BIOLOGY
12th Edition
ISBN: 9781260169614
Author: Raven
Publisher: RENT MCG
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Chapter 16, Problem 6A
In the cell cycle, cyclin proteins are produced in concert with the cycle. This likely involves
a. control of initiation of transcription of cyclin genes, and ubiquitination of cyclin proteins.
b. alternative splicing of cyclin genes to produce different cyclin proteins.
c. RNA editing to produce the different cyclin proteins.
d. transcription/translation coupling.
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Transcription factors such as myoD control gene expression in which of the following ways?
Group of answer choices
A. direct control of transcription of specific genes
B. control of the cell cycle
C. general control of transcription through chromatin modification
D. control of RNA splicing
E. control of translation
A cell is homozygous for a LOF mutation in the histone acetyltransferase (HAT) gene. Which of the following is/are true?
a. Chromatin in the cell will be more condensed than usual, decreasing expression of genes that should be expressed.
b. This will lead to histones with more positive charges on their tails, increasing the interaction between histones and the DNA backbone.
c. This will lead to a decrease in DNA methylation and therefore cause an increase in gene expression.
d. Chromatin in the cell will be less condensed than usual, increasing expression of genes that should not be expressed.
Which of these is NOT true of nucleosomes?
A. Some post-translational modifications to histone proteins serve as binding sites for transcription factors.
B. The position of nucleosomes is fixed on the genome and can not be changed.
C. The position and post-translational modifications of histones can be inherited through mitosis.
D. The binding between DNA and histone proteins can be disrupted by post-translational modifications.
Chapter 16 Solutions
BIOLOGY
Ch. 16.1 - Prob. 1LOCh. 16.1 - Prob. 2LOCh. 16.1 - Prob. 3LOCh. 16.2 - Explain how proteins can interact with base-pairs...Ch. 16.2 - Prob. 2LOCh. 16.3 - Prob. 1LOCh. 16.3 - Prob. 2LOCh. 16.3 - Explain control of gene expression in the trp...Ch. 16.4 - Prob. 1LOCh. 16.4 - Prob. 2LO
Ch. 16.4 - Prob. 3LOCh. 16.5 - Describe at least two kinds of epigenetic mark.Ch. 16.5 - Explain the function of chromatin-remodeling...Ch. 16.6 - Prob. 1LOCh. 16.6 - Prob. 2LOCh. 16.7 - Prob. 1LOCh. 16.7 - Prob. 2LOCh. 16 - Prob. 1DACh. 16 - What advantage might a bacterium gain by linking...Ch. 16 - Prob. 2IQCh. 16 - Prob. 3IQCh. 16 - In prokaryotes, control of gene expression usually...Ch. 16 - Prob. 2UCh. 16 - Prob. 3UCh. 16 - The lac operon is controlled by two main proteins....Ch. 16 - In eukaryotes, binding of RNA polymerase to a...Ch. 16 - In eukaryotes, the regulation of gene expression...Ch. 16 - In the trp operon, the repressor binds to DNA a....Ch. 16 - Prob. 1ACh. 16 - Specific transcription factors in eukaryotes...Ch. 16 - Repression in the trp operon and induction in the...Ch. 16 - Regulation by small RNAs and alternative splicing...Ch. 16 - Eukaryotic mRNAs differ from prokaryotic mRNAs in...Ch. 16 - In the cell cycle, cyclin proteins are produced in...Ch. 16 - A mechanism of control in E. coli not discussed in...Ch. 16 - You have isolated a series of mutants affecting...Ch. 16 - Examples of positive and negative control of...Ch. 16 - What forms of eukaryotic control of gene...Ch. 16 - The number and type of proteins found in a cell...
