Genetics Quiz: Using The Genetic Code

What this quiz covers

This quiz focuses on Using The Genetic Code, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 2
2. 3 (correct answer)
3. 4
4. 5

Explanation: Translation initiates at the first start codon, AUG. The ribosome then reads the sequence in non-overlapping triplets. The sequence is read as 5'-GC(AUG)(CCA)(CGA)(UAG)-3'. Translation starts at AUG. The codons are AUG (Met), CCA (Pro), and CGA (Arg). The next codon is UAG, which is a stop codon. Therefore, translation terminates after the third amino acid, producing a polypeptide with 3 amino acids (Met-Pro-Arg).

Question 2

1. 1
2. 2 (correct answer)
3. 3
4. 4

Explanation: To find the total number of distinct mRNA sequences, we multiply the number of codon options for each amino acid. Methionine (Met) is coded by 1 codon (AUG). Histidine (His) is coded by 2 codons (CAU, CAC). Tryptophan (Trp) is coded by 1 codon (UGG). The total number of unique mRNA sequences is the product of these possibilities: 1 × 2 × 1 = 2.

Question 3

1. The code is degenerate, meaning multiple codons can specify one amino acid.
2. The code is unambiguous, meaning each codon specifies only one amino acid.
3. The code is non-overlapping, meaning codons are read in successive triplets.
4. The code is nearly universal, meaning codons specify the same amino acids across different species. (correct answer)

Explanation: The ability to take a gene from a eukaryote (human) and have it correctly translated into an identical protein by a prokaryote (E. coli) demonstrates that the translational machinery of both organisms interprets the mRNA codons in the same way. This is the definition of the universality of the genetic code. While the other options are also true properties of the code, they are not what is directly demonstrated by this specific experiment.

Question 4

1. Ala-Phe-Thr
2. Arg-Lys-Trp (correct answer)
3. Cys-Phe-Gly
4. Ser-Gln-Pro

Explanation: The tRNA anticodons bind to complementary mRNA codons. We must first determine the mRNA codons corresponding to each anticodon. The anticodon 3'-GCA-5' pairs with the mRNA codon 5'-CGU-3'. The anticodon 3'-UUC-5' pairs with mRNA codon 5'-AAG-3'. The anticodon 3'-ACC-5' pairs with mRNA codon 5'-UGG-3'. Translating these mRNA codons gives Arg (CGU), Lys (AAG), and Trp (UGG).

Question 5

1. Nonsense mutation
2. Missense mutation
3. Frameshift mutation
4. Silent mutation (correct answer)

Explanation: The coding strand sequence is used to determine the mRNA sequence. A triplet 5'-GTC-3' on the coding strand corresponds to a 5'-GUC-3' codon in the mRNA. A triplet 5'-GTT-3' corresponds to a 5'-GUU-3' codon. According to the provided genetic code, both GUC and GUU code for the amino acid Valine (Val). Since the amino acid sequence of the protein is unchanged, this is a silent mutation.

Question 6

1. A polypeptide containing only Serine.
2. A polypeptide containing only Leucine.
3. A polypeptide with alternating Serine and Leucine residues. (correct answer)
4. No polypeptide will be produced due to the lack of a start codon.

Explanation: In a cell-free translation system (and sometimes in vivo), translation can initiate without a canonical AUG start codon, albeit inefficiently. The ribosome will bind and establish a reading frame. If the reading frame starts with the first U, the codons will be UCU, CUC, UCU, CUC..., which translates to Ser-Leu-Ser-Leu.... If the reading frame starts with the first C, the codons will be CUC, UCU, CUC, UCU..., which translates to Leu-Ser-Leu-Ser.... In either case, the resulting polypeptide will consist of alternating Serine and Leucine residues.

Question 7

1. Changing the template triplet for Glu from 3'-CTC-5' to 3'-CTT-5'. (correct answer)
2. Changing the template triplet for Ile from 3'-TAT-5' to 3'-TAC-5'.
3. Changing the template triplet for Ala from 3'-CGT-5' to 3'-CCT-5'.
4. Changing the template triplet for Ile from 3'-TAT-5' to 3'-ATT-5'.

