BIO 1001 Exam 4
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Questions and Answers

The picture below shows nucleosomes along a DNA strand. In what ways will region A and B differ?

  • A and C
  • Region A as an open conformation (correct)
  • A and B
  • Genes in region A will be transcribed at a higher rate. (correct)
  • Genes in region B will be transcribed at a higher rate.

What is the nucleotide sequence of the anticodon of the tRNA that normally carries tryptophan (trp)?

  • 5'-UGG – 3'
  • There is more than one anticodon for trp.
  • 5'-ACC – 3'
  • 3'-ACC – 5' (correct)
  • 3'-UGG – 5'

RNA polymerase reads the ______ strand during transcription.

  • DNA coding strand
  • RNA template strand
  • RNA coding strand
  • DNA template strand (correct)
  • Polypeptide

Based on the lac operon, shown above, what are the current environmental conditions of the cell?

<p>Low glucose, high lactose (C)</p> Signup and view all the answers

The cell becomes exposed to high levels of glucose, would the expression of the operon change and why?

<p>Yes, the activator would become inactive. (A)</p> Signup and view all the answers

The gene XLM1 is expressed in skin cells and muscle cells. This gene undergoes alternative splicing. One exon that is INCLUDED in skin cells is EXCLUDED in muscle cells. Researchers isolate DNA and mRNA from both types of cells. How do you expect the XLM1 gene to differ in those samples?

<p>The gene will be the same length in both cells. The mRNA will be longer in skin cells. (B)</p> Signup and view all the answers

The picture above shows a ribosome in the process of translating mRNA. What is the sequence of the anticodon for the exposed codon in the A-site?

<p>5'-UCA – 3' (C)</p> Signup and view all the answers

After initiation, elongation, and termination of translation, what will be the final polypeptide sequence be?

<p>Met - Gly – Pro – Leu – Ser – Gly (D)</p> Signup and view all the answers

A cluster of genes under the control of one promoter is ______.

<p>an operon</p> Signup and view all the answers

A segment of a chromosome becoming attached to a different chromosome is an example of ______.

<p>Chromosomal translocation (C)</p> Signup and view all the answers

The image to the right shows RNA polymerase (the gray oval) transcribing a region of DNA. The mRNA is identified in the picture by the arrow. For one strand of DNA some of the sequence is shown. What sequence would be found in the RNA for this region?

<p>5'-TTACCCG-3' (B)</p> Signup and view all the answers

A mutation deletes most of the operator sequence in the lac operon. Which option most accurately describes how this would affect the expression of the lac operon? Assume that no glucose is present.

<p>The operon will be expressed in the presence of lactose, and lactose can be metabolized by the cell. (B)</p> Signup and view all the answers

A mutation in which location is least likely to impact gene expression?

<p>Intron (D)</p> Signup and view all the answers

In scenario 1, a(n) ______ is causing ______ in a ______ cell.

<p>Repressor, negative control, prokaryotic cell (E)</p> Signup and view all the answers

Scenario 2 shows the same operon, but in the presence of A. How would you best describe what A is?

<p>Enhancer (D)</p> Signup and view all the answers

In bacteria, an operon encodes several enzymes needed for the synthesis of the amino acid proline. Researchers notice that when proline levels are high in cells, the operon is not expressed. What would be an appropriate hypothesis based on this observation?

<p>Proline acts as a corepressor, binding to a repressor protein to allow the repressor to bind the operator and block transcription of the proline operon.</p> Signup and view all the answers

With mutation #1 (ignore mutation #2), how will the encoded protein be different from the wild-type?

<p>It will have a different primary structure than the wild-type's protein. (C)</p> Signup and view all the answers

Mutation #2 is a ______, and the resulting protein will be ______ the wild-type protein (ignore mutation #1).

<p>Nonsense mutation; shorter than (D)</p> Signup and view all the answers

Events in eukaryotic transcription are listed below. Put them in the order in which they occur. (Note that one of the events does not occur).