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- You add a chemical to a cell that prevents all of the chromatin from uncoiling so that the chromosomes remain highly coiled, even after the completion of mitosis. If you add a transcription factor that triggers the transcription of mRNA for the protein Pitx in this situation what would happen. Group of answer choices A)All transcription of Pitx is stopped by the transcription factor B)Nothing because the Pitx gene is already being transcribed at the maximum ratea C)Nothing, because the Pitx gene cannot be transcribed in this highly coild state D)The amount of transcription of Pitx is greatly increasedtarrow_forwardDefine the following terms: a. promoter b. consensus sequence c. operon d. chromatin-remodeling complex e. general transcription factorsarrow_forwardTranscription factors such as myoD control gene expression in which of the following ways? Group of answer choices ONLY ONE RIGHT ANSWERZ: A. direct control of transcription of specific genes B. control of the cell cycle C. general control of transcription through chromatin modification D. control of RNA splicing E. control of translationarrow_forward
- Which of the following statements correctly describes the structure of chromatin around the promoter of an active eukaryotic gene? O A. It is in the form of heterochromatin and the nucleosomes are tightly wrapped. O B. It is in the form of euchromatin and the nucleosomes are cleared from the region allowing for an open and accessible configuration OC.A repressor protein is bound to the chromatin blocking RNA polymerase. O D.Both B and C.arrow_forwardWhich of the following statements about methylation and acetylation is correct? A. Genes that have hypermethylated (a lot of methylation) promoters are expressed at high levels. B. Deacetylation of histones can lead to a heterochromatin state. C. All nucleotides (CTAG) can be methylated. D. Histones cannot be methylated.arrow_forward3) The tumor suppressor protein Rb regulation of the entry into the S phase of the cell cycle is represented in this diagram. DNA Answer: b) Explain your choice above: Answer: Rb E2F Genes needed for S phase are NOT transcribed Growth factor Ras pathway Cdk-cyclin 30 ATP ADP Phosphorylated Rb protein P Rb E2F Gene transcription a) In hereditary retinoblastoma tumors, Rb is mutated. Among the following mutations, which one is not likely to be found in these tumors. 1) Mutation prevents Rb to bind E2F by modifying the binding site. 2) Mutation prevents Rb to be dephosphorylated and recycled (possibly by prevented phosphorylated Rb to be recognized by the phosphatase that removes its phosphates). 3) Mutation may cause Rb to be misfolded and not have a functional conformation 4) Mutation that prevent Rb to be phosphorylated by cdk-cyclin. 5) Mutation may cause Rb to be unstable and degraded rapidly. c) (4 pts) Human papilloma virus (HPV) infections are the main causes of cervical cancers.…arrow_forward
- 1)A. how do you read a sequence of DNA (template or non-template strand) to convert it an mRNA sequence and to a protein? B.How does chromatin remodeling regulate gene transcription? C. What are the major differences between gene expression in bacteria and eukaryotes D. How are non-coding regions involved in gene transcription? E. Explain how eukaryotic genes sometimes produce multiple protein products?arrow_forwardIdentify the eukaryotic level of gene regulation. Choose the correct answer below. a. Pre- transcriptional control b. transcriptional control c. translational control d. post-translational control 1. in the fruit fly, genes from rRNA can be replicated more or less often compared to the rest of the chromatin depending on the needs of the cell. 2. in the human beta-globin, two introns are spliced out in order to produce the mature mRNA. 3. DNA methylation can change the degree of condensation of the chromatin. 4. The mouse REST gene is under the control of a promoter region that contains alternative promoters.arrow_forwardWhich of the following statements concerning p53 is NOT correct? O a. p53-dependent transcription of target genes leads to cell cycle arrest or cell death (apoptosis). O b. O C. Loss of p53 function results in loss of checkpoint controls. Loss of p53 function creates an environment that is permissive for genome instability--that is, more damaged cells with chromosome aberrations and mutations survive and propagate. d. p53 is a tumor suppressor protein which is largely responsible for protecting cells from cancer- causing DNA-damaging agents.arrow_forward
- What would the direct consequence to a cell be if there was a loss of function of Transcription Factor II D? (Only select processes in which TFIID has a role.) You may answer multiple answers. a. Terminators at the end of the coding regions of genes would not be recognized. b. In the cytoplasm, mRNA would not be protected from enzymes that degrade mRNA. c. Primary RNA transcripts would not be accurately spliced. d. Initiation of transcription would be absent or decreased. mRNAs would be unable to exit the nucleus.arrow_forwardWhich of the following is true of CpG islands? a. They are methylated near promoters of actively transcribed genes. b. They are unmethylated near promoters of actively transcribed genes. c. Acetylation of CpG islands leads to repression of transcription. d. CpG islands code for RNA molecules that activate transcription.arrow_forwardWhich of the following is the most likely effect of a mutation in the gene coding for a DNA repair enzyme? a.The mutated gene will not be transcribed because RNA polymerase cannot transcribe mutated DNA. b.Mutations will accumulate more quickly because the cell will not be able to fix errors in replication. c.The cell will immediately undergo apoptosis so that mutated DNA is not replicated in future rounds of cell division. d.The cell containing the mutation will divide more frequently because the cell cycle checkpoints will not function properly.arrow_forward
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