Genetics Final Review Talar Tfnakjian OH: T, TH 8-9 am Natural Sci I room 2108 [email protected] Emily Ling OH: Mon 9-10 AM SH 149 Wed 3-4 PM SH 149 [email protected].

Genetics Final Review

Talar TfnakjianOH: T, TH 8-9 am Natural Sci I room [email protected]

Emily LingOH: Mon 9-10 AM SH 149Wed 3-4 PM SH [email protected]

Online Evaluationhttps://eee.uci.edu/programs/biotutor/bio97b.php

Talar [email protected]

Emily [email protected]

I will hold 1 extra office hour Monday at 3-3:50PM in SH149

This power point will be uploaded after the review…or at least by the weekend.

Disclaimer Topics discussed in this review are by

no means everything you are expected to know.

We are not in direct contact with the professor. Do not rely solely on this review and expect to get a 100%.

Lecture 1Pathogenetics

Outline What are bacteria What are viruses/phages Antibiotic resistance in bacteria Modes of DNA movement between cells

Lecture 2Cloning

Steps required for cloning

1- remove the nucleus form an egg.

2- inject a nucleus from the organism being cloned into the enucleated egg.

3- inject the egg into a surrogate mother who will birth the organism.

The result is an organism that is genetically identical to the donor of the nucleus.

Summary

Donor

Genetically identical to donor

Cloning a gene

vector

Recombinant DNA

Cloning a gene1. Cut the desired gene with restriction enzyme.2. Cut the vector with the same restriction

enzyme.3. Insert the desired gene into the vector4. Seal the gap with DNA ligase5. This result is a recombinant DNA molecule6. Insert the recombinant DNA into a host for

replication.

Characteristics of a vector A vector should have:a) a replication origin.b)A selectable marker like an antibiotic

resistancy gene.c) Restriction site to insert desired genes.

AmpR

GAATTCCTTAAG

Antibiotic resistance

gene

Replication origin

Useful Restriction

site

Terms to know… Restriction site: A short DNA sequence that

can be cleaved by a restriction enzyme. Restriction Enzyme: A restriction enzyme,

or restriction endonuclease, cleaves DNA at a specific sequence

Tip: review Factor VIII cloning in your notes.

Lecture 3Engineering DNA

Outline Review the structure of the human gene:

regions to know that are related to replication and transciption-translation

Recombinant DNA Genetic Engineering (Transgenic or GM

organism) Uses in therapies for disease and agriculture

Gene therapy

Lecture 4

Linked genesand genetic mapping

What does it mean to be linked? The idea is that we have many

chromosomes containing lots of genes , and not all genes can segregate independently.

Genes located close together on the same gene must be inherited together.

The usual Independent Assortment

A a

B

Stem cellafter chrmreplication

Half of the time

©2000 Lee BardwellA a

B b b

aa

B B

A

b

A

b

aa

b b

A

B

A

B

Half of the time MEIOSIS I

Yellow green

Round wrinkled

A

a

B

b

A

a

B

bA

a

B

b

gametes

Stem cell

Meiosis I + II

Gametes produced with two unlinked genes

©2000 Lee Bardwell

25%

25%

25%

25%

Equal probabilities!

AB/ab

haplotype

haplotype

©2003 Lee Bardwell

Ab/aB

haplotype

haplotype

Recessive alleles are cis

Recessive alleles are in trans

A

a

B

b

A

a

b

B

Yellow,Rounddiploid cell

Yellow,Rounddiploid cell

A

a

B

b

A B

a b

gametes

Two completely linked genes

©1999 Lee Bardwell

Cis

A

a

b

B

A b

a B

gametes

©1999 Lee Bardwell

Yellow

Round

Trans

However the genes are not always a

100% linked

Recombination gives rise to non parental

genotypes

Non-parental chrms

A Bab A b a B a b

©2000 Lee Bardwell

ab/ab x AB/ab

A B a bA b a B

35% : 18% : 16% : 31%

226 : 114 : 102 : 202644 progeny were scored

To figure out the recombination frequencies simply add the recombinant type frequencies.

