BIOL 2905 Chapter Notes - Chapter 2.8: Wild Type, Zygosity, Null Allele

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lavendershark790

13 Oct 2017

School

York University

Department

Biology

Course

BIOL 2905

Professor

Dasantila Golemi- Kotra

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Document Summary

Mutations (cid:272)an (cid:271)e classified as their phenotypi(cid:272) effe(cid:272)t. A null allele is a mutant copy of a gene at a locus that completely lacks that genes normal function. Results in complete absence of gene protein or expression of non-functional gene. Null mutation: a mutation in a gene that leads to it not being transcribed into rna or translated into a non-functional enzyme: hypomorphic mutation: less functional protein- activity reduced, protein doesn"t function well. Described a mutation that causes a partial loss of gene function through reduced expression or reduced functional performance, but not a complete loss. Most loss of function mutations are recessive-why: recessive in terms of phenotypes they encode, ressive and dominant does not mean alleles refers to phenotype. Less than the threshold shows the phenotype: enough activity above threshold for the wildtype phenotype- one copy is enough, if its less than the threshold, it shows mutant phenotype.

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INSTRUCTIONS

Go to the OMIM database on the NCI portal. Search for PAH, select the first record on the list The following page will appear: Read the initial information Phenylketonuria (PKU) is an autosomal recessive inborn error of metabolism resulting from a deficiency of the enzyme PAH (Phenylalanine Hydroxylase). Lack of PAH enzyme activity results in elevated level of phenylalanine. On the other hand, mutations in the PAH gene can also result in a milder disease called hyperphenylalaninemia (HPA) characterized by an elevated level of phenylalanine without the cognitive impairment associated with PKU. QUESTIONS

Question 1: Consider two actual DNA mutations found as 2 PAH mutant alleles. a. Mutation # 0017 changes codon 414 from a tyrosine to a cysteine (TYR414CYS). A homozygous genotype leads to HPA. Indicate the type of mutation for 0017 at the DNA level. Identify the class of mutation at the protein level. Based on the disease phenotype, would you expect the mutated protein to be partially functional or non-functional? (I think it is functional..? http://www.omim.org/entry/612349) Explain your reasoning briefly b. . Mutation #0018 is a mutation that changes the splice site of intron 4 (IVS4). Explain the consequence of mutation 0018 at the level of transcription of the gene, splicing and protein synthesis. Homozygous 0018 leads to PKU phenotype Question 3: Assume an individual heterozygote for 0017/0018. ?Which special term is used to describe a heterozygote with two different mutant alleles - Compound heterozygote Where along the dominance-recessivity scale would the phenotype of this individual falls in relation to the phenotype of the two mutant homozygotes? Explain briefly.

11. You suspect that two genes interact with one another to determine the product of a pathway you are studying. You cross two pure breeding plantsâone with white flowers and one with pinkâto make F1 individuals that have blue flowers. You cross the F1 generation and observe the following numbers of individuals with different phenotypes: Number of individuals Phenotype 91 Blue 30 Pink 39 White Which pathway best explains these results? (Note that arrows indicate enzymatic reactions carried out by products of genes 1 and 2. Assume that buildup of metabolites will determine flower color.)

Question 12 and 13: Consider the biochemical pathway presented below. This pathway determines pigment production in pea flowers. (Note: gene products are required for each reaction to proceed.)

12. Do you expect that genes A and B will interact with one another?

A. Yes B. No C. Insufficient information to determine

13. What phenotype ratio would you expect in offspring produced by crossing AaBb parents?

A. 9:3:3:1 (blue : green : yellow : white) B. 9:3:4 (blue : green : white) C. 9:7 (blue : white) D. 15:1 (blue : white)

14. How can haplosufficiency explain dominance of an allele that codes for a functional protein over a null (loss of function) allele?

A) a single functional copy of the gene produces sufficient gene product for normal biological function B) a single functional copy of the gene produces insufficient gene product for normal biological function C) the null allele produces a product that interferes with proper function of the normal allele D) the null allele produces a product that does not interfere with proper function of the normal allele E) the null allele rapidly mutates to reacquire normal functionality

15. In a plant species with genes A and B, you have isolated recessive mutant alleles at each locus, which you call a and b. You would like to know whether these two genes are linked. To test this hypothesis, you obtain an individual that is heterozygous at both genes (AB/ab) and perform a test cross. Which results would be most consistent with these loci being tightly linked? (Results are counts of the numbers of offspring from the testcross.)

A. Ab/ab 390 aB/ab 410 AB/ab 350 ab/ab 420

B. Ab/ab 80 aB/ab 73 AB/ab 760 ab/ab 754

C. Ab/ab 700 aB/ab 695 AB/ab 56 ab/ab 59

D. Ab/ab 400 aB/ab 65 AB/ab 50 ab/ab 659

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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 2905 Chapter Notes - Chapter 2.8: Wild Type, Zygosity, Null Allele

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