Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN: 9780134580999
Author: Elaine N. Marieb, Katja N. Hoehn
Publisher: PEARSON
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14. Which of the mutations shown will affect the amino acid sequence of the myosin protein?
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- 12arrow_forward4. Why do the dideoxynucleotides stop DNA extension?arrow_forward2. Genetics enables us to understand how proteins interact with one another within the context of a living organism. Different types of mutations can give phenotypes that indicate these protein interactions. a) Why does a dominant negative phenotype indicate a protein may be acting as a homomultimer? b) Does a hypermorphic mutation also indicate this? Why or why not? c) If you have two mutant proteins that show a pseudo-reversion phonotype (second site suppressor) there is a strong possibility that the two proteins interact. Explain in your own words why a mutation in one protein can rescue (at least partially) the phenotype caused by a mutation in a completely different protein. d) Why are second site suppressor mutations allele specific?arrow_forward
- 5. The lacl gene is only transcribed when both glucose and lactose are present. True or False.arrow_forward1arrow_forward6. In the following graph, the dashed line shows the level of mRNA for a certain protein, Prot6, at various positions along the anterior-posterior axis of an insect embryo. The solid lines portray the levels of two regulatory proteins, A and B, that control translation of the Prot6 mRNA. A В Position Biology: How Life Works, Second Edition © 2016 W.H. Freeman and Company If both A and B stimulate translation of the Prot6 mRNA, then which graph, from the ones shown below, best approximates the expected level of Prot6 across the embryo? Briefly justify your answer. M K Q Position Position Position Position Position Protein level MRNA levelarrow_forward
- 2. Consider the scenario where a DNA mutation occurs in a person's skin cells, where the gene p53 becomes damaged, and can no longer be transcribed and translated into a functioning 53kd protein. Why might a mutation of this nature be especially concerning to someone who experiences regular exposure to skin carcinogens, such as UV light? ✓✓✓arrow_forward8. 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 127.73arrow_forward3. Haplopappus is an annual flowering plant that grows in deserts. It is of interest because its 2n number is only four. Cells in the vegetative parts of the plant are not undergoing mitosis and are permanently in G1 of the cell cycle; a. How many DNA double helices do these vegetative cells have in their nucleus? b. During metaphase of mitosis, how many DNA double helices would there be in the cell?arrow_forward
- 7. Explain why a mutant Ras is an oncogene causing many human cancersarrow_forward6. Suppose a particular gene is required for early development and also later for development of a particulartissue, such as the adult nervous system. By generating a homozygous mutant clone in that tissue of a heterozygote, researchers can circumvent the lethalitythat would result if the entire animal is homozygousfor a loss-of-function mutation in that gene.A technique called MARCM (Mosaic Analysiswith a Repressible Cell Marker) was developed to enable Drosophila geneticists to generate homozygousmutant cell clones that are marked by the presence of areporter protein such as GFP. Marker expression enables the investigator to observe clearly the mutantphenotype within a clone of mutant cells. This technique relies on a yeast protein called Gal80 that is anegative regulator of the Gal4 protein described previously in Solved Problem II. Gal80 binds to Gal4 andprevents it from activating transcription. The idea ofMARCM is that Gal4/UASG-driven GFP expression isblocked by Gal80 throughout…arrow_forward11. If pB253 were cleaved with BamHI and SacI, what are the sizes of the two DNA fragments that would result? (show units)arrow_forward
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