Your friend sends you two cancerous cell lines to examine and detérmihe posi mutations. The results are shown below: Cell Line Mutation WT none (wild type DNA) 1 a deletion at the same region on both copies of chromosome 4 a point mutation in a gene on only one copy of chromosome 7 Based on this data, what type of gene is mutated in each of the cell lines? Select all that apply Cell line 2 has a mutation in an oncogene Cell line 1 has a mutation in a tumor suppressor gene Cell line 1 has a mutation in an proto-oncogene Cell line 2 has a mutation in a tumor suppressor gene

Your friend sends you two cancerous cell lines to examine and detérmihe posi mutations. The results are shown below: Cell Line Mutation WT none (wild type DNA) 1 a deletion at the same region on both copies of chromosome 4 a point mutation in a gene on only one copy of chromosome 7 Based on this data, what type of gene is mutated in each of the cell lines? Select all that apply Cell line 2 has a mutation in an oncogene Cell line 1 has a mutation in a tumor suppressor gene Cell line 1 has a mutation in an proto-oncogene Cell line 2 has a mutation in a tumor suppressor gene

Biology: The Unity and Diversity of Life (MindTap Course List)

14th Edition

ISBN:9781305073951

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter10: Control Of Gene Expression

Section: Chapter Questions

Problem 3CT: Almost all calico cats (one is pictured in FIGURE 10.7B) are female. Why? B When this calico cat was...

