Biology: DNA Replication, Transcription and Translation
15 practice questions · 15 flashcards · made from study notes
Terms in this set (15)
- What is the basic structure of DNA?
- A double helix of two antiparallel strands.
- The two antiparallel strands of DNA are held together by ___ bonds between complementary bases.
- hydrogen
- Define complementary base pairing in DNA.
- Adenine pairs with thymine (two hydrogen bonds) and guanine pairs with cytosine (three hydrogen bonds).
- What does it mean for DNA replication to be 'semi-conservative'?
- Each new double helix contains one original strand and one new strand.
- The enzyme responsible for unwinding the DNA helix at the replication fork is ___.
- Helicase
- Why does primase lay down a short RNA primer during DNA replication?
- Because DNA polymerase can only add nucleotides to an existing 3' end.
- Which enzyme synthesizes new DNA in the 5' to 3' direction?
- DNA polymerase III
- What are Okazaki fragments and on which strand are they found?
- Short, discontinuously synthesized DNA fragments found on the lagging strand.
- What is the main difference in base pairing between DNA and RNA during transcription?
- RNA uses uracil in place of thymine.
- In eukaryotes, what are the three main processing steps for pre-mRNA in the nucleus?
- A 5' cap is added, a poly-A tail is added to the 3' end, and introns are removed by the spliceosome while exons are joined.
- Translation occurs at ___ in the cytoplasm.
- ribosomes
- What is the start codon for translation and what amino acid does it code for?
- AUG, which codes for methionine.
- What is the function of tRNA molecules during translation?
- They carry specific amino acids and recognize codons through their anticodons.
- Why is the genetic code described as 'redundant'?
- Most amino acids are specified by more than one codon.
- What type of mutation occurs when an insertion or deletion is not a multiple of three, leading to a change in every codon after it?
- Frameshift mutation
Practice questions (15)
1.In a DNA double helix, if one strand runs in the 5' to 3' direction, the other complementary strand runs in the 3' to 5' direction. This arrangement is known as antiparallel.
- ATrue
- BFalse
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Answer: True
This question directly assesses the student's understanding of the term 'antiparallel' as it applies to the orientation of the two strands in a DNA molecule, a key feature of its structure.
2.What is the primary function of DNA ligase during DNA replication?
- ATo join Okazaki fragments on the lagging strand
- BTo synthesize the RNA primers
- CTo unwind the DNA double helix
- DTo add new nucleotides to the growing DNA strand
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Answer: To join Okazaki fragments on the lagging strand
This question tests the student's knowledge of the specific enzymes involved in resolving the discontinuous synthesis on the lagging strand. DNA ligase's role is to form phosphodiester bonds to seal the nicks between the DNA fragments that have replaced the RNA primers, creating a continuous strand. Understanding this function is key to grasping how the lagging strand is completed.
3.During DNA replication, the lagging strand is synthesized discontinuously as Okazaki fragments because DNA polymerase III can only add nucleotides to an existing 3' end and synthesizes in a 5' to 3' direction.
- ATrue
- BFalse
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Answer: True
This question assesses the student's comprehension of why the leading and lagging strands are synthesized differently. It connects the directional constraint of DNA polymerase III (5' to 3' synthesis) with the antiparallel nature of the DNA template, forcing the lagging strand to be synthesized in fragments away from the replication fork.
4.In the Meselson-Stahl experiment, which demonstrated the semi-conservative nature of DNA replication, what would the composition of DNA molecules be after one round of replication if the process were conservative instead of semi-conservative?
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Answer: If replication were conservative, after one round there would be two distinct types of DNA molecules: one double helix consisting entirely of the original 'heavy' DNA and one double helix consisting entirely of new 'light' DNA. There would be no hybrid molecules.
This question challenges students to apply their understanding of the semi-conservative model by contrasting it with an alternative hypothesis (conservative replication). To answer correctly, they must infer the experimental outcome of a different model, demonstrating a deeper grasp of the logic behind the Meselson-Stahl experiment and the meaning of 'semi-conservative'.
5.Considering that DNA polymerase requires a pre-existing 3' end to begin synthesis, what specific molecule provides this starting point for the synthesis of both the leading strand and each Okazaki fragment on the lagging strand?
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Answer: An RNA primer, synthesized by the enzyme primase, provides the necessary 3' end.
This question focuses on a crucial prerequisite for DNA polymerase action. It requires the student to identify not just the type of molecule (a primer) but its composition (RNA) and the enzyme responsible for its creation (primase). This highlights the fact that DNA synthesis cannot start from scratch and depends on a different type of nucleic acid to get going.
6.During transcription, what is the role of RNA polymerase?
- ATo unwind the DNA double helix and keep the strands separated.
- BTo read the DNA template strand and synthesize a complementary mRNA molecule.
- CTo add a 5' cap and a poly-A tail to the pre-mRNA.
- DTo remove introns and join exons together.
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Answer: To read the DNA template strand and synthesize a complementary mRNA molecule.
This question checks for a fundamental understanding of the central enzyme in transcription. RNA polymerase is responsible for synthesizing the mRNA transcript by reading the DNA template. The other options describe roles of different molecules (helicase, processing enzymes, spliceosome).