Explanation: We need to find the mutation with the least impact. A) The original template for Glu (GAG) is 3'-CTC-5'. Changing it to 3'-CTT-5' results in the mRNA codon 5'-GAA-3'. Both GAG and GAA code for Glutamic acid (Glu), so this is a silent mutation and the protein is unchanged. B) The template for Ile (AUA) is 3'-TAT-5'. Changing it to 3'-TAC-5' creates the mRNA codon 5'-AUG-3', which codes for Methionine. This is a missense mutation. C) The template for Ala (GCA) is 3'-CGT-5'. Changing it to 3'-CCT-5' creates the mRNA codon 5'-GGA-3' (Glycine). This is a missense mutation. D) The template for Ile (AUA) is 3'-TAT-5'. Changing it to 3'-ATT-5' creates the mRNA codon 5'-UAA-3', which is a stop codon. This nonsense mutation would truncate the protein. A silent mutation (A) is the most likely to preserve protein function.

Question 8

1. 3'-AGA AAT-5'
2. 3'-TCG GAC-5'
3. 3'-AGG AAG-5' (correct answer)
4. 3'-AGC GAA-5'

Explanation: To solve this, we must transcribe each DNA template strand (given 3' to 5') into its corresponding mRNA codon pair (read 5' to 3') and check if it codes for Ser-Leu. A) 3'-AGA AAT-5' gives mRNA 5'-UCU UUA-3', which codes for Ser-Leu. B) 3'-TCG GAC-5' gives mRNA 5'-AGC CUG-3', which codes for Ser-Leu. C) 3'-AGG AAG-5' gives mRNA 5'-UCC UUC-3'. While UCC codes for Serine, UUC codes for Phenylalanine (Phe), not Leucine. Thus, this DNA could not code for Ser-Leu. D) 3'-AGC GAA-5' gives mRNA 5'-UCG CUU-3', which codes for Ser-Leu.

Question 9

1. A missense mutation replaces Aspartic acid with Glycine.
2. A missense mutation replaces Aspartic acid with Valine. (correct answer)
3. A silent mutation occurs, with no change to the amino acid sequence.
4. A nonsense mutation results in premature termination.

Explanation: First, determine the original peptide sequence. The DNA coding strand is 5'-TGG GAC TTA-3'. The mRNA sequence will be identical except that T is replaced by U: 5'-UGG GAC UUA-3'. The codons are UGG (Trp), GAC (Asp), and UUA (Leu). The peptide is Trp-Asp-Leu. Next, identify the mutation. A transversion is a substitution of a purine (A, G) for a pyrimidine (C, T) or vice versa. The fourth nucleotide is a G (a purine). A transversion would change it to a C or a T. Let's assume G becomes T. The new DNA coding strand is 5'-TGG TAC TTA-3'. The new mRNA is 5'-UGG UAC UUA-3'. This codes for Trp-Tyr-Leu. Asp is replaced by Tyr. Let's assume G becomes C. New DNA: 5'-TGG CAC TTA-3'. New mRNA: 5'-UGG CAC UUA-3'. This codes for Trp-His-Leu. Asp is replaced by His. None of these match the options. Let's re-read the DNA sequence and check the options. Maybe the transversion is G->T, making the codon GTC instead of GAC. mRNA would be GUC instead of GAC. GAC codes for Asp. GUC codes for Val. This is a missense mutation changing Asp to Val. This matches option B. Let's check the other possibility. Transversion G->C. DNA codon becomes GCC. mRNA is GCC (Ala). This would be an Asp -> Ala change. Since Asp -> Val is an option, the G->T transversion is the intended scenario.

Question 10

1. Met-Gln-Gln-Gly
2. Met-Gln (correct answer)
3. Met-Gln-STOP
4. Met-Gln-Tyr-Gly

Explanation: The question requires translating an mRNA sequence using the standard genetic code, because the gene is being expressed in E. coli, not the ciliate. The information about the ciliate's special code is context to explain the origin of the gene, but it's a distractor for the translation task itself. The mRNA sequence is 5'-AUG CAG UAA GGU-3'. The bacterial ribosome will translate this as follows: AUG codes for Methionine (Met). CAG codes for Glutamine (Gln). UAA is a stop codon in the standard genetic code. Therefore, translation will terminate after the Glutamine residue is added. The resulting peptide will be Met-Gln.

Question 11

1. 1
2. 2 (correct answer)
3. 3
4. 4

Explanation: This problem requires navigating the genetic code table to find the shortest path between codons for two different amino acids. First, identify the codons. Tyrosine (Tyr) is encoded by UAU and UAC. Isoleucine (Ile) is encoded by AUU, AUC, and AUA. Let's find the shortest path. Start with a Tyr codon, for example, UAU. We want to reach an Ile codon (AUU, AUC, or AUA) with the minimum number of changes. UAU -> AUU requires two changes (U->A at position 1, U->U at position 2, U->U at position 3). Let's check a path through an intermediate. UAU -> AUU (2 steps: U->A at pos 1, U->U at pos 3). Path: UAU -> AAU (Asn) -> AUU (Ile). This is two single-base substitutions. Can we do it in one step? Let's check all single-step mutations from UAU: AAU (Asn), GAU (Asp), UUU (Phe), UCU (Ser), UGU (Cys), UAC (Tyr), UAG (Stop). None of these code for Isoleucine. Therefore, a minimum of two substitutions is required.

