BIOL-108 Lecture Notes - Lecture 11: Bond Length, Electronegativity, Oxoacid

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14 Apr 2018

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Arrhenius defined an acid as any substance that increases the concentration of hydrogen ions when dissolved in water. He defined a base as any substance that increases the concentration of hydroxide ions when dissolved in water. Br nsted and lowry realized that arrhenius" definition was a little to narrow. For instance, it relies on the substances being dissolved in water. The defined an acid as any substance that is capable of donating a proton. A base is any substance that is capable of accepting a proton. When viewed in this context the acid and bases appear in chemical reactions as conjugate acid-base pairs. The difference in the pairs is one proton. Lewis realized that even the definition of br nsted and lowry was too narrow. He defined an acid a any substance that is capable of accepting an electron pair. A base is any substance that is capable of donating an electron pair.

Related Questions

#1 You decide to identify the CFTR mutation by analyzing the genomic DNA of your patients compared to healthy individuals. You specifically are looking to see whether a specific 3' gene truncation has occurred in the patients. You will determine this using hybridization techniques with samples from healthy and CF patients. Which of the following will allow you to accomplish this?

A) Using an RNA probe complementary to the region not removed by the truncation

B) Using an RNA probe complementary to the region removed by the truncation.

C) Using an DNA probe complementary to the region not removed by the truncation.

D) Using an DNA probe complementary to the region removed by the truncation

E) Using an antibody complementary to the region not removed by the truncation.

F) Using an antibody complementary to the region removed by the truncation.

#2To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?

A) As both are composed of nucleic acids, using either would result in identical results.

B) An RNA probe because RNA has uracil bases.

C) A DNA probe because it is more stable than RNA.

D) A DNA probe because RNA cannot bind to DNA.

#3Which of the following will lower the Tm of a given DNA strand?

A)Increasing the percentage of GC base pairs

B) Raising the pH of the solution from extremely acidic to neutral

C) Decreasing the buffer concentration from 50mM NaCl to 5mM NaCl.

#4 Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment. Would radioactivity still have been found in the pellet?

A) No, because the RNA genome would not enter the bacteria upon infection

B) No, because while DNA and RNA nucleotides are similar, they are not identical

C) Yes, because DNA and RNA nucleotides are similar

D) Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly.

#5 Griffith and Avery's transformation experiments allowed us to identify that DNA is our genetic information. Which of the following scenarios would result in bacterial cells that are capable of killing mice upon injection?

A) Heat killed non-virulent bacteria is added to a live virulent bacteria strain

B) Heat killed virulent bacteria is added to a heat killed non-virulent bacteria strain.

C) Heat killed mouse cells are added to a non-virulent bacteria.

#6 The human genome consists mostly of non-coding DNA. Which of the following are benefits of this?

A) Random DNA mutations generally won't affect RNA and protein function

B) It is faster to duplicate the genome when these are present.

C) The existence of introns can lead to multiple variations of proteins encoded by a single gene.

D) It is unlikely transposons would exist in the genome if there was too much protein coding DNA.

harlequineel21

Cystic Fibrosis (CF) is an autosomal recessive genetic disease that is characterized by the secretion of thick, abnormal mucous in the lungs. The mutated gene that leads these symptoms is the CF transmembrane conductance regulator (CFTR gene). The encoded protein is a chloride channel that moves chloride ions out of epithelial cells. In a homozygous individual, the lack of this chloride channel function in a mutant slows the flow of water into extracellular spaces. This lack of water export leads to the heavy mucus in the lungs of CF patients. Although children with CF would die in childhood, current treatments provide life expectancies into the 30âs.

