ACHIEVE:INTRO TO GENETIC ANALYSIS 1TERM
12th Edition
ISBN: 9781319401399
Author: Griffiths
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Question
Chapter 2, Problem 12P
Summary Introduction
To determine: The way in which mutant could have a phenotypic effect in a green area.
Introduction: Mutations can be inherited and hence transferred on from one person to another. If a variation produces a new
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
You are working with a fly hair cell developmental system.
This Notch/Delta-regulated system results in clusters of cells where the central one differentiates
into a specialized hair cell.
To better understand this system you have tagged the C-terminal cytoplasmic domain of Notch
with GFP.
You have done a forward genetic screen to look for mutants that have unusual phenotypes in this
Notch system.
The first one is a mutation in Notch itself. This mutant is in the ADAM10 cleavage site and blocks
proteolysis. Draw the expected outcome for such a mutant: GFP localization and developmental
outcome 24hr after differentiation.
WT before differentiation
WT 24 hours after differentiation
In the module, you have learned about P-element mediated transgenesis in Drosophila and the concept of using transgenes to rescue mutant phenotypes. In the figure below, you will see a wild type fly with its natural eye colour and three mutants with their eye colours changed to vermillion, white and rosy, respectively. A schematic of P-element mediated transgenesis (as shown in the lectures) is also included in the figure. Please inspect the schematic carefully and choose which of the following statements is true:
I.
Injection of the white experimental transgene into the vermillion mutant embryo will not change the vermillion mutant phenotype
II.
Injection of the white experimental transgene in the rosy mutant embryo will change rosy eye colour to red (wild type)
III.
Injection of the white experimental transgene in the white mutant embryo will not change the white mutant phenotype
IV.
Injection of the white experimental transgene in the rosy mutant…
The genes described below are part of the yeast mating signal transduction pathway that signals the presence of extracellular alpha pheromone to the nucleus. These genes
are listed in the order they function in the pathway.
STE3 encodes alpha pheromone receptor
STE4 encodes the ß subunit of a heterotrimeric G protein
STE7 encodes a protein kinase
STE12 encodes a transcription factor
Loss-of function-mutations in any of these genes are sterile. In addition, gain-of-function mutations can be isolated in these genes. For which gene would a loss of function
mutation suppress gain-of-function mutations in any of the other three genes.
STE3
STE7
STE12
STE4
Chapter 2 Solutions
ACHIEVE:INTRO TO GENETIC ANALYSIS 1TERM
Ch. 2 - Prob. 1PCh. 2 - Prob. 2PCh. 2 - Prob. 3PCh. 2 - Prob. 4PCh. 2 - Prob. 5PCh. 2 - Prob. 6PCh. 2 - Prob. 7PCh. 2 - Prob. 8PCh. 2 - Prob. 9PCh. 2 - Prob. 10P
Ch. 2 - Prob. 11PCh. 2 - Prob. 12PCh. 2 - Prob. 13PCh. 2 - Prob. 14PCh. 2 - Prob. 15PCh. 2 - Prob. 17PCh. 2 - Prob. 18PCh. 2 - Prob. 19PCh. 2 - Prob. 20PCh. 2 - Prob. 21PCh. 2 - Prob. 22PCh. 2 - Prob. 23PCh. 2 - Prob. 24PCh. 2 - Prob. 25PCh. 2 - Prob. 26PCh. 2 - Prob. 27PCh. 2 - Prob. 28PCh. 2 - Prob. 29PCh. 2 - Prob. 30PCh. 2 - Prob. 31PCh. 2 - Prob. 32PCh. 2 - Prob. 33PCh. 2 - Prob. 34PCh. 2 - Prob. 35PCh. 2 - Prob. 36PCh. 2 - Prob. 37PCh. 2 - Prob. 38PCh. 2 - Prob. 39PCh. 2 - Prob. 40PCh. 2 - Prob. 41PCh. 2 - Prob. 42PCh. 2 - Prob. 43PCh. 2 - Prob. 44PCh. 2 - Prob. 45PCh. 2 - Prob. 46PCh. 2 - Prob. 47PCh. 2 - Prob. 48PCh. 2 - Prob. 49PCh. 2 - Prob. 50PCh. 2 - Prob. 51PCh. 2 - Prob. 52PCh. 2 - Prob. 53PCh. 2 - Prob. 54PCh. 2 - Prob. 55PCh. 2 - Prob. 56PCh. 2 - Prob. 56.1PCh. 2 - Prob. 56.2PCh. 2 - Prob. 56.3PCh. 2 - Prob. 