CH 461 Lecture Notes - Lecture 33: Pyrimidine Dimer, Hydroxylamine, Mutagen

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silvercicada472

12 Jul 2018

School

University of Alabama

Department

Course

CH 461

Professor

Dunkel

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Document Summary

Dna information is subject to loss or degradation. After all, dna is a chemical, and chemicals can undergo reactions. Damage to a dna leads to a loss or alteration of its information content, a process known as mutation. Compounds or agents that cause mutations are called mutagens. causes the loss of the amino group of deoxycytosine in dna, converting it to. Several types of mutations exist, depending on the changes to the sequence information of the chromosome. Transition mutations are defined as the conversion of. G c to a t or a t to g c base pairs. When replicated, deoxyuracil is a template for deoxy a incorporation. Hydroxylamine is a monodirectional mutagen, only converting g c a t base pairs. Transversion mutations convert purines to pyrimidines or vice versa. For example, conversion of an a t base pair to a t a or c g base pair is a transversion. Ultraviolet light can cause transversions, although not exclusively.

Related Questions

GENETICS QUESTION: For this experiment we focused on forward genetics: First, his- mutants were identified and then complementation analysis was used in order to determine which of the 7 his genes was mutated. ( The exact data can be seen at the bottom). In this experiment we screened for his- mutants that were generated via a random mutation process resulting from errors in DNA replication or induced by EMS aka Ethyl methanesulfonate. We compared a mutagenized culture that was treated with EMS and an un-mutagenized culture that was not treated with EMS. However, both cultures went trhough an enrichment process to increase the frequency of his- mutants.

Point mutations involve the substitution of one type of base pair for another. If the substitution replaces a purine with a purine (or a pyrimidine with a pyrimidine), it is called a transition. If the substitution replaces a purine with a pyrimidine (or a pyrimidine with a purine), it is called a transversion. Many mutagenic chemicals cause this type of mutation by alkylation of a nucleotide base. The chemically altered base can then be misread during replication or repair. For example the mutagen, ethyl methanesulfonate (EMS), adds an ethyl (-CH2-CH3) group to both guanine or thymine bases causing the modified guanine to pair with thymine and the modified thymine to pair with guanine.

1. Why might one gene be more susceptible to EMS mutagenesis than another gene?

2. Why might we have expected to find a higher frequency of polar mutants (his4A^-B^-C^-) in the unmutagenized culture compared to the mutagenized culture?

Table 2. Experiment 1B class data summary (number of his^-mutants by section) Unmutagenized
Section	1	2	3	*4ABC*	4A	4B	4C	5	6	7	*Total his^-* mutants**
Tuesday	7	1	5	4	2	1	9	2	3	1	35
Wednesday	2	1	14	0	4	2	2	4	8	2	39
Total	9	2	19	4	6	3	11	6	11	3	74
% each mutant	12.1	2.7	25.7	5.4	8.1	4.1	14.9	8.1	14.9	4.1	~100%

Mutagenized
Section	1	2	3	4ABC	4A	4B	4C	5	6	7	*Total his^-* mutants**
Tuesday	2	5	4	2	6	5	5	0	4	1	34
Wednesday	9	1	15	0	9	1	2	2	9	4	52
Total	11	6	19	2	15	6	7	2	13	5	86
% each mutant	12.8	7.0	22.1	2.3	17.4	7.0	8.1	2.3	15.1	5.8	~100%

13. The arms of a human and the wing of a bat are _________________ structures, composed of the same bones that have been modified for different functions.

analogous

vestigial

c.	homologous

d.	convergent

14. Xeroderma pigmentosum is an autosomal recessive genetic disorder. People who are homozygous for the disease-causing allele have skin that is damaged by exposure to ultraviolet radiation; this can lead to skin cancer at an early age. The mutation that causes this disorder is in a gene that functions:

a.	to repair DNA by excising (removing) nucleotides damaged by UV light.

b.	in lysosomes, to break down lipids that otherwise accumulate in the nervous system.

c.	to produce the UV blocking pigment melanin in the skin cells

d.	in red blood cells to carry oxygen to the skin.