Question 12

1. The protein will be shorter due to premature termination.
2. Translation will not be affected because another stop codon is nearby.
3. The protein will have an altered C-terminus and be longer than normal. (correct answer)
4. A silent mutation will occur, leaving the protein unchanged.

Explanation: The original mRNA sequence near the end is 5'-...GCA UAG GCU...-3'. The codon GCA codes for Alanine (Ala), and UAG is a stop codon, so translation normally terminates after Alanine. The mutation is a deletion of the U from the UAG stop codon. The new sequence becomes 5'-...GCA AGG CUG...-3' (assuming the next base after the shown segment is a G, for example). The deletion causes a frameshift. The ribosome no longer reads UAG as a stop codon. Instead, it reads the new codons in the shifted frame: AGG (Arginine), CUG (Leucine), and so on. Translation will continue until a new stop codon is encountered in this new reading frame, or until it reaches the end of the transcript. This results in a protein with an altered C-terminal sequence (Ala is replaced by Arg, Leu, etc.) and that is longer than the original protein.

Question 13

1. An aspartic acid residue will be replaced by a methionine.
2. A leucine residue will be replaced by a methionine. (correct answer)
3. A methionine residue will be replaced by a leucine.
4. A valine residue will be replaced by a methionine.

Explanation: This is a multi-step problem. First, determine which mRNA codon the tRNA normally recognizes. The anticodon is 3'-GAU-5'. It pairs with mRNA codons in an antiparallel fashion, so it recognizes the codon 5'-CUA-3'. Second, determine the amino acid normally encoded by this codon. Using the genetic code table, 5'-CUA-3' codes for Leucine (Leu). Third, understand the consequence of misacylation. The tRNA is carrying Methionine (Met) instead of its correct amino acid, Leucine. Therefore, when the ribosome encounters a CUA codon, this faulty tRNA will bind and incorporate Methionine. This results in a Leucine residue being replaced by a Methionine residue.

Question 14

1. Substitution of the 5th base (A) to G.
2. Substitution of the 9th base (U) to C.
3. Substitution of the 11th base (G) to A. (correct answer)
4. Substitution of the 6th base (G) to A.

Explanation: The original mRNA 5'-AUG GAG UAU UGC-3' codes for Met-Glu-Tyr-Cys. We need to evaluate the impact of each mutation. (A) 5th base A -> G: Codon GAG becomes GGG. GAG is Glutamic acid (acidic, polar). GGG is Glycine (nonpolar, small). This is a significant missense mutation. (B) 9th base U -> C: Codon UAU becomes UAC. Both UAU and UAC code for Tyrosine. This is a silent mutation, causing no change. (C) 11th base G -> A: Codon UGC becomes UGA. UGC is Cysteine. UGA is a stop codon. This is a nonsense mutation, which truncates the protein to Met-Glu-Tyr. This is almost always the most drastic change. (D) 6th base G -> A: Codon GAG becomes GAA. Both GAG and GAA code for Glutamic acid. This is a silent mutation. Comparing a missense mutation (A) with a nonsense mutation (C), the nonsense mutation causing premature termination is the most drastic alteration to the protein structure.

Question 15

1. Tryptophan (correct answer)
2. Glycine
3. Threonine
4. Cysteine

Explanation: The anticodon sequence is 3'-ACC-5'. It pairs with an mRNA codon in an antiparallel fashion. The standard base pairing would be with the codon 5'-UGG-3'. The codon 5'-UGG-3' codes for Tryptophan (Trp). The wobble position is the first base of the anticodon (3' end of codon, 5' end of anticodon). In this case, the first base of the anticodon is 3'-A...-5', which corresponds to the 5'-...C-3' base of the anticodon. C at the wobble position of the anticodon pairs strictly with G at the third position of the codon. Therefore, this tRNA only recognizes the UGG codon. The amino acid is Tryptophan. A common mistake is to think 3'-ACC-5' is the codon, which would be Threonine.