In this case 18% + 16%

Knowing the recombination frequency is useful for mapping

genes along a chromosome

©2001 Lee Bardwell

ab/ab x AB/ab: 39% AB 11% Ab 11% aB 39% ab

ad/ad x AD/ad: 42% AD 8% Ad 8% aD 42% ad

bd/bd x BD/bd: 46% BD 4% Bd 4% bD 46% bd

Ar Be

Ar Du

Be Du

22 cM

16 cM

8 cM What’s the order of the genes?

solution

0rBe ArDu BeAr Du

Lecture 5Human Gene CloningAnd more molecular genetics

Outline Cloning recombinant DNAs and creating

a library Analyzing DNA Polymorphism (brief)

Lecture 6

Mutation and Repair of DNA

Fine-Scale Mutations -

Involve less than 1000 base pairs

Often, just a single base pair is changed

Include substitutions, deletions, insertions, duplications

©1999 Lee Bardwell

5’----------GAATTC---------3’3’----------CTTAAG---------5’Insertion:5’----------GAACTTC---------3’3’----------CTTGAAG---------5’Duplication:5’----------GAATATTC--------3’3’----------CTTATAAG--------5’

5’----------GAATTC---------3’3’----------CTTAAG---------5’Substitutions:5’----------GATTTC---------3’3’----------CTAAAG---------5’5’----------GAGTTC---------3’3’----------CTCAAG---------5’Deletion:5’----------GAATC----------3’3’----------CTTAG----------5’

Substitutions• Silent- TGT (Cys)--> TGC (Cys)

GCA (Ala)--> GCN (Aladoes no result in a different amino acid (N = any)

• Missense-TGT (Cys)--> TGG (Trp)

• Nonsense- TGT (Cys)--> TGA (STOP)

Spontaneous MutationsI. Mistakes

Replication errors Polymerase sometimes incorporates the wrong

base.It has a proofreadingactivity that takes care of the misincorporations. Howeever some mistakes may not be corrected

Recombination errors Unequal crossing over, etc.

Between similar seqs on non-homologous chrms --> translocation

Between similar seqs within a chrm --> inversion or deletion

Between tandemly repeated seqs on homologous chrms --> duplication or deletion

Refer to lecture slides for more details

II. endogenous DNA Damage

Spontaneous base damage Deaminations, depurinations

-Byproducts of metabolism Oxygen radicals that damage DNA

Deamination of CytosineNH

H

H

H

O N

N

2

H*

H

H

O N

N

O

*Thymine has CH3 here

deNHn

Cytosine Uracil

©2000 Lee Bardwell

H

Endogenous DNA damage

Induced MutationsCause by outside

sources

Cause by: -Chemicals

Natural In foods, e.g. aflatoxin

Man-made/man-increasedNitrogen Mustard - WWI nerve gas

Benzopyrene - smoke from coal, autos, cigs -Ultraviolet (UV) Radiation (from sun)>> Pyrimidine dimerIonizing radiation

Natural: radon gas, cosmic raysMan-made: x-rays, nuclear tests

what happens to the Damage DNA? =(

-Tolerated (ignored)-Repaired-Can kill the cell or cause the cell to kill itself -Can become fixed, resulting in a mutation

Note fixed≠ repaired

Repair mechanisms

Uracil DNA glycosylase

An enzyme that removes Uracil from DNA

Resulting abasic site is filled in by polymerase

Uracil in DNA comes mainly from deamination of cytosine

That may be why DNA uses thymine instead of uracil

If the uracil isn’t removed, it will pair with A, causing C/G --> T/A transition.

©2007 Lee Bardwell

5’-GAATTU-3’3’-CTTAAG-5’

5’-GAATT_-3’3’-CTTAAG-5’

5’-GAATTC-3’3’-CTTAAG-5’

Nucleotide Excision Repair

Carried out by a multi-protein complexRemoves bulky adducts from DNA, e.g.