15. The Amish are a religious sect descended from a small group of colonists who came to the United States over 200 years ago. Their religious beliefs have kept them isolated from society. The Amish have a much higher than average incidence of both polydactyly (extra fingers and toes) and dwarfism in their communities. This is most likely explained by:

a.	inheritance of acquired characters.

b.	the founder effect.

c.	disruptive selection.

d.	natural selection.

16. In the structure of DNA that Watson and Crick proposed, the sides of the "ladder" of the DNA molecule are always the same distance apart, making the molecule a constant width. This occurs because when the bases hydrogen bond to form the rungs of the "ladder", a:

a.	purine always pairs with a pyrimidine.

b.	purine always pairs with a purine.

c.	pyrimidine always pairs with a pyrmidine.

d.	The structure of the double helix is not determined by how the bases pair.

17. The RNA transcript of DNA that travels to the cytoplasm, carrying the instructions to make a protein, is called:

rRNA.

tRNA.

mRNA.

d.	RNA polymerase.

18. When DNA is replicated, the error rate is approximately one error for every 10,000 nucleotides copied. However, that error rate is reduced to only 1 error for every 1 billion nucleotides. The DNA sequence is "corrected" by:

a.	repeating S phase to see if the error rate is lower the second time around.

b.	enzymes that proofread the DNA and repair errors.

c.	messenger RNA during the process of transcription.

d.	Any errors made during DNA replication cannot be corrected. If errors are detected after S phase, the cell is destroyed.

19. Mendel's principle of segregation says that:

a.	when gametes are formed, each gamete receives only one allele for a particular gene.

b.	some genes are dominant to others.

c.	a testcross must be used to determine the genotype of an organism with a dominant phenotype.

d.	all of the above

20. Tay-Sachs disease is lethal before reproductive age, but the allele persists in Louisiana French Canadians because:

a.	it is a dominant allele.

b.	new mutations causing this disease are common in that population.

c.	it is not expressed in the phenotype of heterozygous individuals.

d.	the disease is contagious and can be transmitted by nonheritable means.

21. Physical features that are similar in organisms which are not closely related to each other are most likely a result of ____________. A good example of this is the similarities seen between some marsupial and placental mammals.

a.	natural selection.

b.	adaptation to similar environments

c.	convergent evolution

d.	All of the above phrases can be used to correctly fill this blank.

22. In humans, which of the following sex chromosome compliments could be found in males?

23. The proteins that are associated with eukaryotic chromosomes:

a.	help to package the DNA within the nucleus.

b.	are involved in gene expression, through binding with the DNA.

c.	may be replicating or transcribing the DNA.

d.	All of these are true of proteins that are associated with eukaryotic chromosomes.

24. A sequence of DNA nucleotides that contains the information to produce a single protein is a(n):

codon.

gene.

c.	polypeptide.

d.	anticodon.

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Evaluation of PCR product electrophoresed on a 0.8% agarose gelshows non-specific bands. The appropriate modification for the nextPCR reaction is to

a )increase the concentration of Taq polymerase.

b) increase the number of PCR cycles.

c) decrease the template denaturation temperature.

d) reduce the concentration of primers.

In the Sanger sequencing method, chain termination is the resultof

a)misincorporation of dUTP instead of dTTP.

b)incorporation of ddGTP.

c)secondary structures in the template strand.

d)sequencing through repeats.

The purpose of EDTA in a DNA storage buffer is to

a)maintain single strands.

b)bind magnesium

c)activate DNases.

d)dissolve DNA into solution.

The thermal cycler ramp refers to the

a)degree of incline of the machine on the laboratory bench.

b)gradient between denaturation and annealing temperature.

c)programming key to set number of PCR cycles.

d)amount of time required to reach the desired temperature.

Evaluation of a blood sample indicates the presence of a clot.Corrective action to salvage this specimen for analysis would mostlikely require

a)digestion with proteinase K.

b)homogenizing the sample under UV irradiation.

c)removal of the clot with urea.

d)requesting a new sample immediately.

Incorporation of biotin into template DNA by nick-translationrequires

a)DNase.

b)ligase.

c)random hexanucleotides.

d)3 temperatures.