Question 16

1. The protein will end with Selenocysteine instead of Tryptophan.
2. The protein will be truncated, ending before the Tryptophan position.
3. The protein will end with Tryptophan, but with additional amino acids attached.
4. The protein will end with Cysteine, followed by other amino acids. (correct answer)

Explanation: The original mRNA sequence is 5'-...UGG UGA...-3'. This codes for Tryptophan (Trp), followed by a stop codon (UGA), so the protein ends with Trp. The mutation is a deletion of the final G in the UGG codon. This causes a frameshift. The new mRNA sequence will be 5'-...UGU GA...-3'. The ribosome will now read the sequence in this new frame. The first codon in the new frame is UGU. UGU codes for Cysteine (Cys). The next codon will be formed from the 'GA' and whatever nucleotide follows. The original UGA stop codon is no longer in frame and will not be read as a stop codon or as a codon for Selenocysteine. Translation will continue in the new reading frame, adding Cysteine and then other amino acids until a new stop codon is encountered. Therefore, the protein will now end with Cysteine, followed by an extended, altered C-terminus.

Question 17

1. Lys-Phe-Gly
2. Lys-Phe-Pro-Gly (correct answer)
3. Lys-Leu-Pro-Gly
4. Phe-Lys-Pro-Gly

Explanation: First, determine the original polypeptide. The DNA coding strand is 5'-AAG TTC GGT-3'. The mRNA is 5'-AAG UUC GGU-3'. The codons translate to Lys-Phe-Gly. Next, apply the mutation. The trinucleotide 5'-CCA-3' is inserted after the TTC codon. The new DNA coding strand is 5'-AAG TTC CCA GGT-3'. Since a full trinucleotide (a codon) was inserted in-frame, the reading frame is not shifted. The new mRNA sequence is 5'-AAG UUC CCA GGU-3'. Translating the new sequence: AAG codes for Lysine (Lys), UUC codes for Phenylalanine (Phe), CCA codes for Proline (Pro), and GGU codes for Glycine (Gly). The resulting polypeptide is Lys-Phe-Pro-Gly.

Question 18

1. 5'-UUA AGU-3'
2. 5'-CUC UCA-3' (correct answer)
3. 5'-CUA AGC-3'
4. 5'-UUG UCG-3'

Explanation: The goal is to find the sequence with the most opportunities for silent mutations. We need to analyze the codons for Leu and Ser. A silent mutation occurs when changing a nucleotide doesn't change the amino acid. (A) 5'-UUA AGU-3': UUA (Leu) can change to UUG (1 silent mutation). AGU (Ser) can change to AGC (1 silent mutation). Total = 2. (B) 5'-CUC UCA-3': CUC (Leu) can change to CUU, CUA, or CUG (3 silent mutations). UCA (Ser) can change to UCU, UCC, or UCG (3 silent mutations). Total = 6. (C) 5'-CUA AGC-3': CUA (Leu) can change to CUU, CUC, or CUG (3 silent mutations). AGC (Ser) can change to AGU (1 silent mutation). Total = 4. (D) 5'-UUG UCG-3': UUG (Leu) can change to UUA (1 silent mutation). UCG (Ser) can change to UCU, UCC, or UCA (3 silent mutations). Total = 4. Option B offers the highest number of potential silent mutation sites with 6 total possibilities.

Question 19

1. Gly-Leu-Glu
2. Lys-Stop-Ala
3. Lys-Ile-Ala (correct answer)
4. Pro-Val-Lys

Explanation: This is a multi-step problem involving transcription after a DNA inversion. First, find the peptide from the original sequence to have a distractor. Original template: 3'-GGC AAT CTT-5'. Original mRNA: 5'-CCG UUA GAA-3'. Original peptide: Pro-Leu-Glu. Now, process the inverted sequence. The new template strand is 3'-TTC TAA CGG-5'. We must transcribe this to find the new mRNA. The mRNA will be complementary and antiparallel: 5'-AAG AUU GCC-3'. Finally, translate this new mRNA using the genetic code table. AAG codes for Lysine (Lys). AUU codes for Isoleucine (Ile). GCC codes for Alanine (Ala). The new amino acid sequence is Lys-Ile-Ala.

Question 20

1. Position 3 (the G of AUG)
2. Position 5 (the U of GUC)
3. Position 6 (the C of GUC) (correct answer)
4. Position 9 (the G of UGG)

Explanation: The question asks for a position where any of the three possible substitutions would be a silent mutation. Let's analyze the codons: 1) AUG codes for Methionine; it is the only codon for this amino acid, so any change would alter the polypeptide. 2) UGG codes for Tryptophan; it is also the only codon, so any change at position 9 would alter the polypeptide. 3) GUC codes for Valine. The codons for Valine are GUU, GUC, GUA, and GUG. At position 6, the original nucleotide is C. If it's changed to U (GUU), A (GUA), or G (GUG), the codon still codes for Valine. Therefore, position 6 is the correct answer.