CF is the most frequently inherited genetic disorder among American children. It affects 1 of every 2,500 children; 30,000 currently have the disease. Although its frequency is higher in European populations, it affects all human populations. Heterozygous carrier frequencies in the US are estimated to be 1 out of 29 Caucasians, 1 out of 46 Hispanics, 1 out of 61 African Americans and 1 out of 90 Asians. There are genetic screens for carriers of CF but they are not routinely performed despite this high frequency of carriers. At least all states in the US proscribe testing of newborns to identify children with CF early so that treatment can be started. Newborn tests for CF: http://www.cff.org/AboutCF/Testing/NewbornScreening/ScreeningforCF/ . If CF is such a common genetic disease, and such a simple Mendelian trait, why is genetic screening of parents not done more frequently? Is it just economic, a pain to do, or is it more complex such that the results of genetic testing are not just positive or negative?

Answer the following questions:
1. If two parents are heterozygous carriers of CF, show a pedigree with four children showing the incidence of carriers and afflicted children. If each unafflicted child (carrier or not) mates with another person, who has a 1 out of 30 chance of being a carrier of CF, what is the expected incidence CF disease in the third generation (frequency of CF in any one of grandchildren of the original couple)?

2. Pleiotropy of Cystic Fibrosis. Watch the two videos on the left at:

http://www.cftrscience.com/MOD-animation
or, if you can not view these, read the short NIH Genetics Home Reference for Cystic Fibrosis http://ghr.nlm.nih.gov/condition/cystic-fibrosis

Connect the description with one of the following videos, or related ones (there are many) you see:
video (10 minutes) of daily life of a child with Cystic Fibrosis http://www.youtube.com/watch?v=WUsuGKzdDg8
Lauraâs (20 year old) video diary of life with CF http://www.youtube.com/watch?v=Pgrf8qJXhpc&feature=related

In your own words, describe the range of CF phenotypes. List the other organs besides lungs that are affected and how does it affects those organs.

3.The DNA sequence of the CFTR gene is 170,000 base pairs in length and encodes the CFTR protein that is 1480 amino acids long. Open the accompanying text file (in Canvas/Files/Discussion folder) âCFTR gene DNA sequenceâ. You will see an alignment of the NIH human genome sequence of the CFTR gene â called âReferenceâ rather than âwild-typeâ- against a gene sequence containing a common mutation in CFTR. Look at the alignment and determine how the DNA sequence differs between them. What is the position number where they differ? How many and what nucleotides do they differ in? A â-â mark in one sequence in the alignment means that that base sequence is missing the corresponding bases in the other sequence. The sequence is a long one but the differences are in bolded bright red text. Matching sequence is marked with a | linking the two sequences. Look near the sequence AAATATCATC if you use FIND tool in Word.

In another file, you will an alignment of the reference translated into a protein sequence, compared to the mutant allele translated into protein sequence. The differences are again in red whereas the areas where the proteins match are in black. Look at the alignment. How do these two proteins differ? Again, find the position number and describe how they differ at that location. A â-â mark in the alignment means that amino acid is missing when compared to the other sequence.

4. Given that the CFTR gene can be analyzed for a mutation that causes CF, couples interested in being tested as carriers face several choices given the results.
What could be a consequence of a false positive result (couple is identified as carriers but they would not produce any children with symptoms)?

What could be a consequence of a false negative result (a test indicates the couple is not both carriers but they would produce some children with CF)?

5. The sensitivity of the test can vary with ethnic group. The following is a chart from the NCBI Genetic testing site. A sensitivity less than 100% would be likely to produce false negative results from an inability to determine that someone is carrier.

CF Carrier Test Sensitivity by Ethnic Group http://www.ncbi.nlm.nih.gov/projects/GeneTests/static/concepts/primer/ptscarr5.shtml
Ethnic Group ¹	Population Frequency of CF	Sensitivity of Carrier Detection ²
Caucasian	1 in 29	90%
Ashkenazi Jewish	1 in 29	97%
Native American	1 in 32	94%
Latino	1 in 46	57%
African-American	1 in 60	75%
Asian	1 in 90	30%

Why might different ethnic populations differ in the ability of the genetic test to identify carriers of CF?

