13. Two transcription factors, A and B, are required to bind an enhancer, E, in order for Gene X to be transcribed (see diagram) 00 Gene X Let a and b represent recessive amorphic alleles of the transcription factors and E represent a deletion of the enhancer. Assuming Gene X IS haplosufficient, what phenotype will the following individuals have? Individual 1: Ala B/b E/E Individual 2: A/A B/B E/E- a. Individual 1: Wildtype; Individual 2: Wildtype b. Individual 1: Mutant; Individual 2: Wildtype c. Individual 1: Wildtype; Individual 2: Mutant d. Individual 1: Mutant; Individual 2: Mutant

13. Two transcription factors, A and B, are required to bind an enhancer, E, in order for Gene X to be transcribed (see diagram) 00 Gene X Let a and b represent recessive amorphic alleles of the transcription factors and E represent a deletion of the enhancer. Assuming Gene X IS haplosufficient, what phenotype will the following individuals have? Individual 1: Ala B/b E/E Individual 2: A/A B/B E/E- a. Individual 1: Wildtype; Individual 2: Wildtype b. Individual 1: Mutant; Individual 2: Wildtype c. Individual 1: Wildtype; Individual 2: Mutant d. Individual 1: Mutant; Individual 2: Mutant

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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Q1. (Shown below is the structure of a Drosophila gene, divided into 10 segments, designated A-J. The gene contains three exons, two introns, a promoter, and a site in I for poly(A) addition. Exon 1 Exon 2 Exon 3 Promoter Intron 1 Intron 2 AATAA A в CDE FG|H a) What segment or segments of the gene will be represented in the initial RNA transcript (primary RNA)? List the appropriate letter or letters. b) What segment or segments of the gene will be found in the completely processed transcript (mature RNA)? c) What segment or segments of the gene in the processed transcript will have additional nucleotides added to them? d) If the segment... . is methylated, this could block the transcription's initiation as RNA polymerase will not be able to bind to this segment.
8. Below is a diagram of the MYC gene, match the arrows with the correct description. 127.73M Genes (Comprehensive set... MYC-201 protein coding - MYC-209 protein coding DO MYC-203> protein coding 0- protein coding 2 - MYC-208> protein coding O MYC-202 protein coding O MYC-205> protein coding MYC-207> protein coding D 0 MYC-204 processed transcript 1 MYC-211> protein coding ← MYC-206> protein coding 2 i. Arrow 1 is pointing at ii. Arrow 2 is pointing at A. polyadenylation site B. 3' untranslated region C. transcription start site D. intron -27-52 kb 127.74M - MYC-212 > arotein codina E. 5' untranslated region F. promoter region G. transcription stop site H. protein coding region 127.73
1. Histone methylation can have many different effects on gene expression. In some cases, histone methylation is associated with activation of transcription, whereas in other cases it can trigger the formation of heterochromatin and a decrease in transcription. If histone methylation has been detected in the region of gene YFG in yeast, describe an experiment that could distinguish whether the methylation is important to activate or repress transcription of gene YFG. al 6 2. An E. coli strain of chromosomal genotype lacl* lacP" lacO* lacz* lacY*. You wish to transform this strain into a wild-type lac operon by the addition of an extra piece of DNA (plasmid). a. What regulatory region(s), gene or genes would you add on this extra DNA that would make this strain express the lac operon genes as wild type? b. How would you design this plasmid if you want the strain of E. coli to express the lac operon genes even if lactose is absent in the medium (constitutively)?
