PARA 438 Study Guide - Final Guide: Inbreeding, Memory T Cell, Macrophage

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rubymuskrat11

23 Dec 2015

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Describe an experiment that demonstrates the genetic basis for. Histocompatibility & histoincompatability , and that the genes responsible for this are co-dominantly expressed. Begin with two unrelated, inbred donor mice, one from each mouse strain a and mouse strain b. A transplant between one mouse of strain a to another genetically identical mouse of strain a is syngeneic, and will not be rejected. Transplanting from mouse strain b to mouse strain a leads to a rejection of the graft even though it"s allogenic. The children were able to receive skin from either parent (either strain a or strain. The parents were not able to receive skin from their children. You can conclude that there is a genetic component because you"re able to transfer the a to the axb, but not the axb to the a, for example. The alleles both had to be there = co-dominantly expressed. In mice, there are 3 loci for mhci and.

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#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

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

tealcockroach755

Case Study Section 4 â Nitrogen Metabolism Based on âA Sickeningly Sweet Baby Boy: A Case Study on Autosomal Recessive Inheritanceâ by Jacqueline Washington and Anne Zayaitz (National Center for Case Study Teaching in Science) The Patient: Mathew Miller is a 7-day-old infant that is rushed to the emergency room by his parents Emma and Jacob Miller, Mennonites from Lancaster County, PA. Emmaâs pregnancy and delivery with Mathew was normal but he started having trouble nursing and now has completely stopped feeding. By the time they reach the emergency room Mathewâs limbs were rigid and he had had a seizure. The initial examination showed no infection and his x-rays were normal. The family history collected by the Doctor revealed that Emma and Jacob had had a previous son, Samuel that died at 9 days-of-age. There also is a history of unexplained childhood mortality on both sides of the family; Emmaâs mother had two sisters who died in the first year of their lives and Jacobâs father had a sister who died at 7 months of age. Emma also points out to the Doctor that Mathewâs diaper has a funny smell to it. Blood and urine samples were taken for testing and skin biopsies from Mathew, Emma, and Jacob were taken and tested for enzyme activity. The tests results showed that:

â¢ Mathewâs urine had elevated levels of the branched chain amino acids and their Î±-keto acid derivatives (which caused the sweet smell in his diapers)

â¢ Both Emma and Jacobâs skin biopsies had nearly normal levels of branched chain amino acid metabolism/enzyme activity

â¢ Mathewâs enzyme activity was less than 2% of normal

Question 1: What is the most likely diagnosis for Mathew?

A. Argininemia

B. Diabetes mellitus '

C. Maple syrup urine disease

D. Phenylketonuria

Question2: What enzyme did the skin biopsy test measure?

A. Carbamoyl phosphate synthetase I

B. Alanine aminotransferase

C. Phenylalanine hydroxylase

D. Branched-chain Î±-keto acid dehydrogenase complex

Question 3: This enzyme functions VERY similarly to two other enzymes that we have discussed, what are they?

A. Phosphofructokinase-1 and fructose 1,6-bisphosphatase (FBPAse-1)

B. Pyruvate dehydrogenase complex and Î±-ketoglutarate dehydrogenase complex

C. Urease and arginase

D. Isocitrate dehydrogenase and malate dehydrogenase

Question 4: How did Mathew (and presumably Samuel) inherit the disease if both Emma and Jacob have normal enzyme activity levels?

A. Emma and Jacob must both be carriers for the disease but their one normal gene is enough to maintain enzyme activity levels.

B. The disease can happen spontaneously, there is no genetic link.

C. Jacob is not really the father of the boys.

D. The disease can skip generations.

Question 5: What would be the best diet for Mathew to follow to control his disease?

A. Atkinâs diet (low carbohydrates, high fat, and high protein)

B. Low protein

C. Low fat

D. No special diet will control the disease

Over the next 4 years Mathewâs metabolic disease is controlled by his special diet but he still suffers approximately 3 metabolic crises a year. These occur when amino acids accumulate in his blood leading to the swelling of his brain. Even something as simple as a cold or the flu can affect his amino acid levels and send his metabolism into crisis. Jacob and Emma decide that âWe cannot continue to live in constant fear that a minor infection or a simple cold or ear infection could kill our son. Even though we are doing everything we are supposed to, he is still getting sick and we are afraid we may lose Mathew.â They learn that some children with Mathewâs disease have undergone liver transplants and are effectively cured.

Question 6: How/why would a liver transplant effectively cure Mathewâs condition?

A. A new liver would be able to spread itâs genes throughout the body and correct the genetic disorder.

B. A new liver would more effectively respond to metabolic hormonal control.

C. A new liver would express the functional enzyme needed to metabolize the amino acids causing Mathewâs condition.

D. A new liver would produce more hemoglobin to transport oxygen more effectively.

The family is directed to Dr. Morton of Childrenâs Hospital of Pittsburgh, where transplant experts agree to list Mathew for a liver transplant. Jacob and Emma learn that children who receive a liver transplant have to take strong immunosuppressive drugs for the rest of their lives. It was also clear that there was a 40% possibility that Mathew could reject the liver and need a second transplant (which also might be rejected) or he could die from surgical complications.

Question 7: Assuming the liver transplant was successful and Mathew did not reject the new liver, what type of diet would Mathew be able to/need to eat?

A. He would still need to eat his special diet but he would be at less risk of a metabolic crisis

B. He could eat a normal diet

C. He would need to change to a high protein diet like Atkinâs

D. He would be on an all liquid diet

Question 8: Emma and Jacob decide they want to try to have another child. What would be the chance that this child would also have the disorder that Mathew and Samuel have/had?

A. 0 %

B. 25%

C. 50%

D. 75%

E. 100%

Question 9: If Mathew grew up, got married, and decided to have children, what would be the chance that his children would have the disorder? A. 0%, the liver transplant would have cured him

B. 25%

C. 50%

D. 100%

E. It would depend on his wife, if she was normal than there would be a 0% chance of his children having the disorder but 100% would be carriers, if she was a carrier than there would be a 50% chance of his children having the disorder and a 50% chance that they would be a carrier.

Question 10: The disorder is found in 1 in 200,000 infants throughout the US but 1 in 200 in the Amish and Mennonites of Lancaster County, PA. Why is there such a difference in the prevalence of the disease?

A. There are environmental factors associated with being Amish or Mennonites that increase the prevalence.

B. There is a large amount of genetic similarity within these populations because they tend to marry within the population so recessive disorders such as this are more likely.

C. The gene mutation is preferentially carried in Caucasians so when you average the prevalence across all races in the US the prevalence drops.

D. It is unknown why the prevalence is so much greater in Amish and Mennonites.

PARA 438 Study Guide - Final Guide: Inbreeding, Memory T Cell, Macrophage

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