Question 1

1 pts

Cystic fibrosis is caused by nonsense and missense mutations in the CFTR gene, which encodes for a chloride channel. You are studying cystic fibrosis patients to determine what mutation they possess in the CFTR gene. The difference between the mutant and wild type CFTR genes can be uncovered by examining the CFTR:

	DNA
	mRNA
	protein
	tRNA

Question 2

1 pts

You decide to identify the CFTR mutation by analyzing the genomic DNA of your patients compared to healthy individuals. You specifically are looking to see whether a specific 3' gene truncation has occurred in the patients. You will determine this using hybridization techniques with samples from healthy and CF patients. Which of the following will allow you to accomplish this?

	Using an RNA probe complementary to the region not removed by the truncation.
	Using an RNA probe complementary to the region removed by the truncation.
	Using an DNA probe complementary to the region not removed by the truncation.
	Using an DNA probe complementary to the region removed by the truncation.

Question 3

1 pts

You would like to ensure that this experiment (to determine whether patients have a specific CFTR gene truncation using hybridization) is properly controlled. Which of the following samples must you test?

	The genomic DNA of a healthy individual who does not have cystic fibrosis.
	The genome of a CFTR patient known to have the specific truncation you are trying to identify.
	The genome of a CFTR patient with a missense mutation but full length gene.
	The genome of a healthy individual married to a CFTR patient with the specific truncation you are trying to identify.
	The genome of a patient with muscular dystrophy, which can be due to a trucation in the dystrophin gene.

Question 4

1 pts

To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?

	As both are composed of nucleic acids, using either would result in identical results.
	An RNA probe because RNA has uracil bases.
	An RNA probe because it could also be used in a translation experiment.
	A DNA probe because it is more stable than RNA.
	A DNA probe because RNA cannot bind to DNA.

Question 5

1 pts

Which of the following will lower the Tm of a given DNA strand?

	Increasing the percentage of GC base pairs.
	Raising the pH of the solution from neutral to basic.
	Decreasing the buffer concentration from 50mM NaCl to 5mM NaCl.
	None of the above.

Question 6

1 pts

One step of the Hershey/Chase experiment involved blending the virus/cell mixture before centrifugation and probing the pellet for radioactivity. Why was the blending step necessary?

	To collect the bacteria at the bottom of the tube.
	To break open the bacteria to release the genome.
	To separate the bacteria from the bacteriophages.
	To be able to detect the radioactivity.

Question 7

1 pts

Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment. Would radioactivity still have been found in the pellet?

	No, because only DNA can be labeled with radioactivity.
	No, because the RNA genome would not enter the bacteria upon infection.
	No, because while DNA and RNA nucleotides are similar, they are not identical.
	Yes, because DNA and RNA nucleotides are similar.
	Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly.

Question 8

1 pts

Griffith and Avery's transformation experiments allowed us to identify that DNA is our genetic information. Which of the following scenarios would result in bacterial cells that are capable of killing mice upon injection?

	Heat killed non-virulent bacteria is added to a live virulent bacteria strain.
	Heat killed virulent bacteria is added to a heat killed non-virulent bacteria strain.
	A heat killed virulent bacteria that is treated with a nuclease, is then added to a non-virulent bacteria strain.
	Heat killed mouse cells are added to a non-virulent bacteria.

Question 9

1 pts

The human genome consists mostly of non-coding DNA. Which of the following are benefits of this?

	Random DNA mutations generally won't affect RNA and protein function.
	It is faster to duplicate the genome when these are present.
	The existence of introns can lead to multiple variations of proteins encoded by a single gene.
	It is unlikely transposons would exist in the genome if there was too much protein coding DNA.

Question 10

1 pts

Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (âade) and histidine (âhis). The plasmid she is currently working with consists of an origin of replication and the Ade gene.

Following her transformation of the plasmid into her yeast, what media will the cells be plated on to select for cells that have picked up the plasmid?

	Media containing histidine
	Media containing adenine
	Media lacking adenine
	Media lacking histidine

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BIOL-108 Lecture Notes - Lecture 11: Bond Length, Electronegativity, Oxoacid

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