6). To investigate the genetic control of locomotive behavior in the diploid nematode, Caenorhabditis elegans, eight "wiggly" mutants were isolated, which wiggle in an uncoordinated way rather than smoothly gliding. These eight mutants were crossed together in all possible pairwise combinations, yielding the results summarized in the table above (+ means that all the progeny were wild type, and m means that all the progeny showed the wiggly mutant phenotype). Organize the eight mutations into con lementation groups, each consisting of a group of mutations affecting the same gene. 1 2 3 4 5 6 7 8 1 E N+E 2 3+1 m + m 4 5 + m m E+ + + EE 6 ++E E+ +EE m m + + + + + + + + + m 7 8 ++4 ∞+E +E m m ++ E
1. You are studying a variant in a gene controlling wing development, Wng, in Drosophila. Using analysis software, you can see that the variant changes the protein product’s amino acid sequence, but homozygotes for this variant express a WNG protein that functions normally. Which term best describes the effect of this variant? conditional mutation gain of function mutation loss of function mutation neutral mutation synonymous mutation 2. In base excision repair, Photolyase breaks bonds between pyrimidines & Purines Single-stranded binding proteins are used to stabilize the unwound helix. Mut H recognizes the demythylated state, and removes the base DNA Glycolase removes the damaged nitrogenous base 3. Aneuploidy is a change in the number of individual chromosomes. Which of the following is not an example of aneuploidy? Nullisomy Trisomy Monosomy Polysomy Tetrasomy
Geneticists have been exploring ways to suppress the expression of the extra chromosome 21 in individuals with Down syndrome in hopes of preventing the medical problems and intellectual disability of individuals with trisomy 21. One approach involves modifying a gene that is already present in human cells and using it to suppress the expression of the extra copy of chromosome 21. What approach, do you think, are they taking, and what may be some of the challenges for using it in patients?
The length of dogs’ legs varies widely across species, with the Nova Scotia duck tolling retriever representing a short-legged breed. Genome-wide association studies (GWAS) linked this phenotype to a locus on Canine familiaris (CFA) chromosome 12 with strong similarity to the FGF4 gene on chromosome 18. Further work demonstrated that both loci are transcribed. Note: CpG islands (green in diagram) often mark active promoters; we will discuss in a later lecture. Based on the diagram below, what do you think is the relationship between these two loci? The genome sequence shows another locus on chromosome 10 that is almost identical to the one depicted above on chromosome 12, from 5’UTR through 3’UTR. The gene does have one polymorphism, though, in exon 2, that distinguishes it from the Chr. 12 gene. You have developed an assay to easily distinguish mRNA derived these two variants. However, when you assay the Chr. 10-derived mRNA, you don’t detect anything at all.…
8. Dioxin (in the human diet!) and ethidium bromide (used in labs to see DNA using UV light) are classified as because they insert themselves in between adjacent DNA bases. 9. UV light can cause two adjacent pyrimidines to form a (Fig 18.21). 10. Study Figure 18.26. What is the difference between black, green and red grapes relative to expression of genes that produce anthocyanin pigments? Possible answers: retrotransposon prevents pigment expression; partial removal of retrotransposon allows partial pigment expression; absence of retrotransposon allows normal pigment expression. a. Black: b. Green : C. Red:
Looking at Figure 11-9, why were partial diploids essential for establishing the trans-acting nature of the Lacrepressor? Could one distinguish cis-acting from transacting genes in haploids?
2. Which ONE of these statements on chromatin modification is UNTRUE? O DNA is methylated on the cytosine of CpG sequences in the DNA O Histone acetyl transferases transfer an acetyl group to lysines in histones O Bromodomain proteins bind to acetylated histones O Histone lysine methyltransferases (KMTS) are epigenetic erasers O Epigenetic changes can contribute to cancer evolution NE
18. Two eukaryotic genes (A and B) are involved in two different metabolic pathways, and each has a control element sequence (enhancer) upstream of their coding sequence. The interaction of these enhancers with specific transcription factors determines the level of gene expression. These two genes are expressed differentially in two different tissues (tissue 1 and tissue 2) according to the table below. Tissue 1 Tissue 2 Gene A OFF ON Gene B ON OFF What can explain this pattern of gene expression? A. Gene A and B have different control element sequences and the cells of tissue 1 and tissue 2 express the same transcription factors (at the same concentration and at the same time). B. Gene A and B have the same control element sequences and the cells of tissue 1 and tissue 2 express different transcription factors. C. Gene A and B have the same control element sequences and the cells of tissue 1 and tissue 2 express the same transcription factors (at the same concentration and at the same…