An isolated DNA sample is viscous, difficult to pipet, and willnot resuspend in TE buffer after rehydration at 37C for 24 hours.The most appropriate step is to

a)digest the DNA with a frequent cutting restriction enzyme.

b)re-extract the original sample with higher volumes ofreagents.

c)increase the rehydration temperature to 95C.

d)add more TE buffer.

Excess glycerol in a restriction digestion reaction resultsin

a)enzyme degradation.

b)diminished enzyme activity.

c)increased temperature required for digestion.

d)decreased enzyme buffer pH.

Primer dimers will most likely be a result of PCR when

a)3' and 5' primers are designed as 50% GC.

b)DNA template concentration is very low.

c)enhancing agents are added to the master mix.

d)template DNA is initially denatured completely.

Correct operation of a pH meter requires

a)calculation of the electrode millivolt equivalents.

b)standardization of the meter with a range of buffers.

c)documentation of absorbance and standard deviation.

d)verification of pH readings with calf thymus DNA.

The uracil DNA glycosylase pre-PCR decontamination procedurefailed. The most likely cause of the failure is the

a)enzyme solution was heated prior to PCR.

b)enzyme buffer solution was alkaline pH.

c)N-glycosidic DNA bonds cleaved after enzyme treatment.

d)PCR was performed with dTTP instead of dUTP.

The real-time PCR analysis does not show the characteristicmelting curve for the patient or control samples. The fluorimetryreadings of the PCR products indicate the appropriate concentrationof template was added to the real-time PCR reaction mix. Which ofthe following reagents was likely missing from the PCR master mixto account for this finding?

a)SYBR Green I

b)template DNA

c)DMSO

d)5X Cresol Red

Which of the following conditions favor enhanced specificity ofPCR?

a)decrease in the concentration of dNTPs

b)increase in the sodium chloride concentration

c)increase in the concentration of Taq polymerase

d)decrease in the ramp speed and temperature

Which of the following compounds is an inhibitor of Taqpolymerase?

a)magnesium

b)hemoglobin

c)DMSO

d)betaine

Which of the following compositions of gels would allow for theefficient separation of linear DNA molecules in the range of 100 to300 base pairs in length?

a)0.3% agarose

b)1.0% agarose

c)1.0% polyacrylamide

d)8.0% polyacrylamide

4.92*10^4 cpm of a random primed 32P dCTP (3000 Ci/mmol) labeledoligonucleotide probe were precipitated from a 50 uL standardreaction by TCA and 5.28*10^4 cpm is the total cpm in the sample.What is the percent incorporation of radioactivity for thislabeling reaction?

a)1.86%

b)46.50%

c)93.00%

d)98.40%

The stock solution of HCl is 1M. How much stock solution isrequired to prepare 250 mL of 50 mM HCl?

a)0.2 mL

b)5 mL

c)12.5 mL

d)125 mL

Spectroscopy measurement of a DNA sample isolated by saltextraction technique gave the following absorbance readings: A2600.4032 A280 0.3765

This sample is:

a)acceptable for molecular testing.

b)contaminated with salt.

c)contaminated with protein.

d)contaminated with high amounts of RNA.

What is the dilution factor if 50 uL of blood is diluted with950 uL of saline?

a)19

b)20

c)40

d)50

Which of the following are used to prepare a 0.9% agarosegel?

a)0.9 g agarose + 100 mL distilled water

b)0.9 g agarose + 100mL TAE

c)9 g agarose + 100mL TBE

d)90 g agarose + 100 mL buffer

The concentration of double-stranded DNA with an optical densityreading of 1.0 is

a)0.5 ug/ul

b)1.0 ug/ul

c)10 ng/ul

d)50 ng/ul

How much Tris base (MW 121.1) would be required to make 4 litersof Tris-EDTA buffer that is 15 mM Tris?

a)0.7266 grams

b)0.1211 grams

c)1.8165 grams

d)7.266 grams

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CH 461 Lecture Notes - Lecture 33: Pyrimidine Dimer, Hydroxylamine, Mutagen

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