Introduction To Genetic Analysis
12th Edition
ISBN: 9781319114787
Author: Anthony J.F. Griffiths, John Doebley, Catherine Peichel, David A. Wassarman
Publisher: W. H. Freeman
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 3, Problem 27P
Summary Introduction
To determine: The cells that could have been used for the measurements.
Introduction: The length of a DNA molecule is often measured in "base pairs," or bp that is, the number of rungs in the ladder, and sometimes, this unit of measurement is shortened simply to "bases."
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
A 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?
Imagine that you need to run an experiment that requires you to alter the current genetic make of your organism. You have two options: using nitrite or acridine orange. Which one would you choose considering the effectiveness of the treatment and why? State in no more than 3 sentences.
A unique aquatic plant was discovered from a lagoon in El Nido,
Palawan. To determine the protein content of the plant, an
adequate amount of the plant extract was acquired, and Bradford
assay was performed. Determine the total protein concentration,
in µg/mL, of the acquired plant extract. A bovine serum albumin
(BSA) stock solution with a concentration of 250 µg/mL was used
and mixtures with the following compositions and absorbance
readings were prepared:
Volume of BSA
(mL)
Volume of water
(mL)
Absorbance
Tube #
at 595 nm
1
0.00
2.00
0.000
2
0.20
1.80
0.112
3
0.40
1.60
0.225
4
0.60
1.40
0.318
5
0.80
1.20
0.432
6
1.00
1.00
0.551
The absorbance reading of the plant extract is 0.275.
Chapter 3 Solutions
Introduction To Genetic Analysis
Ch. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Prob. 3PCh. 3 - Prob. 4PCh. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 8PCh. 3 - Prob. 9PCh. 3 - Prob. 10P
Ch. 3 - Prob. 11PCh. 3 - Prob. 12PCh. 3 - Prob. 13PCh. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - Prob. 18PCh. 3 - Prob. 19PCh. 3 - Prob. 20PCh. 3 - Prob. 21PCh. 3 - Prob. 22PCh. 3 - Prob. 23PCh. 3 - Prob. 24PCh. 3 - Prob. 25PCh. 3 - Prob. 26PCh. 3 - Prob. 27PCh. 3 - Prob. 28PCh. 3 - Prob. 29PCh. 3 - Prob. 30PCh. 3 - Prob. 31PCh. 3 - Prob. 32PCh. 3 - Prob. 33PCh. 3 - Prob. 34PCh. 3 - Prob. 35PCh. 3 - Prob. 36PCh. 3 - Prob. 37PCh. 3 - Prob. 38PCh. 3 - Prob. 39PCh. 3 - Prob. 40PCh. 3 - Prob. 41PCh. 3 - Prob. 42PCh. 3 - Prob. 43PCh. 3 - Prob. 43.1PCh. 3 - Prob. 43.2PCh. 3 - Prob. 43.3PCh. 3 - Prob. 43.4PCh. 3 - Prob. 43.5PCh. 3 - Prob. 43.6PCh. 3 - Prob. 43.7PCh. 3 - Prob. 43.8PCh. 3 - Prob. 43.9PCh. 3 - Prob. 43.10PCh. 3 - Prob. 43.11PCh. 3 - Prob. 43.12PCh. 3 - Prob. 43.13PCh. 3 - Prob. 43.14PCh. 3 - Prob. 43.15PCh. 3 - Prob. 44PCh. 3 - Prob. 45PCh. 3 - Prob. 46PCh. 3 - Prob. 47PCh. 3 - Prob. 48PCh. 3 - Prob. 49PCh. 3 - Prob. 50PCh. 3 - Prob. 51PCh. 3 - Prob. 52PCh. 3 - Prob. 53PCh. 3 - Prob. 54PCh. 3 - Prob. 55PCh. 3 - Prob. 56PCh. 3 - Prob. 57PCh. 3 - Prob. 58PCh. 3 - Prob. 59PCh. 3 - Prob. 61PCh. 3 - Prob. 62PCh. 3 - Prob. 63PCh. 3 - Prob. 64PCh. 3 - Prob. 65PCh. 3 - Prob. 66PCh. 3 - Prob. 67PCh. 3 - Prob. 70PCh. 3 - Prob. 1GSCh. 3 - Prob. 2GSCh. 3 - 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
- A plant isolate was subjected to qualitative tests to determine the presence of macromolecules. Below is a summary of results obtained: Biuret Ninhydrin Schiff's Test for Test for test test test Deoxyribose Phosphat Violet Violet Pink Blue Yellow solution solution solution solution precipitat Conclusion: Chromosomal DNA was possibly extracted from the plant. O True O Falsearrow_forwardYou were tasked to conduct a population genetic survey of a diploid insect population. You obtained tissue samples from 20 individuals, ran a starch gel electrophoresis, and stained the gel for lactate dehydrogenase. Below is the result of your gel electrophoresis. You found that there are three alleles and you decided to call them F, M, and S for their fast, medium, and slow mobility on the gel. please explain this in deep. 1) What is the M allele frequency in this population? A) 0.05 b) 0.25 c) 0.30 d) 0.4 e) 0.5 2)Following up the previous question, what is the observed SS genotype frequency in this population? a) 0.1 b) 0.2 3) 0.3 4) 0.4 5) 0.5arrow_forwardIf you want to make a colorful chromatograph, which type of plant leaves would you use, pure green coleus leaves or variegated Coleus leaves, and why?arrow_forward