56.4PCh. 2 - Prob. 56.5PCh. 2 - Prob. 56.6PCh. 2 - Prob. 56.7PCh. 2 - Prob. 56.8PCh. 2 - Prob. 56.9PCh. 2 - Prob. 56.10PCh. 2 - Prob. 56.11PCh. 2 - Prob. 56.12PCh. 2 - Prob. 56.13PCh. 2 - Prob. 56.14PCh. 2 - Prob. 56.15PCh. 2 - Prob. 57PCh. 2 - Prob. 58PCh. 2 - Prob. 59PCh. 2 - Prob. 60PCh. 2 - Prob. 61PCh. 2 - Prob. 62PCh. 2 - Prob. 63PCh. 2 - Prob. 64PCh. 2 - Prob. 65PCh. 2 - Prob. 67PCh. 2 - Prob. 68PCh. 2 - Prob. 69PCh. 2 - Prob. 70PCh. 2 - Prob. 71PCh. 2 - Prob. 72PCh. 2 - Prob. 73PCh. 2 - Prob. 74PCh. 2 - Prob. 75PCh. 2 - Prob. 76PCh. 2 - Prob. 77PCh. 2 - Prob. 78PCh. 2 - Prob. 79PCh. 2 - Prob. 80PCh. 2 - Prob. 81PCh. 2 - Prob. 82PCh. 2 - Prob. 83PCh. 2 - Prob. 84PCh. 2 - Prob. 85PCh. 2 - Prob. 86PCh. 2 - Prob. 87PCh. 2 - Prob. 88PCh. 2 - Prob. 89PCh. 2 - Prob. 90PCh. 2 - Prob. 91PCh. 2 - Prob. 1GSCh. 2 - Prob. 2GSCh. 2 - Prob. 3GS
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- After isolating six Neurospora mutations, you mate A and a representatives of each in all pairwise combinations and determine which mutations complement. Table 1gives the results, with a (+) indicating complementation, and a (-) representing lack of complementation. With the exception of mutant 1, how many different genes are represented in this experiment, or can you tell from these experiments?arrow_forwardIn a particular strain of Neurospora, a poky mutation exhibits biparental inheritance, whereas poky mutations in other strains are inherited only from the maternal parent. Explain these results.arrow_forwardConsidering Figure 2-13, if you had a homozygous double mutant m3/m3 m5/m5, would you expect it to be mutant in phenotype? (Note: This line would have two mutant sites in the same coding sequence.)arrow_forward
- In a haploid, a nonsense suppressor su1 acts on mutation1 but not on mutation 2 or 3 of gene P. An unlinked nonsense suppressor su2 works on P mutation 2 but not on 1or 3. Explain this pattern of suppression in regard to thenature of the mutations and the suppressors.arrow_forwardCx is a member of the family of connexin genes that encode the proteins of gap junction intercellular channels. Cx proteins in one cell form hemi-channels in the plasma membrane. Hemi-channels in adjacent cells dock to form complete intercellular channels through which ions and small molecules diffuse from cell to cell. Distinct Cx mutations were identified in three different families, F1, F2 and F3, affected by the same disease. To study their functional properties, normal (wild type, wt) and mutant (m) Cx proteins were expressed in cultured cells. Translation of the proteins was checked (Fig. 3). A extracellular EC 1 SM TM 1 membrane 2 3 4 F10 intracellular N F2 EC 2 F3 oricand c B 42 kDa C 42 kDa 35 kDa Control wt m-F1 m-F2 PM C PM C PM C PM C Western blot anti-Cx Control wt PM C m-F1 PM C PM C = Metabolically labelled m-F2 PM C m-F3 PM C m-F3 PM C WSEY Fig. 3 mont (A). Membrane topology of Cx protein indicating positions of mutations. Cx is an integral membrane protein with 4…arrow_forwardThe transcription factor Pax6 is required continually during the life of a mouse (or a human) for the development of the retina. Homozygous Pax6 knockout mice die soon after birth because Pax6 protein is also required in essential organs, such as the pancreas. a) In order to study the role of Pax6 in eye development a researcher wants to generate a mouse that expresses Pax6 everywhere except in its eyes. Describe how you could construct such a mouse by floxing the gene. Is it possible to achieve the same end with a transgene? (Hint: think about using cDNA and RNAI) b) Suppose you want to create a mouse similar to that in part (a), but one where the eye cells from Pax6 function has been removed and now express a gene that specifies a green fluorescent protein (GFP). Marking the cells in this way will allow the investigators to see the shapes of the Pax6- eye cells more easily than if they did not express GFP. Diagram a Pax6 gene construct that would enable the researcher to do this…arrow_forward
