DOC PREVIEW
ISU BBMB 405 - Final Exam Study Guide
Type Study Guide
Pages 8

This preview shows page 1-2-3 out of 8 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 8 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 8 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 8 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 8 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

BBMB 405 1st EditionExam #4 Study Guide Lectures: 34—41 Chapter 31: Lectures 34 – 361. Predict the effects of deleting the following regions of DNA:(a) The gene encoding lac repressor(b) The lac operator(c) The gene encoding CAPa. The gene encoding lac repressor: When the gene for the lac repressor is deleted, the lac repressor protein can not be produced. When the lac repressor protine doesn’t bind to the operator, the lac operon will not be regulated. The cell espresses beta-galactosidase,lac permease, and thiogalactidasetransacetylase gene even in the absence of lactose.b. The lac operator: The gene for the lac operator is deleted. The repressor protein cannot bind to it and the lac operon is not regulated. Cell expresses beta-galactosidase, lac permease, and thiogalacotidasetransacetylase genes in absence of lactose.c. The gene encoding Catabolite Activator Protein (CAP): When glucose is at a low level in the cell, levels of cAMP rise and activate CAP. CAP binds to CAP binding site in lac operon and induces transcription of lac operon. When gene encoding CAP is deleted, the levels of catabolic enzymes will remain low even at low levels of glucose.4. The lac repressor and the pur repressor are homologous proteins with very similar three-dimensional structures, yet they have different effects on gene expression. Describe two important ways in which the gene-regulatory properties of these proteins differ.a. In the presence of an inducer the lac repressor doesn’t bind to DNA because the inducerbinds to the repressor and doesn’t allow it to bind to DNA. The pur repressor binds to DNA only in the presence of a corepressor which is a small molecule that binds to the repressor and helps it bind to DNAb. E. coli genome contains only one site for binding of lac repressor, but there are many sites for binding of pur repressor.9. What is the effect of an increased Cro concentration on the expression of the gene for the λ repressor? Of an increased concentration of λ repressor on the expression of the Cro gene? Of an increased concentration of λ repressor on the expression of the λ repressor gene?A: Cro and lambda repressor share operators Or1, Or2, Or3 and are regulated through these operators. When [Cro] increases, it binds to Or3 operator and prevents lambda repressor transcription. When lambda repressor concentration increases, it binds to Or1 and Or2 operators and prevents Cro transcription. Lambda repressor will continue to be produced untilit reaches a higher concentration, where the lambda repressor will bind to Or3 operator preventing its own transcription.Chapter 32: Lectures 36 – 383. Assuming that 145 base pairs of DNA wrap around the histone octamer 1 3/4 times, estimate the radius of the histone octamer. Assume 3.4 Å per base pair and simplify the calculation by assuming that the wrapping is in two rather than three dimensions and neglecting the thicknessof the DNA.a. Calculate length of DNA wrapped around histone octamer by multiplying number of base pairs with given length per base pair: 145 bp x 3.4 A per bp = 493 Ab. Determine circumference by dividing length by number of times DNA wraps around octomer: 493 A/1.75 = 282 Ac. Divide circumference (C) by 2*pi to give radius: r= C/2pi r=282/2pi = 44.8 A4. Growth of mammalian cells in the presence of 5-azacytidine results in the activation of some normally inactive genes. Propose an explanation.A: 5-Azacytidine cannot be methylated. The explanation for the activation of normally inactive genes in the presence of 5-Azacytidine is that they normally are repressed by methylation and they are active.5. A protein domain that recognizes 5-methylcytosine in the context of double-stranded DNA has been characterized. What role might proteins containing such a domain play in regulating gene expression? Where on a double-stranded DNA molecule would you expect such a domain to bind?A: Proteins containing these domains will be targeted to methylated DNA in repressed promoterregions, regulating expression. The domains would bind to the major groove of DNA where methyl groups are situated.6. Through recombinant DNA methods, a modified steroid hormone receptor was prepared thatconsists of an estrogen receptor with its ligand-binding domain replaced by the ligand-bindingdomain from the progesterone receptor. Predict the expected responsiveness of gene expression for cells treated with estrogen or with progesterone.A: Steroid hormone receptors are activated by a specific ligand. The receptor contains a ligand-binding domain for progesterone so the receptor will only be activated with progesterone and will not respond to estrogen.The receptor contains signal transduction machinery that are from original estrogen receptor. The receptor can be activated by progesterone, it will signal gene production naturally caused by estrogen activation.7. What is the effect of acetylation of a lysine residue on the charge of a histone protein? Of lysine methylation?A: Acetylation of a lysine residue results in the loss of the residue’s positive charge and it becomes neutral. Methylation of a lysine residue will not change the positive charge.8. The following amino acid sequence of one of the four transcription factors is used to generateiPS cells:This transcription factor belongs to one of the three structural classes discussed in Section 32.2. Identify the class.A: The sequence doesn’t contain many leucine residues so it is not a basic residue leucine zipper. Sequence is rick in cysteine and histidine residues so the sequence contains zinc-finger domains.10. What effect would you expect from the addition of an IRE to the 5 end of a gene that is not ′normally regulated by iron levels? To the 3 end?′A: Iron response elements (IREs) are used to regulate translation of mRNA sequences following the element based on absence or presence of iron. When IRE is added to 5’-end of mRNA sequence, transcript would not be translated without iron. IRE should not affect translation of mRNA sequence if placed after transcript at 3’-end.11. Suppose that you have identified amiRNA that has the sequence 5 -′GCCUAGCCUUAGCAUUGAUUGG-3 .′ Propose a strategy for identifying mRNA that might be regulated by this miRNA, given the sequences of all mRNAs encoded by the human genome.A: miRNA regulate mRNA sequences by base pairing to miRNA transcript to block translation. To determine which mRNA sequences are targeted by this


View Full Document

ISU BBMB 405 - Final Exam Study Guide

Type: Study Guide
Pages: 8
Documents in this Course
Load more
Download Final Exam Study Guide
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Final Exam Study Guide and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Final Exam Study Guide 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?