<p>III, I, IV (C)</p> Signup and view all the answers

Where do transcription, translation, and splicing occur in a prokaryotic cell?

<p>Transcription and translation in the cytosol, splicing does not happen. (A)</p> Signup and view all the answers

In gene expression, the input of a gene is ______, and the final output is ______.

<p>DNA, protein</p> Signup and view all the answers

In eukaryotic cells, the small ribosomal subunit recognizes the ______, once attached it will scan the mRNA and ______.

<p>16s rRNA, tRNAMet will recognize the start codon. (A)</p> Signup and view all the answers

Researchers identify a mutation (a deletion of 10 base pairs) about 250 bases upstream of the transcription start site of the VRN4 gene of mice. The result of the mutation is that transcription and mRNA levels of the VRN4 gene are much lower in the mutant than in wildtype mice. What can you conclude about the mutation?

<p>It deletes the promoter sequence. (D)</p> Signup and view all the answers

It has been noted that individuals with colorectal cancer have a mutation in the KRAS gene increasing its activity and leading to excess cell proliferation. The KRAS gene is an example of what?

<p>Oncogene (A)</p> Signup and view all the answers

How would the expression of a eukaryotic gene change if an enzyme adds methyl groups to the core promoter region of the gene?

<p>Transcription will decrease, resulting in less protein. (B)</p> Signup and view all the answers

What is the mRNA transcript of the above DNA strand?

<p>5'-UUAGCACCCACUAUGAAACAUUAA AUA – 3' (D)</p> Signup and view all the answers

What is the polypeptide sequence that would result if this DNA strand was expressed?

<p>Met- Phe – His – Ser – Gly – Cys (B)</p> Signup and view all the answers

A mutation results in the loss of the aminoacyl-tRNA synthetase responsible for attaching methionine (met) to tRNAmet. What would be the most likely effect of this mutation? (Note: Met = Methionine)

<p>Translation would not be initiated, and no protein would be made. (A)</p> Signup and view all the answers

Researchers develop a technique to isolate ribosomes from the cytoplasm of eukaryotic cells. Which of the following molecules would NEVER be found in one of these samples?

<p>DNA (A)</p> Signup and view all the answers

Which answer shows the mRNA transcript that the DNA coding strand below would make if it was transcribed?

<p>5'- UAC AUG AGU ACA CCA GCA AUC – 3' (E)</p> Signup and view all the answers

Which of the following mutations would be expected to produce the largest change to the structure and function of a protein? (Assume all mutations occur in the coding region of the gene.)

<p>A nonsense mutation in the 5' end of the gene. (E)</p> Signup and view all the answers

Which of the following options describes conditions when the trp operon will be expressed?

<p>Absence of tryptophan with an inactive repressor. (A)</p> Signup and view all the answers

It is common to get a DNA sample from cheek cells collected by swabbing the inside of the mouth. Would it be possible to determine a person's hair color from the cheek cell's DNA? If not, why?

<p>No, cheek cells contain genomes that only contain genes that the cheek cell needs to function. (D)</p> Signup and view all the answers

Wonka is a type of cancer that involves the loss-of-function mutation of the gene, chocolate, a tumor suppressor gene. What is the function of this gene when it is working normally?

<p>The chocolate gene acts as a tumor suppressor gene that normally prevents cells from dividing uncontrollably. When the gene is mutated, cancer can develop.</p> Signup and view all the answers

Which of the following statements is NOT TRUE of the proteins synthesized by a polysome.

<p>They will have the same amino acid at the C-terminus. (C)</p> Signup and view all the answers

Prokaryotic genes have a promoter, coding region, and terminator. ______ plays an important part in RNA polymerase binding to DNA, and allowing RNA polymerase to transcribe which of these sequences (promoter, coding region, terminator)?

<p>Sigma factor, all three are transcribed. (C)</p> Signup and view all the answers

The formation of a stem-loop is important to which of the following processes?