Pyrimidine dimers caused by UV Benzopyrene-DNA adducts

Nearby nucleotides are also excisedResulting single-strand gap is filled in

by polymerase

©2000 Lee Bardwell

Lecture 7Cancer

Outline Cancer: a loss of growth regulation

Disease of somatic cells Oncogenes and Tumor Suppressor

Genes Spontaneous and Induced Mutations Malignant and Benign 6 Factors of a Successful Cancer Cell Hereditary Cancers

Lecture 8

Population Genetics

Key words to knowGenotype frequency = proportion

of individuals in a population with a specific genotype

Allele frequency = proportion of alleles in a population

Population = group of organisms of the same species living in the same geographical area

Gene pool = all alleles in population

The big picture…

Relationship between allele frequencies and genotype frequencies

How to predict what AF’s and GF’s will be in future generations

Hardy & Weinberg

Developed a simple mathematical model of the transmission of alleles from generation to generation

Based on this equation, if we know the allele frequencies in a population then we can calculate the allele frequency in the next

generation

(p+q)2 = p2 + 2pq + q2 = 1 This gives you the genotype frequencies.

To count alleles in the next generation either actually count the alleles in the gene pool Or use the shortcut.

freq(A) = freq(AA) + 1/2 freq(Aa)

freq(a) = freq(aa) + 1/2 freq(Aa)

Hardy –weinberg condition Allele frequencies will remain unchanged under

assumptions of hardy-Weinberg

Mating is random Allelic frequencies are the same in males and

females All genotypes have equivalent viability and fertility Mutation does not occur Migration into the population is absent Population is large so that allelic variations do not

occur by chance

Things to watch out for

If we know GF’s in a generation of a pop, then we can always calculate AF’s, whether or not HW equil.

If we know AF’s, we can only calculate GF’s (in this or the next generation) if the populations is in HW equilibrium

Let’s walk through an Example… An autosomal recessive disease affects

1/100 people In a population. Calculate the frequency of carriers.

So the info we are given is:genotype frequency = q2

we need to get the allele frequency first q= 1/10=.1P=.9To calculate carrier frequency simply do 2pq= 2(.1)(.9)= .18

X-linked With X- linked things are a little different

only with males. The genotype frequencies for males is

the same as the allele frequency.

Let’s work on another problem - One out of five females are

affected by a certain X- linked disease which causes them to have a shy personality. Calculate the heterozygote frequency in males and females:

Solution in this case q2 is given > q= (1/5)

~0.45 p= 0.55 Males don’t have a carrier because they

only have one copy of X Female carriers: 2pq= 2x .45x .55= .50

Forces that can cause deviation from HW ratios1. Gene flow (e.g., migration)2. Genetic drift3. Mutation4. Natural Selection5. Non-random mating (e.g.,

inbreeding)

©2005 Lee Bardwell

Lecture 9Population Genetics II

Outline What keeps unfavorable genes in gene

pool: When Bad alleles can be good as well Slow selection against recessive alleles

in diploids Mutation-selection equilibrium Heterozygote superiority

You made it!

Study Hard.

GOOD LUCK TO YOU ALL.

Outline What are bacteria What are viruses/phages Antibiotic resistance in bacteria Modes of DNA movement between

cells

A straight forward lecture(s).

Other Topics to Understand Characteristics of Bacteria Characteristics of Viruses & Phages How Antibiotics Work

Targets components that are different from infected host (like humans) Best ones target structures ONLY

associated with the micro-organism (like a cell wall of a bacteria)

Modes of DNA Transfer Conjugation – F-plasmid Transformation – DNA from environment Transduction – via Virus/Phage

Plasmids

A Question… 13. Antibiotic resistance genes are

typically found in: A. plasmids B. transposable elements C. viruses D. A and B E. A, B and C

D. A and B

A Question… 12. The process in which recipient

cells acquire genes from free DNA molecules in the surrounding medium is called A. transduction B. ligation C. transformation D. conjugation E. transposition

C. Transformation

Lecture 2Questions

A Question… 14. If you wanted to reproductively clone

your cat Trixie, you would need to take take a haploid/diploid/polyploid nucleus from a somatic cell/gamete of Trixie, and put it into an enucleated egg. A. haploid, somatic cell B. haploid, gamete C. diploid, somatic cell D. diploid, gamete E. polyploid, gamete

C. Diploid, Soma

An 8kb plasmid is digested with EcoRI (E) and/or BamHI (B), and the digests are run on an agarose gel and stained. The results are shown below; molecular size standards are shown on the left.