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WHIM syndrome is a disease of the immunesystem resulting in warts and frequent infections.The disease is caused by a dominant gain-of-functionmutation in a gene on chromosome 2 called CXCR4.A 38-year-old woman suffering with WHIM syndrome her entire life was suddenly and mysteriouslycured. Genome analysis of her blood precursor cells(stem cells) revealed that many of these cells had achromosome 2 that had undergone chromotripsis—arare (and poorly understood) process where a chromosome is “shattered” into small pieces that are subsequently stitched back together in random order, resultingin many deletions and inversions. Explain howchromotripsis of chromosome 2 in a blood stem cellcould have cured the woman of WHIM syndrome.
A temperature-sensitive mutant yeast strain stops dividing when shifted from 25°C to 37°C. These cells are analyzed at different temperatures by a machine that measures the amount of DNA they contain, and the following graphs are obtained. Cells in G1 phase should have ___ unit(s) amount of DNA. Which of the following would explain the behavior of your mutant? Mark yes or no for each of the following choices. Inability to initiate DNA replication Defect in chromosome condensation Defect in centrosome duplication Defect in cytokinesis
Let’s suppose an X-linked gene in mice exists as two alleles, whichwe will call B and b. X-chromosome inactivation, a process inwhich one X chromosome is turned off, occurs in the somatic cellsof female mammals . Allele B encodes an mRNA that is 900 nucleotides long, whereas allele b contains a small deletion that shortens the mRNA to a length of 825 nucleotides.Draw the expected Northern blot that will be obtained usingmRNA isolated from somatic tissue of the followingmice:Lane 1. mRNA from an XbY male mouseLane 2. mRNA from an XbXb female mouseLane 3. mRNA from an XBXb female mouse.Note: The sample taken from the female mouse is not from aclone of cells. It is from a tissue sample, like the one shown.
An individual has a deletion of part of the short arm of chromosome 5. This individual also has a normal copy of chromosome 5. Would you expect this individual to have Cri-du-chat syndrome? Why or why not? Question 7 options: A) Yes, because they will inactivate their normal chromosome 5 B) Yes, because they only contain one copy of some of their genes C) No, because this is a terminal deficiency D) No, because they still have at least one copy of all the genes on chromosome 5
A person with cancer walks into the classroom. As a biology class, we ask if we can examine some of the cancerous cells and find that many of these cancerous cells have an extra chromosome 10 (trisomy) leading to more cyclin for forming the MPF being made, leading to uncontrollable cell division (i.e. tumor and progression to cancer). What would cause extra chromosomes in cells (be it somatic or gametic cells)? Select one: a. None of these choices are linked to how sister chromatids separate and would lead to extra chromosomes in daughter cells of cell division. b. APC not functioning correctly to add ubiquitin to the cohesins, thus sister chromatids do not split correctly during cell division c. MPF itself would cause more chromosome replication d. APC is overly functioning adding cohesins to the sister chromatids
Look at the two yeast colonies in Figure Q4–3. Eachof these colonies contains about 100,000 cells descendedfrom a single yeast cell, originally somewhere in the mid-dle of the clump. A white colony arises when the Ade2 geneis expressed from its normal chromosomal location. Whenthe Ade2 gene is moved to a location near a telomere, itis packed into heterochromatin and inactivated in mostcells, giving rise to colonies that are mostly red. In theselargely red colonies, white sectors fan out from the middleof the colony. In both the red and white sectors, the Ade2 gene is still located near telomeres. Explain why white sec-tors have formed near the rim of the red colony. Based onthe patterns observed, what can you conclude about thepropagation of the transcriptional state of the Ade2 genefrom mother to daughter cells in this experiment?
A cell from a human female has just undergone mitosis. For unknown reasons, the centromere of chromosome 7 failed to divide. Describe the chromosomal contents of the daughter cells.
Almost all calico cats (one is pictured in FIGURE 10.7B) are female. Why? B When this calico cat was an embryo, one of the two X chromosomes was inactivated in each of her cells. The descendants of the cells formed her adult body, which is a mosaic for expression of X chromosome genes. Black fur arises in patches where genes on the X chromosome inherited from one parent are expressed; orange fur arises in patches where genes on the X chromosome inherited from the other parent are expressed. FIGURE 10.7 Animated X chromosome inactivation.
Prader-Willi syndrome and Angelman syndrome are both caused by deletion of a set of genes onchromosome 15. The symptoms of Prader-Willi syndrome are short stature, small hands andfeet, hypotonia (floppiness), hypogonadism, mild mental retardation, and an uncontrollabledesire to eat (polyphagia). Angelman syndrome is characterized by an intellectual anddevelopmental delay, sleep disturbances, seizures, hand-flapping and other jerky movements,frequent laughter or smiling, and usually a happy demeanor.Please explain how the deletion of the same set of genes can result in such different diseases. Inyour answer, be sure to discuss the role of genetic imprinting and epigenetics.
. Early in development, most human cells turn off expression of an essential component of telomerase, the enzyme responsible for addition of telomere repeat sequences (5’-TTAGGG) to the ends of chromosomes. Thus, as our cells proliferate their telomeres get shorter and shorter, but are normally not lost over the course of a lifetime. If cells are removed from the body and grown in culture, however they ultimately enter a state of replicative senescence and stop dividing when their telomeres get too short. By contrast, most human tumor cells express active telomerase, allowing them to maintain their telomeres and grow beyond the normal limit imposed by senescence - good for them, bad for us. Anticipating a universal cure for cancer, you set up a company to screen chemical ‘libraries’ for telomerase inhibitors. The company share price takes a dive, however, when a rival group generates a strain of telomerase-knockout mice. These mice breed happily for several generations, but…
In humans, chromosome 16 sometimes has a heavily stained area in the long arm near the centromere. This feature can be seen through the microscope but has no effect on the phenotype of the person carrying it. When such a “blob” exists on a particular copy of chromosome 16, it is a constant feature of that chromosome and is inherited. A couple conceived a child, but the fetus had multiple abnormalities and was miscarried. When the chromosomes of the fetus were studied, it was discovered that it had three copies of chromosome 16 (it was trisomic for chromosome 16), and that two of the three chromosome 16s had large blobs. Both chromosome 16 homologs in the mother lacked blobs, but the father was heterozygous for blobs. Which parent experienced nondisjunction, and in which meiotic division did it occur?
The dominant condition elliptocytosis causes red blood cells to become misshapen into oval-shaped cells. One of the genes responsible for the abnormal shape encodes the band 4.1 protein that together with ankyrin and other scaffold proteins creates and maintains the spherical concave shape of a normal red blood cell. The gene for band 4.1 protein, EPB41, is found on the p arm of chromosome 1. This is very close to the gene encoding the red blood cell Rhesus (Rh) blood type, either phenotype + (dominant) or - (recessive), with a recombination frequency of 2%. This means that 98% of the time alleles for these two genes are linked and are transmitted together. Diane and Jack are siblings, and both have elliptocytosis and Rh+ blood type. Due to the elliptocytosis, both had emergency splenectomies after having severe anemia. Their younger brother, Devonté, has not yet shown signs of elliptocytosis, but has Rh- blood. André, their dad, also has elliptocytosis and Rh+ blood; while their…