<p>A and C (A)</p> Signup and view all the answers

If gene P has a mutation, the cell could be rescued when supplemented with...

<p>D or E (A)</p> Signup and view all the answers

All but one of the following statements are true of gene regulation in both prokaryotes and eukaryotes. Identify the one statement that is not true of both.

<p>Gene regulation is influenced by how tightly packed the chromatin is. (B)</p> Signup and view all the answers

Flashcards

Nucleosomes and gene expression

Regions of DNA with tighter nucleosome packing (closed conformation) are less accessible to transcription factors, leading to lower gene expression. Conversely, open conformations allow for more gene expression.

Anticodon of tryptophan tRNA

The anticodon of the tRNA that carries tryptophan is 3'-ACC-5'. This sequence is complementary to the tryptophan codon on mRNA (5'-UGG-3').

Strand read by RNA polymerase

RNA polymerase reads the template strand of DNA during transcription. This strand is complementary to the coding strand, which has the same sequence as the resulting mRNA (except for thymine replaced by uracil).

Lac operon conditions

The figure shows an active lac operon, indicating high lactose and low glucose levels. The repressor is inactive (not bound to the operator), and the CAP-cAMP complex (activator) is bound to the promoter, promoting transcription of lac genes.

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Glucose effect on lac operon

High glucose levels will inactivate the CAP-cAMP complex, thus reducing the expression of the lac operon. This occurs because glucose is the preferred energy source over lactose.

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Alternative splicing and DNA/RNA length

Alternative splicing can result in different mRNA transcripts from the same gene. In the example, the XLM1 gene has an exon included in skin cells but excluded in muscle cells. This means the XLM1 mRNA will be longer in skin cells and shorter in muscle cells. However, the DNA sequence of the gene itself remains unchanged.

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Anticodon in the A-site

The exposed codon in the A-site of the ribosome is 5'-GGA-3'. The anticodon of the tRNA that will bind to this codon is 3'-CCU-5'.

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Polypeptide sequence

The final polypeptide sequence translated from the mRNA shown is Met-Gly-Pro-Leu-Ser-Gly. This sequence is determined by the order of codons on the mRNA and the corresponding amino acids.

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Cluster of genes under one promoter

An operon is a cluster of genes under the control of a single promoter. This allows for coordinated expression of these genes, often involved in a specific metabolic pathway.

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Chromosomal translocation

Chromosomal translocation involves a segment of one chromosome breaking off and attaching to a different chromosome. This can lead to altered gene expression and potentially disease.

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DNA sequence and RNA transcript

The RNA transcript would be 5'-UUACCCG-3'. This is because RNA polymerase uses the template strand to build the complementary mRNA, replacing thymine (T) with uracil (U).

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Operator mutation and lac operon

A mutation deleting most of the operator sequence in the lac operon would result in constitutive (constant) expression of the operon, regardless of the presence of lactose. This occurs because the repressor can no longer bind to the operator to block transcription.

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Least likely mutation to impact gene expression

Mutations within introns are least likely to impact gene expression because introns are non-coding regions that are removed during splicing. Mutations in promoters, operators, splice sites, or translational regulatory elements are more consequential.

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Repressor, negative control, prokaryotic

Scenario 1 depicts a repressor molecule engaging in negative control of gene regulation in a prokaryotic cell. The repressor binds to the operator, blocking RNA polymerase and thus inhibiting transcription.

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Activator in scenario 2

A is a small effector molecule that binds to the activator protein, changing its conformation to allow it to bind to DNA and promote transcription. The activator functions in positive control of gene regulation.

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Proline and operon hypothesis

The hypothesis is that proline acts as a small effector molecule, binding to a repressor protein. This binding activates the repressor, allowing it to bind to the operator and block transcription of the proline operon. This is a form of negative feedback regulation.

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Mutation #1 and protein effect

Mutation #1 is a missense mutation, resulting in a different amino acid being incorporated into the protein. This change can affect the protein's structure and function.