Answer: B

Based on the results of the gel, which plasmid map looks the most correct? Fill in answer E if you think none of them are correct. The light grey lines are to help you gauge distance.

A B C D E

A Question… 5. Which of the following statements are

TRUE?: A. All eggs produced by the same woman are

genetically identical to each other. B. All sperm produced by the same man are

genetically identical to each other. C. Both A and B are true. D. All the gametes produced by a person are

genetically identical to a somatic cell of that individual.

E. None of the statements (A-D) are true.

E. None True

Lecture 3Engineering DNA

Outline Brief: review the structure of the human

gene: regions to know that are related to replication and transcription-translation

Recombinant DNA Genetic Engineering (Transgenic or GM

organism) Uses in therapies for disease and agriculture

Gene therapy

Lecture 5Human Gene CloningAnd more molecular genetics

Outline Cloning recombinant DNAs and creating

a library Analyzing DNA Polymorphism (brief)

A Question… 16. If you wanted to express a cloned human

gene in the bacterium E. coli, so that you could grow up the bacteria in large batches and make the corresponding human protein, you would need to: A. Remove the human introns B. Swap the human promoter/enhancer for a

bacterial promoter C. Change the codons because otherwise the wrong

amino acids would be specified D. A and B E. B and C

D. A and B

A Question…29. Your uncle had a heart attack at an early age. Since early onset heart disease is a multifactorial trait, one important piece of information you need to know to assess your risk is to know the fraction of alleles you share with your uncle. The fraction of your alleles that you share with your uncle is...

Answer: ¼ or 25%

Lecture 7Cancer

Outline Cancer: a loss of growth regulation

Disease of somatic cells Oncogenes and Tumor Suppressor

Genes Spontaneous and Induced Mutations Malignant and Benign 6 Factors of a Successful Cancer Cell Hereditary Cancers

What are some of these genes that lead to cancer when mutated?

Oncogenes Gas pedal for cell proliferation Mutation --> Gas pedal stuck down

Tumor suppressor genes Brakes for cell division Mutation --> Brakes don’t work

©2001 Lee Bardwell

The Difference

The 6 Steps

A Question… 7. Fill in the blanks. A gene that encodes a

protein required for the repair of certain types of DNA damage is likely to be a _______. A gene that encodes a protein required to relay a signal telling a non-dividing cell to divide is likely to be a _______. A. oncogene; tumor-suppressor gene B. tumor-suppressor gene; oncogene C. oncogene; oncogene D. tumor-suppressor gene; tumor-suppressor gene E. malignant, hemizygous

B. TSG; Oncogene

Lecture 9Population Genetics II

Outline What keeps unfavorable genes in gene

pool: When Bad alleles can be good as well Slow selection against recessive alleles

in diploids Mutation-selection equilibrium Heterozygote superiority

A Question…29. Your uncle had a heart attack at an early age. Since early onset heart disease is a multifactorial trait, one important piece of information you need to know to assess your risk is to know the fraction of alleles you share with your uncle. The fraction of your alleles that you share with your uncle is...

An uncle shares 50% of the same DNA as his brother (the father). 50% of the genes from the father is transferred to you. Multiplication rule: the event: “What is the chance that the 50% similarity between your Uncle and your father is passed to you?

0.50 x 0.50 = .25 = 25%

Reasoning

17 &18.THIS PROBLEM IS WORTH 2 POINTS – FILL IN BOTH NUMBERS ON YOUR SCANTRON(Note – this is a tough problem meant to challenge the best-prepared students)

A polymorphic region in the human genome can be detected by a marker called 9Q17. As such, the polymorphic region is named the "9Q17 polymorphism", or 9Q17P for short. There are 4 common alleles of 9Q17P, named after the size of the bands seen on a Southern blot when EcoRI-digested genomic DNA is probed with radioactively-labeled 9Q17 marker DNA. 9Q17P is very tightly linked to the Huntington gene, the gene associated with the autosomal dominant, delayed-age of onset disease Huntington's disease. Three generations of a family, some members of which have Huntington's disease, are typed for 9Q17P. Alice has the 2 kb and 4 kb alleles, and her husband Ben has the 3 kb and 5 kb alleles. Alice and Ben have a son named Chuck, who has the 2 kb and 3 kb alleles. Chuck is married to Doris, who has the 4 kb and 5 kb alleles, and they have a 10-year-old son named Ed. Ben and Chuck have the symptoms of Huntington's disease.

