glu: GAA, GAG Val: GUU, GUC, GUA, GUG Met: AUG Trp: UGG Phe: UUU, UUC Asp: GAU, GAC These sequences are complementary to the sequences in DNA. a. Give the corresponding sequences in DNA for the amino acids listed above. b. Give a DNA sequence that would code for the peptide trp–glu–phe–met. c. How many different DNA sequences can code for the tetrapeptide in part b? d. What is the peptide that is produced from the DNA sequence T–A–C–C–T–G–A–A–G? e. What other DNA sequences would yield the same tripeptide as in part d?

Short Answer

Expert verified
a. DNA sequences: CTT, CTC (Glu); CAA, CAG, CAT, CAC (Val); TAC (Met); ACC (Trp); AAA, AAG (Phe); CTA, CTG (Asp) b. One sequence: ACC-CTT-AAA-TAC (Trp-Glu-Phe-Met) c. 4 different DNA sequences d. Peptide: Met-Asp-Phe e. Other DNA sequences: TAC-CTA-AAA, TAC-CTA-AAG, TAC-CTG-AAA

Step by step solution

01

mRNA and DNA are complementary. For each base in mRNA, there is a corresponding base in DNA: - Adenine (A) in mRNA corresponds to Thymine (T) in DNA - Uracil (U) in mRNA corresponds to Adenine (A) in DNA - Cytosine (C) in mRNA corresponds to Guanine (G) in DNA - Guanine (G) in mRNA corresponds to Cytosine (C) in DNA #Step 2: Convert mRNA sequences to DNA sequences#

Using the complementarity rule from Step 1, we can convert the given mRNA sequences to the corresponding DNA sequences: For Glu (GAA, GAG): The corresponding DNA sequences are: CTT, CTC For Val (GUU, GUC, GUA, GUG): The corresponding DNA sequences are: CAA, CAG, CAT, CAC For Met (AUG): The corresponding DNA sequence is: TAC For Trp (UGG): The corresponding DNA sequence is: ACC For Phe (UUU, UUC): The corresponding DNA sequences are: AAA, AAG For Asp (GAU, GAC): The corresponding DNA sequences are: CTA, CTG #b. Give a DNA sequence that would code for the peptide trp–glu–phe–met.# #Step 3: Determine DNA sequences for the given peptide#
02

We need to find the DNA sequence corresponding to the peptide trp–glu–phe–met. Using the DNA sequences found in part a: Trp (ACC) – Glu (either CTT or CTC) – Phe (either AAA or AAG) – Met (TAC) #Step 4: Choose one possible DNA sequence#

One possible DNA sequence could be: ACC - CTT - AAA - TAC #c. How many different DNA sequences can code for the tetrapeptide in part b?# #Step 5: Count the possibilities for each amino acid#
03

We need to find the number of different DNA sequences that can code for the tetrapeptide trp-glu-phe-met. For this, we need to count the number of possibilities for each amino acid: Trp: 1 possibility (ACC) Glu: 2 possibilities (CTT, CTC) Phe: 2 possibilities (AAA, AAG) Met: 1 possibility (TAC) #Step 6: Calculate the number of different DNA sequences#

The number of different DNA sequences can be calculated using the multiplication rule: 1 (Trp) × 2 (Glu) × 2 (Phe) × 1 (Met) = 4 different DNA sequences #d. What is the peptide that is produced from the DNA sequence T–A–C–C–T–G–A–A–G?# #Step 7: Divide the DNA sequence into triplets#
04

To determine the peptide produced from the given DNA sequence (TACCTGAAG), we need to divide the sequence into triplets: TAC - CTG - AAG #Step 8: Identify the amino acids corresponding to the triplets#

Using the DNA sequences found in part a, we can identify the amino acids corresponding to each triplet: TAC: Met CTG: Asp AAG: Phe Thus, the peptide produced is Met-Asp-Phe. #e. What other DNA sequences would yield the same tripeptide as in part d?# #Step 9: Find other DNA sequences that code for the same amino acids#
05

We need to find other DNA sequences that can produce the same tripeptide (Met-Asp-Phe) found in part d. Using the DNA sequences found in part a for each amino acid: Met: TAC (1 possibility) Asp: CTA, CTG (2 possibilities) Phe: AAA, AAG (2 possibilities) #Step 10: Calculate the total number of possible DNA sequences for the tripeptide#

The total number of possible DNA sequences for the tripeptide can be calculated by multiplying the possibilities for each amino acid: 1 (Met) × 2 (Asp) × 2 (Phe) = 4 possible DNA sequences However, since the original DNA sequence (TACCTGAAG) is already known, we need to find the remaining 3 other DNA sequences that yield the same tripeptide: - TAC - CTA - AAA - TAC - CTA - AAG - TAC - CTG - AAA

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Most popular questions from this chapter

Reagents such as \(\mathrm{HCl}\) , HBr, and $\mathrm{HOH}\left(\mathrm{H}_{2} \mathrm{O}\right)$ can add across carbon-carbon double and triple bonds, with \(\mathrm{H}\) forming a bond to one of the carbon atoms in the multiple bond and \(\mathrm{Cl}\) , Br, or OH forming a bond to the other carbon atom in the multiple bond. In some cases, two products are possible. For the major organic product, the addition occurs so that the hydrogen atom in the reagent attaches to the carbon atom in the multiple bond that already has the greater number of hydrogen atoms bonded to it. With this rule in mind, draw the structure of the major product in each of the following reactions.

How would you synthesize each of the following? a. 1,2-dibromopropane from propene b. acetone (2-propanone) from an alcohol c. tert-butyl alcohol (2-methyl-2-propanol) from an alkene (See Exercise 68.) d. propanoic acid from an alcohol

Alcohols are very useful starting materials for the production of many different compounds. The following conversions, starting with 1-butanol, can be carried out in two or more steps. Show the steps (reactants/catalysts) you would follow to carry out the conversions, drawing the formula for the organic product in each step. For each step, a major product must be produced. (See Exercise \(68 . )\) (Hint: In the presence of \(\mathrm{H}^{+},\) an alcohol is converted into an alkene and water. This is the exact reverse of the reaction of adding water to an alkene to form an alcohol.) $$ \begin{array}{l}{\text { a. } 1 \text { -butanol } \longrightarrow \text { butane }} \\ {\text { b. } 1 \text { -butanol } \longrightarrow 2 \text { -butanone }}\end{array} $$

Polychlorinated dibenzo-p-dioxins (PCDDs) are highly toxic substances that are present in trace amounts as by-products of some chemical manufacturing processes. They have been implicated in a number of environmental incidents—for example, the chemical contamination at Love Canal and the herbicide spraying in Vietnam. The structure of dibenzo-pdioxin, along with the customary numbering convention, is The most toxic PCDD is 2,3,7,8-tetrachloro-dibenzo-p-dioxin. Draw the structure of this compound. Also draw the structures of two other isomers containing four chlorine atoms.

Structural and optical isomers can be drawn having the formula $\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{F}$ . Give examples to illustrate these types of isomerism for \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{F}\) . Why can't \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{F}\) exhibit geometrical isomerism?
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