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Mutation #2 and protein effect

Mutation #2 is a nonsense mutation that introduces a premature stop codon. This results in a shorter protein than the wild type because translation terminates prematurely at the stop codon.

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Order of eukaryotic transcription events

The correct order of eukaryotic transcription events is: III, I, IV. 1. Transcription factors bind to the promoter.2. The 5' cap is added.3. The poly-A tail is added.RNA polymerase II recognizes the start codon is not a step in eukaryotic transcription.

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Transcription, translation, and splicing in prokaryotes

In prokaryotic cells, transcription and translation occur in the cytoplasm, and splicing doesn't happen. Prokaryotes lack a nucleus, so all steps occur in the same compartment.

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Input and output of gene expression

The input of gene expression is DNA, and the final output is protein.

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Small ribosomal subunit recognition

In eukaryotic cells, the small ribosomal subunit recognizes the 5' cap of the mRNA. It then scans the mRNA until it encounters the start codon, AUG, where translation initiation begins.

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VRN4 mutation and conclusion

The mutation upstream of the VRN4 transcription start site likely affects an enhancer sequence. This sequence regulates gene expression by enhancing RNA polymerase binding and transcription. Deletion would result in reduced mRNA levels and thus lower VRN4 protein production.

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KRAS gene and cancer

A mutated KRAS gene with increased activity is classified as an oncogene. Oncogenes promote uncontrolled cell proliferation, contributing to cancer development.

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Methylation and gene expression

Adding methyl groups to the core promoter region of a gene is associated with gene silencing, reducing transcription and ultimately protein production.

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mRNA transcript

The mRNA transcript of the DNA strand is 5' AUGAAACAUUAAAUA 3'. This sequence mirrors the coding strand (except for thymine replaced by uracil), and represents the RNA molecule that will be translated.

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Polypeptide sequence from DNA

The polypeptide sequence is Met-Lys-Leu. This is determined by the mRNA sequence and the genetic code.

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Aminoacyl-tRNA synthetase mutation effect

The loss of the methionine (Met) aminoacyl-tRNA synthetase would prevent translation initiation. This is because tRNAmet must be charged with Met to recognize the start codon AUG, which is essential for initiating protein synthesis.

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Molecules found in ribosomes

Ribosomes contain rRNA (ribosomal RNA), tRNA (transfer RNA), and proteins. They do not contain DNA, which is located in the nucleus of eukaryotic cells.

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DNA coding strand and mRNA transcript

The mRNA transcript would be 5' UAC AUG AGU ACA CCA GCA AUC 3'. The coding strand and the mRNA have the same sequence, except for thymine (T) replaced by uracil (U) in the mRNA.

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Largest change in protein structure/function

A nonsense mutation in the 5' end of the gene is most likely to produce the largest change in protein structure and function. This is because it results in premature termination of translation, producing a truncated and likely non-functional protein.

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Trp operon expression

The trp operon is expressed in the absence of tryptophan with an inactive repressor. This means that when tryptophan levels are low, the repressor is unable to bind to the operator, allowing RNA polymerase to transcribe the operon and synthesize tryptophan.

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Hair color from cheek cell DNA

It is possible to determine a person's hair color from the cheek cell DNA. Cheek cells contain the complete genome, including the genes that determine hair color.

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Tumor suppressor gene function

When functioning normally, the chocolate tumor suppressor gene would stop the cell cycle when DNA damage is detected. This prevents cells with damaged DNA from replicating, thus reducing cancer risk.

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Polysome protein differences

Although polysome-synthesized proteins have the same primary structure, they can differ in length and amino acid sequence. This difference arises from differences in post-translational modifications and polypeptide processing.

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Sigma factor and transcription

Sigma factor plays a crucial role in the initiation of transcription in prokaryotes. It assists RNA polymerase in recognizing and binding to the promoter region on DNA. This process allows proper initiation of transcription of the coding region and terminator.