(i). If Ed has the 3 kb and 4 kb alleles, is it likely that Ed will get HD?(ii). If Ed has the 2 kb and 5 kb alleles, is it likely Ed will get HD?

A. (i) = yes; (ii) = NOB. (i) = yes; (ii) = yesC. (i) = NO; (ii) = yesD. (i) = NO; (ii) = NO

E. (i) = NO; (ii) = maybe

A. Yes, No

17 &18.THIS PROBLEM IS WORTH 2 POINTS – FILL IN BOTH NUMBERS ON YOUR SCANTRON(Note – this is a tough problem meant to challenge the best-prepared students)

A polymorphic region in the human genome can be detected by a marker called 9Q17. As such, the polymorphic region is named the "9Q17 polymorphism", or 9Q17P for short. There are 4 common alleles of 9Q17P, named after the size of the bands seen on a Southern blot when EcoRI-digested genomic DNA is probed with radioactively-labeled 9Q17 marker DNA. 9Q17P is very tightly linked to the Huntington gene, the gene associated with the autosomal dominant, delayed-age of onset disease Huntington's disease. Three generations of a family, some members of which have Huntington's disease, are typed for 9Q17P. Alice has the 2 kb and 4 kb alleles, and her husband Ben has the 3 kb and 5 kb alleles. Alice and Ben have a son named Chuck, who has the 2 kb and 3 kb alleles. Chuck is married to Doris, who has the 4 kb and 5 kb alleles, and they have a 10-year-old son named Ed. Ben and Chuck have the symptoms of Huntington's disease.

(i). If Ed has the 3 kb and 4 kb alleles, is it likely that Ed will get HD?(ii). If Ed has the 2 kb and 5 kb alleles, is it likely Ed will get HD?

A. (i) = yes; (ii) = NOB. (i) = yes; (ii) = yesC. (i) = NO; (ii) = yesD. (i) = NO; (ii) = NO

E. (i) = NO; (ii) = maybe

Reasoning

21-24. In a certain population of students at Hardy-Weinberg equilibrium, one of every twenty-five individuals is affected by an autosomal recessive condition called ʻsleepyheadnessʼ, which causes them to fall asleep during important lectures.

21. What is the frequency of heterozygous carriers of ʻsleepyheadnessʼ in this population?

22. What is the allele frequency p of the dominant SH allele? What is the allele frequency q of the recessive SH allele? Your answer should take the form “p = __, q = __”.

23. Assume the population consists of 100 students. One day, the Professor kicks all four sleeping students out of the class, so now the population consists of 96 students. What are the allele frequencies of the dominant SH allele (p) and the recessive sh allele (q) in the new population? Your answer should take the form “p = __, q = __”.

24. Is the new population is Hardy-Weinberg equilibrium (yes or no)?

The Question

21-24. In a certain population of students at Hardy-Weinberg equilibrium, one of every twenty-five individuals is affected by an autosomal recessive condition called ʻsleepyheadnessʼ, which causes them to fall asleep during important lectures.

21. What is the frequency of heterozygous carriers of ʻsleepyheadnessʼ in this population? 0.32 (or 8/25)

22. What is the allele frequency p of the dominant SH allele? What is the allele frequency q of the recessive sh allele? Your answer should take the form “p = __, q = __”. p = 0.8 (or 4/5), q = 0.2 (or 1/5)

23. Assume the population consists of 100 students. One day, the Professor kicks all four sleeping students out of the class, so now the population consists of 96 students. What are the allele frequencies of the dominant SH allele (p) and the recessive sh allele (q) in the new population? Your answer should take the form “p = __, q = __”. p = 0.833 (or 5/6), q = 0.167 (or 1/6).

24. Is the new population is Hardy-Weinberg equilibrium (yes or no)? No

Answers