7.In eukaryotic cells, the initial mRNA transcript, or pre-mRNA, must be processed before it can be translated. What are the three main processing steps that occur in the nucleus?
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Answer: The three main processing steps are: the addition of a 5' cap, the addition of a poly-A tail to the 3' end, and the removal of introns by the spliceosome, which joins the exons.
This question requires students to recall the specific modifications that eukaryotic mRNA undergoes. This is a key difference between prokaryotic and eukaryotic gene expression and is crucial for understanding how a functional mRNA molecule is produced.
8.The process of translation terminates when the ribosome encounters a stop codon (UAA, UAG, or UGA) on the mRNA strand.
- ATrue
- BFalse
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Answer: True
This question assesses the student's knowledge of how the translation process is terminated. Understanding the role of stop codons is fundamental to grasping how a polypeptide chain of the correct length is synthesized.
9.What is the function of transfer RNA (tRNA) in the process of translation?
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Answer: Transfer RNA (tRNA) molecules act as adaptors by carrying specific amino acids and recognizing corresponding mRNA codons via their anticodons. This ensures the correct amino acid is added to the growing polypeptide chain.
This question probes the student's understanding of the crucial role tRNA plays in decoding the mRNA message into a protein. It connects the genetic code (codons) to the building blocks of proteins (amino acids).
10.A point mutation in a DNA sequence changes a codon from UGU to UGC. Both of these codons specify the amino acid Cysteine. What type of mutation is this?
- AFrameshift mutation
- BMissense mutation
- CNonsense mutation
- DSilent mutation
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Answer: Silent mutation
This question assesses the student's understanding of silent mutations. Because the genetic code is redundant, a change in a single nucleotide does not always lead to a change in the amino acid sequence. Recognizing this specific outcome helps differentiate silent mutations from other point mutations.
11.A mutation that introduces a UAG codon in the middle of a gene's mRNA sequence is a nonsense mutation.
- ATrue
- BFalse
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Answer: True
This question tests the student's knowledge of nonsense mutations and the specific stop codons. Knowing that UAG is a stop codon is crucial for understanding how this type of mutation can truncate a protein, which usually results in a loss of function.
12.Explain why an insertion of two nucleotides in a gene's coding sequence is likely to be more detrimental to the resulting protein than a missense mutation. Refer to the concept of a reading frame in your answer.
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Answer: An insertion of two nucleotides causes a frameshift mutation. This alters the reading frame for all subsequent codons, changing every amino acid from the point of the insertion onward and likely leading to a premature stop codon. In contrast, a missense mutation only changes a single amino acid, which may or may not significantly impact the protein's function.
This question challenges the student to compare the effects of two different mutation types (frameshift vs. missense) and analyze the severity of their impact. It requires them to connect the concept of the genetic code's reading frame to the functional consequences for the protein, demonstrating a deeper level of analysis.
13.Consider a gene where a specific G-C base pair undergoes a point mutation, resulting in an A-T base pair. If this change leads to the substitution of a polar amino acid with a nonpolar one in a critical region of the resulting protein, what type of mutation occurred and what is a likely consequence for the protein's function?
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Answer: This is a missense mutation. The substitution of a polar amino acid with a nonpolar one can disrupt the protein's tertiary structure, which is determined by interactions between amino acid side chains. This change could alter the protein's folding, stability, and ultimately its biological function, potentially rendering it inactive.
This question requires students to analyze a scenario and predict its outcome. They must identify the mutation type (missense) and then apply their knowledge of protein structure to infer the consequences. It challenges them to think about how a single molecular change can cascade into a significant functional impact, moving from definition to application.
14.During DNA replication, topoisomerase works to relieve the tension created by the unwinding of the DNA double helix by the helicase. What would be the most likely immediate consequence for the replication process if topoisomerase were absent or non-functional?
- AThe DNA strands would not separate at the replication fork.
- BThe RNA primers would not be replaced with DNA nucleotides.
- CThe replication fork would stall and DNA synthesis would stop due to supercoiling.
- DThe Okazaki fragments on the lagging strand would not be joined together.
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Answer: The replication fork would stall and DNA synthesis would stop due to supercoiling.
This question tests the understanding of the specific role of topoisomerase. As helicase unwinds DNA, it introduces torsional stress and supercoiling ahead of the replication fork. Without topoisomerase to relieve this strain, the replication machinery would be physically unable to proceed, leading to a halt in DNA synthesis.
15.Alternative splicing of pre-mRNA allows for a single gene to produce multiple different proteins. How does this process challenge the traditional 'one gene, one polypeptide' hypothesis?
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Answer: The 'one gene, one polypeptide' hypothesis proposed a direct, one-to-one relationship between a gene and the protein it codes for. Alternative splicing challenges this by demonstrating that a single gene's pre-mRNA can be processed in different ways, by treating different segments as introns or exons. This results in different mature mRNA molecules, which are then translated into distinct proteins with potentially different functions, showing that one gene can, in fact, code for many polypeptide
This question requires students to connect a specific mechanism (alternative splicing) to a foundational, historical concept in genetics. It challenges them to analyze how a molecular process leads to a more complex understanding of gene expression, moving beyond a simplified model to appreciate the versatility of the eukaryotic genome.
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