- Which mRNA codon could this tRNA molecule successfully bind to in addition to 5'-CCG-3'?
Answer: 5'-CCU-3'
This question tests the 'wobble' hypothesis. The anticodon is given as 3'-GGC-5'. It pairs with an mRNA codon read 5' to 3'. The first two bases pair conventionally: the 3'-C pairs with 5'-G, and the middle G pairs with middle C. The wobble occurs at the third position. The 5'-G of the anticodon pairs with the 3'-C of the codon 5'-CCG-3' (standard Watson-Crick pairing). According to wobble rules, the guanine (G) at the 5' (wobble) position of the anticodon can also form a non-standard base pair with uracil (U) at the 3' position of the codon. Therefore, this tRNA can also recognize the codon 5'-CCU-3'.
- What is the primary function of the poly-A tail added to the 3' end of eukaryotic pre-mRNA during post-transcriptional modification?
Answer: It helps prevent the degradation of the mRNA by exonucleases in the cytoplasm and facilitates its export from the nucleus.
The poly-A tail is a long chain of adenine nucleotides added to the 3' end of the mRNA. It has several functions, including increasing the stability of the mRNA in the cytoplasm by protecting it from enzymatic degradation and playing a role in the export of the mRNA from the nucleus to the cytoplasm.
- A human gene containing introns and exons is inserted into a bacterial plasmid. The bacterium successfully transcribes the gene into pre-mRNA. However, the resulting polypeptide is non-functional and much longer than the human protein. What is the best explanation for this outcome?
Answer: Bacteria lack the spliceosome machinery to remove introns, so the intron sequences are translated.
Eukaryotic genes contain non-coding sequences called introns that are removed from the pre-mRNA by a process called splicing. Prokaryotic cells like bacteria do not have introns and therefore lack the necessary enzymes and machinery (spliceosomes) for splicing. As a result, the introns are not removed and are translated along with the exons, leading to a long, non-functional polypeptide.
- A segment of an mRNA molecule has the sequence 5'-CCG AUG CUC AAG-3'. If translation starts at the appropriate initiation codon, what is the sequence of the second and third amino acids in the polypeptide?
Answer: Leucine, Lysine
Translation initiates at the start codon AUG. In the given sequence 5'-CCG AUG CUC AAG-3', the ribosome begins translation at AUG. Therefore, the first amino acid incorporated is Methionine (Met). The second codon is CUC, which codes for Leucine (Leu). The third codon is AAG, which codes for Lysine (Lys). The question asks for the sequence of the second and third amino acids, which are Leucine and Lysine.
- A functional eukaryotic protein consists of 200 amino acids. What is the minimum number of nucleotides that must be present in the coding sequence of the mature mRNA molecule required to produce this protein?
Answer: 603
Each amino acid is coded by a three-nucleotide codon. Therefore, 200 amino acids require 200 x 3 = 600 nucleotides. In addition, a stop codon (also three nucleotides long) is required to terminate translation. The start codon (AUG) is included in the count as it codes for methionine. Therefore, the minimum number of nucleotides is 600 + 3 = 603.
- Which of the following describes the relationship between a gene's DNA coding strand and its corresponding mRNA transcript?
Answer: The coding strand has a sequence identical to the mRNA transcript, with thymine replacing uracil.
The DNA template strand is complementary to the mRNA. The DNA coding strand, by definition, has the same sequence as the mRNA transcript, except that DNA contains thymine (T) where RNA contains uracil (U). Both the coding strand and the mRNA are synthesized antiparallel to the template strand, giving them the same 5' to 3' sequence (with T/U substitution).
- The structure often observed as multiple ribosomes simultaneously translating a single mRNA molecule is known as a polysome. What is the primary functional advantage of this arrangement?
Answer: It significantly increases the rate of production of a specific polypeptide from a single mRNA.
A polysome (or polyribosome) allows for the mass production of a single type of protein. By having multiple ribosomes work on the same mRNA transcript at the same time, a cell can generate a large number of polypeptide copies in a much shorter time than if a single ribosome had to complete translation before another could start.
- The original DNA coding strand for a gene segment is 5'-TCA GCT CTT-3'. A mutation occurs which inserts a single guanine (G) nucleotide after the first cytosine (C). What is the most likely effect on the polypeptide produced?
Answer: A frameshift will occur, leaving the first amino acid unchanged but altering all subsequent amino acids.
The original DNA coding strand 5'-TCA GCT CTT-3' corresponds to mRNA 5'-UCA GCU CUU-3' (Ser-Ala-Leu). The mutated DNA is 5'-TCG AGC TCT T-3', corresponding to mRNA 5'-UCG AGC UCU U-3' (Ser-Ser-Ser-...). The first codon (TCA -> UCA) is unaffected. The insertion of G shifts the reading frame for all subsequent codons. Thus, the first amino acid (Serine) is correct, but all following amino acids are changed.
- Which statement accurately distinguishes protein synthesis in prokaryotes from that in eukaryotes?
Answer: In prokaryotes, translation can begin on an mRNA molecule that is still being transcribed.
In prokaryotic cells, there is no nuclear membrane separating the DNA and ribosomes. This allows for coupled transcription and translation, where ribosomes can attach to the 5' end of the mRNA and begin translation while the 3' end is still being synthesized by RNA polymerase. This is not possible in eukaryotes due to compartmentalization.
- A point mutation in a gene's coding sequence changes a codon from UGU to UGA. What is the most likely consequence for the resulting polypeptide chain?
Answer: Translation will be terminated prematurely, resulting in a truncated and likely non-functional polypeptide.