- Using Excel produce a plot that shows the decay of D over time (for 50 generations) for four different recombination frequencies: r = 0.5 (independent assortment) r = 0.25 r = 0.1 r = 0.01 *show graph and plot *arrow_forwardBoth the isolation buffer and assay buffer that was used in isolation of mitochondria contained 0.3M mannitol. What was the purpose of including mannitol at this concentration? Type your answer here:arrow_forwardIn spectrophotometry, we measure the concentration of an analyte by its absorbance of light. A low concentration sample was prepared, and nine replicate measurements gave absorbances of 0.0047, 0.0054, 0.0062, 0.0060, 0.0046, 0.0056, 0.0052, 0.0044, 0.0058. Nine reagent blanks gave values of 0.0006, 0.0012, 0.0022, 0.0005, 0.0016, 0.0008, 0.0017, 0.0010, and 0.0011. a) What is the absorbance detection limit? This is related to both the scatter of the blank and the scatter in the measured samples. b) The calibration curve is a graph of absorbance versus concentration. Absorbance is a dimensionless quantity. The slope of the calibration curve is m = 2.24 x 104 M-1. Find the concentration detection limit. c) Find the lower limit of quantitation.arrow_forward
- In spectrophotometry, we measure the concentration of an analyte by its absorbance of light. A low concentration sample was prepared, and nine replicate measurements gave absorbances of: 0.0047, 0.0054, 0.0062, 0.0060, 0.0046, 0.0056, 0.0052, 0.0044, 0.0058. Nine reagent blanks gave values of: 0.0006, 0.0012, 0.0022, 0.0005, 0.0016, 0.0008, 0.0017, 0.0010, and 0.0011. a) What is the absorbance detection limit? This is related to both the scatter of the blank and the scatter in the measured samples.arrow_forwardSeveral cell culture lines of epithelial cells, called “Cell Line A, B, or C,” are incubated with 104 infectious particles of influenza virus, and viral titers in the culture media are measured 2 days later. Looking at the results of this experiment, it is apparent that the three lines do not all show the same response to the virus. To investigate these differences, mixing experiments are performed, where cells from two different cell lines are mixed together at a 1:1 ratio before the Influenza infection. Based on the results shown in the figure below, propose an explanation for these data.arrow_forwardDNA samples were then run in agarose gel electrophoresis against a molecular standard of 100 bp and 1Kb Ladder. The figure is analyzed and labelled manually and recorded with the used product ladder information. Figure 2A is labeled with corresponding lane numbers and with M1 and M2 as markers used. Write an interpretation of the documented gelresult (Figure 2). Note: Agarose gel of isolated DNA (A) from plant (lanes 1 & 2), chicken (lanes 3 & 4), and E. coli (lanes 5 & 6) with Vivantis* product information of 100bp (B) and 1kb ladders (C).arrow_forward
- what is the full name of the TISSUSE TYPE at the end of the arrow for letter F? F1) Does this look like any other lettered structure on this photomicrograph? If yes, which one and how are they similar.arrow_forwardWhat is the purpose of adding two drops of acetic acid in thin layer chromatography of plant pigment?arrow_forwardA cross was performed between a yeast strain that requires methionine and lysine for growth (met− lys−)and another yeast strain, which is met+ lys+. One hundred asci were dissected, and colonies were grownfrom the four spores in each ascus. Cells from thesecolonies were tested for their ability to grow on petriplates containing either minimal medium (min), min+ lysine (lys), min + methionine (met), or min + lys+ met. The asci could be divided into two groupsbased on this analysis:Group 1: In 89 asci, cells from two of the four spore colonies couldgrow on all four kinds of media, while the other two spore coloniescould grow only on min + lys + met.Group 2: In 11 asci, cells from one of the four spore colonies couldgrow on all four kinds of petri plates. Cells from a second one ofthe four spore colonies could grow only on min + lys plates andon min + lys + met plates. Cells from a third of the four sporecolonies could only grow on min + met plates and on min + lys+ met. Cells from the…arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Case Studies In Health Information ManagementBiologyISBN:9781337676908Author:SCHNERINGPublisher:Cengage
Case Studies In Health Information Management
Biology
ISBN:9781337676908
Author:SCHNERING
Publisher:Cengage