Introduction to Genetic Analysis
11th Edition
ISBN: 9781464109485
Author: Anthony J.F. Griffiths, Susan R. Wessler, Sean B. Carroll, John Doebley
Publisher: W. H. Freeman
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Chapter 2, Problem 15P
Summary Introduction
To determine: The number of distinct DNA bands that would be visible in each species if saparated by gel electrophoresis.
Introduction: Gel electrophoresis is a technique for detachment and study of macromolecules and fragments, based on the size and charge.
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Below are two pictures of perithecium on microscope slides that have been squished under cover slips so that the asci inside have spread
out. In the asci, we can clearly see the tan ascospores (light colored) and wildtype ascospores (dark colored). We will perform our tetrad
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Ascospore pattern of
non-recombinant asci
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You will count the total number of non-recombinant and recombinant asci in the two pictures below. You will only count the asci that
have a star next to them because these asci can be clearly identified as non-recombinant or recombinant. Select the correct number
below after you are done.
Total number of non-recombinant asci [Select]
Ascospore pattern of…
Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction).
Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine.
All three yeast strains are homozygous for the underlying alleles.
When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation;
when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine.
After crossing the F1 generation of the cross between mutant strains 1 and 3, you count and determine the phenotypes of 1,000 colonies (here a colony is equivalent to an individual): 563 colonies that can grow on minimal medium alone; 437 colonies that require adenine…
Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction).
Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine.
All three yeast strains are homozygous for the underlying alleles.
When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation;
when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine.
A. What conclusions can you make about the alleles of mutant strains 1, 2, and 3 and their relationships with each other?
B. What phenomenon is occurring in the cross between mutant strains 1 and 3?
After crossing the F1 generation of the cross between mutant strains 1…
Chapter 2 Solutions
Introduction to Genetic Analysis
Ch. 2 - Prob. 1PCh. 2 - Prob. 2PCh. 2 - Prob. 3PCh. 2 - Prob. 4PCh. 2 - Prob. 6PCh. 2 - Prob. 7PCh. 2 - Prob. 8PCh. 2 - Prob. 9PCh. 2 - Prob. 10PCh. 2 - Prob. 11P
Ch. 2 - Prob. 12PCh. 2 - Prob. 13PCh. 2 - Prob. 14PCh. 2 - Prob. 15PCh. 2 - Prob. 16PCh. 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. 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. 44.1PCh. 2 - Prob. 44.2PCh. 2 - Prob. 44.3PCh. 2 - Prob. 44.4PCh. 2 - Prob. 44.5PCh. 2 - Prob. 44.6PCh. 2 - Prob. 44.7PCh. 2 - Prob. 44.8PCh. 2 - Prob. 44.9PCh. 2 - Prob. 44.10PCh. 2 - Prob. 44.11PCh. 2 - Prob. 44.12PCh. 2 - Prob. 44.13PCh. 2 - Prob. 44.14PCh. 2 - Prob. 44.15PCh. 2 - Prob. 45PCh. 2 - Prob. 47PCh. 2 - Prob. 48PCh. 2 - Prob. 49PCh. 2 - Prob. 50PCh. 2 - Prob. 51PCh. 2 - Prob. 52PCh. 2 - Prob. 53PCh. 2 - Prob. 56PCh. 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. 66PCh. 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. 79P
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- Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. A. What conclusions can you make about the alleles of mutant strains 1, 2, and 3 and their relationships with each other? B. What phenomenon is occurring in the cross between mutant strains 1 and 3?arrow_forwardA technique called fluorescence in situ hybridization (FISH) is described. In this method, a labeled piece of DNA is hybridized to a set of chromosomes. Let’s suppose that you cloned a piece of DNA from G. pubescens and used it as a labeled probe for in situ hybridization. What would you expect to happen if this DNA probe were hybridized to the G. speciosa or G. tetrahit chromosomes? Describe the expected results.arrow_forwardA blood stain from a crime scene and blood samples from four suspects were analyzed by PCR using fluorescent primers associated with three STR loci: D3S1358, vWA, and FGA. The resulting electrophoretograms are shown below. The numbers beneath each peak identify the allele (upper box) and the height of the peak in relative fluorescence units (lower box). Solve, (a) Since everyone has two copies of each chromosome and therefore, two alleles of each gene, what accounts for the appearance ofonly one allele at some loci? (b) Which suspect is a possible source of the blood? (c) Could the suspect be identifi ed using just one of the three STR loci? (d) What can you conclude about the amount of DNA obtained from Suspect 1 compared to Suspect 4?arrow_forward
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