The codon UGU codes for cysteine, while UGA is a stop codon. This type of mutation is a nonsense mutation, which signals the ribosome to terminate translation. This results in a shorter, or truncated, polypeptide that is typically non-functional.
- What is the most likely fate of this defective mRNA molecule?
Answer: It will be rapidly degraded in the nucleus or cytoplasm and will likely not be translated successfully.
The poly-A tail is crucial for mRNA stability and export from the nucleus. Without it, the mRNA is highly susceptible to degradation by exonucleases present in both the nucleus and cytoplasm. While export may also be impaired, its rapid degradation is the most immediate and certain fate, preventing significant translation.
- A genetic disorder is caused by a defective aminoacyl-tRNA synthetase that attaches the amino acid arginine to tRNAs with the anticodon for alanine (3'-CGA-5'). What is the direct consequence of this defect on protein synthesis?
Answer: The amino acid arginine will be incorporated into proteins at positions coded for by alanine.
The ribosome recognizes the anticodon of the tRNA, not the amino acid it carries. The anticodon 3'-CGA-5' pairs with the mRNA codon 5'-GCU-3', which codes for alanine. If this tRNA is incorrectly charged with arginine, the ribosome will incorporate arginine at positions where alanine should be, leading to the synthesis of faulty proteins.
- A researcher attempts to express a mature eukaryotic mRNA in a bacterial system. Translation is highly inefficient. What is a plausible reason for this failure?
Answer: The bacterial ribosomes cannot initiate translation using the eukaryotic 5' cap and require a Shine-Dalgarno sequence.
Prokaryotic and eukaryotic translation initiation differ. Eukaryotic ribosomes typically recognize and bind to the 5' cap of the mRNA to start scanning for the first AUG codon. Bacterial ribosomes bind to a specific nucleotide sequence called the Shine-Dalgarno sequence, located just upstream of the start codon. Since mature eukaryotic mRNA lacks this sequence, bacterial ribosomes cannot efficiently initiate translation.
- Which molecule is directly responsible for providing the energy required for the translocation step during translational elongation, where the ribosome advances one codon along the mRNA?
Answer: GTP
During the elongation cycle of translation, the hydrolysis of Guanosine triphosphate (GTP) provides the energy for two key steps: the binding of a new aminoacyl-tRNA to the A site and the translocation of the ribosome along the mRNA after the peptide bond is formed. ATP is primarily used to charge tRNA molecules with amino acids.
- A mutation changes the stop codon of a gene from UGA to UGG. The codon UGG codes for the amino acid Tryptophan. What is the most probable outcome for the protein synthesized from this altered gene?
Answer: The polypeptide will be longer than the original, with Tryptophan at one position and additional amino acids appended.
The stop codon (UGA) signals the end of translation. By mutating it to a sense codon (UGG for Tryptophan), translation will not terminate at that position. The ribosome will incorporate Tryptophan and continue translating along the mRNA until it encounters the next in-frame stop codon. This results in a polypeptide that is longer than the normal version.
- Based on this observation, which specific process in translation is most directly inhibited by the toxin?
Answer: Translocation, the movement of the ribosome one codon down the mRNA.
Translocation is the process where the ribosome moves one codon down the mRNA. This movement shifts the tRNA from the A site to the P site and the P site tRNA to the E site. The description indicates this movement is blocked, which is the definition of inhibiting translocation.
- During transcription, RNA polymerase reads the DNA template strand in the 3' to 5' direction. What is the direct consequence of this for the synthesis of the mRNA molecule?
Answer: The mRNA is synthesized in the 5' to 3' direction, antiparallel to the template strand.
The synthesis of all nucleic acids (both DNA and RNA) proceeds by adding new nucleotides to the 3' hydroxyl group of the growing chain. This means synthesis always occurs in the 5' to 3' direction. Since the RNA polymerase reads the template strand in the 3' to 5' direction, the new mRNA molecule is synthesized in an antiparallel fashion, growing from its 5' end to its 3' end.
- Alternative splicing is a process in eukaryotes that contributes to protein diversity. Which statement provides the most accurate description of this process?
Answer: A single pre-mRNA transcript can be processed into different mature mRNAs by including or excluding certain exons.
Alternative splicing occurs after transcription and before translation. It involves the differential removal of introns and the joining of exons from a single pre-mRNA molecule in various combinations. This results in the production of multiple, distinct mature mRNA molecules from a single gene, each of which can be translated into a different protein isoform.
- During the elongation phase of translation, a polypeptide is constructed using tRNA molecules with the following anticodon sequence (read 3' to 5'): UAC, CGA, UUG. What was the sequence of the DNA coding strand that produced this message?
Answer: 5'-ATG GCT AAC-3'
First, determine the mRNA codons by finding the complementary bases to the tRNA anticodons: Anticodons 3'-UAC CGA UUG-5' pair with mRNA codons 5'-AUG GCU AAC-3'. Second, the DNA coding strand has the same sequence as the mRNA, but with Thymine (T) instead of Uracil (U). Therefore, the DNA coding strand sequence is 5'-ATG GCT AAC-3'.
- How many distinct nucleotide sequences of mature mRNA could code for this specific polypeptide segment?
Answer: 8
To find the total number of possible mRNA sequences, multiply the number of codon options for each amino acid. Methionine (Met) has 1 codon (AUG). Phenylalanine (Phe) has 2 codons (UUU, UUC). Proline (Pro) has 4 codons (CCU, CCC, CCA, CCG). The total number of distinct sequences is 1 × 2 × 4 = 8.