- Based on the partial diploid sketched shown in Figure 3, where cross-hatched rectangles and the red bar indicate mutations that inactivate that particular genetic element of the operon, state if expression of Lacl (Lac repressor) and LacZ (Beta-Galactosidase) is (select one of the options below for each genetic element and briefly justify your answer): constitutively ON (in the presence/absence of lactose) constitutively OFF (in the presence/absence of lactose) • inducible by lactose • not enough information given to conclude • none of the above FIGURE 3: lacl laco lacz- lacY lacl laco lacz lacYarrow_forwardWhat is the correct answer, and why?arrow_forwardIn a qPCR experiment, you find that the mRNA for the gene CDK1 goes down (decreased expression) when you treat your cells with doxorubicin. You wonder whether this is due to epigenetic changes on the gene for CDK1. Name two methods that you might use to check whether there are changes in epigenetic modifications (e.g. DNA methylation or histone tail modifications) on the CDK1 DNA.arrow_forward
- A gene, which we will call gene C, can be epigenetically modified in such a way that its expression in some cells is permanently silenced. Describe how you could conduct cell-fusion experiments to determine if a cis- or a trans-epigenetic mechanism is responsible for maintaining the silencing of gene C.arrow_forwardFollowing a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. What is the genotype of the F1 (progeny of Females 3 x Males 4). Select only one answer. 1. C/c D/d 2. C/C d/d 3. C/C D/D 4. c/c D/d…arrow_forwardFollowing a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. Which is the genotype of the rhythmic flies resulting from the testcross? Select only one answer 1. C/C d/d 2. C/c D/d 3. C/c d/d 4. c/c…arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Human Anatomy & Physiology (11th Edition)BiologyISBN:9780134580999Author:Elaine N. Marieb, Katja N. HoehnPublisher:PEARSONBiology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStaxAnatomy & PhysiologyBiologyISBN:9781259398629Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa StouterPublisher:Mcgraw Hill Education,
- Molecular Biology of the Cell (Sixth Edition)BiologyISBN:9780815344322Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter WalterPublisher:W. W. Norton & CompanyLaboratory Manual For Human Anatomy & PhysiologyBiologyISBN:9781260159363Author:Martin, Terry R., Prentice-craver, CynthiaPublisher:McGraw-Hill Publishing Co.Inquiry Into Life (16th Edition)BiologyISBN:9781260231700Author:Sylvia S. Mader, Michael WindelspechtPublisher:McGraw Hill Education
Human Anatomy & Physiology (11th Edition)
Biology
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:PEARSON
Biology 2e
Biology
ISBN:9781947172517
Author:Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher:OpenStax
Anatomy & Physiology
Biology
ISBN:9781259398629
Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa Stouter
Publisher:Mcgraw Hill Education,
Molecular Biology of the Cell (Sixth Edition)
Biology
ISBN:9780815344322
Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter Walter
Publisher:W. W. Norton & Company
Laboratory Manual For Human Anatomy & Physiology
Biology
ISBN:9781260159363
Author:Martin, Terry R., Prentice-craver, Cynthia
Publisher:McGraw-Hill Publishing Co.
Inquiry Into Life (16th Edition)
Biology
ISBN:9781260231700
Author:Sylvia S. Mader, Michael Windelspecht
Publisher:McGraw Hill Education
What are Mutations and what are the different types of Mutations?; Author: Science ABC;https://www.youtube.com/watch?v=I16YlE8qTBU;License: Standard youtube license