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Stem-loop formation importance

Stem-loop formation is essential for termination of transcription in prokaryotes. It is also vital for the structure and function of tRNA molecules.

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Order of enzymes in metabolic pathway

Based on the data provided, the correct order of enzymes in the metabolic pathway is: Enzyme 1, Enzyme 4, Enzyme 3, Enzyme 2. This order is determined by the rescue results with different supplements.

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Gene P mutation and rescue

If gene P has a mutation, the cell could be rescued when supplemented with E only. This indicates that Enzyme 2 is responsible for converting D to E.

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Gene regulation difference between prokaryotes and eukaryotes

Chromatin structure, the way DNA is packed in eukaryotic cells, plays a significant role in gene regulation. Prokaryotic DNA is not organized into chromatin, and therefore this factor is absent in prokaryotes.

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Study Notes

Exam 4 - Answer Key

  • Exam is for BIOL1001 (ABRAMOVICH) FA’24
  • Answer key provided for multiple-choice questions.
  • Please bubble in answers on scantron.
  • Discussion section and TA name required if desired
  • A codon table is on the last page of the exam; use as needed.
  • Picture of nucleosomes shows nucleosomes along a DNA strand.
  • Region A is in an open conformation. Genes in region A transcribe at a higher rate than region B.
  • The anticodon sequence for tryptophan (trp) tRNA is 5'-UGG-3'.
  • RNA polymerase reads the DNA template strand during transcription.
  • Environmental conditions: High glucose, low lactose.
  • Expression of lac operon:
  • High glucose, expression of the operon remains unchanged.
    • Repressor is already inactive in the presence of high glucose.
  • XLM1 gene: differs in length of mRNA in skin and muscle cells.
  • In skin cells the XLM1 gene's mRNA is shorter than the muscle cell mRNA.
  • The ribosome is translating mRNA, the anticodon for the exposed codon in the A-site is 5'-UCA-3'.
  • Final polypeptide sequence: Met-Gly-Pro-Leu-Ser-Arg-Val-Ser
  • A cluster of genes under the control of one promoter is an operon.
  • A segment of a chromosome becoming attached to a different chromosome is a chromosomal translocation.
  • RNA polymerase directs the sequence of the mRNA strand via the DNA template sequence during transcription.
  • Mutation in the operator sequence of a lac operon, results in the operon functioning more frequently.
  • A mutation is a change in nucleotide sequence of a DNA molecule.
  • Mutation #1 alters the protein sequence in a way that determines premature termination of translation.
  • Mutation #2 results in a frame shift mutation which will result in a longer protein than the wild-type.
  • Events in eukaryotic transcription (out of order, some not applicable): Addition of the 5' cap. RNA polymerase recognizes the start codon. Transcription factors bind to the promoter. Addition of the poly-A tail.
  • Transcription, translation, and splicing occur in the cytosol of a prokaryotic cell, without a nucleus.
  • The input to gene expression is DNA and the final output is protein.
  • The small ribosomal subunit recognizes the 16s rRNA to initiate translation.
  • Mutation in VRN4 upstream of the transcription site lead to transcription decline.
  • Mutation increased cell proliferation as a KRAS oncogene mutation.
  • Methylation of the core promoter region of eukaryotic genes decrease transcription.
  • A polysome is formed from multiple ribosomes translating the same mRNA.
  • The correct order for a metabolic pathway is Enzyme 2, Enzyme 1, Enzyme 4, Enzyme 3.
  • A mutation or loss of aminoacyl-tRNA will prevent protein production, as the amino acid is not properly attached to the tRNA.
  • Changes in the nucleotide sequence lead to changes of the final protein product.
  • If the operator region of the lac operon is mutated, the operon will be frequently expressed even in the absence of lactose
  • The environment (presence of glucose, lactose etc) directly influences the gene expression
  • The function of a tumor suppressor is normally to stop cell division in the event of DNA damage.

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