Home
Up

 

 

Happy Holidays!

Final Exam Review Sheet 2007

Exam date Wednesday December 19th 2007 2:00-3:50

Exam format:

20 fill-in the black  questions  (5 from each section --exam 1, 2, 3 and new material) 1 point each
16 short answer questions (4 from each section) 3 points each
5 problems/longer questions (1 each from exam 1, 2 and 3 material, 2 on new stuff) 6 points each
10 multiple choice (genetic disease questions)1 point each

Total possible points = 108

The exam will be cumulative.  Approximately 15-25% of the exam will be on new material that we have covered since the last exam.   The remaining 75-85% will be on material from previous exams. 

I have made a review outline for the new material and have edited the review sheets for the old material to exclude some topics.  If you see something that was on an old review sheet that is now missing, it means you are not responsible for that particular material.  Do try to keep in mind what are some of the bigger concepts that we have covered during the course.   Use your exams to get an idea of the type of questions I may ask.  They may be very similar or even the same.  Remember what I said way back in September about wanting students to know the answers to a few basic questions (I posed these questions on the first day, see chapter 1) by the time the course was over.  I could very well ask those questions.  One thing that I know I will not ask you to do is solve a three point testcross.  I may ask why or how one is done, but won't ask you to solve the problem.  The one type of problem I am sure to ask is a dihybrid cross.  Do remember that 15-25% of the questions will be on the relatively small amount of material that we have covered since exam 3, and that these questions will be more challenging.  Hope this is helpful.  As usual, feel free to contact me with any questions you may have.  

The old stuff:

Exam 1

Chapter 2 Mitosis and Meiosis

bulletWhat are chromosomes?  What is the difference between chromatin and chromosomes? 
bulletBe able to distinguish between chromosome, sister chromatids and nonsister chromatids. 
bulletWhat is a karyotype?
bulletHumans have 46 chromsomes, 23 pairs.  22 pairs of autosomes and 1 pair of sex chromsomes. 
bulletBe sure to know what a homologous pair of chromosomes is.  
bulletReview the Cell cycle.  When does replication of DNA occur?
bulletReview the stages of mitosis
bulletReview the stages of meiosis.  (don't worry about sub stages of prophase 1)
bulletWhen does crossing over occur?
bulletWhat is the purpose of  Meiosis? 
bulletIn what cells does meiosis occur? 

Chapter 3 Mendelian Genetics

What are gametes
Understand the terms: monohybrid cross, dihybrid cross, P generation, F1 generation and F2 generation.
Know the terms homozygous or heterozygous.
What are alleles?  How many alleles does an individual have?  How many alleles does a gamete have? How many alleles can there be in a population of organisms?
Understand the difference between dominant and recessive traits.
Know the terms genotype and phenotype.  How do they differ?  

Monohybrid cross This is a  cross involving a single pair of contrasting traits (monohybrid)

P (parental) PP X pp (crossing truebreeding parents)
F1 Genotype: Pp
Phenotype: purple
F2 ¼ PP + ½ Pp + ¼ pp Genotypes
¾ purple and ¼ white Phenotypes
A standard monohybrid cross results in a 1:2:1 genotypic ratio and a 3:1 phenotypic ratio

Dihybrid cross This is a  cross involving two pairs of contrasting traits (dihybrid)

PP = purple BB = short
Pp = purple Bb = short
pp = white bb = tall

P (parental) PPBB X ppbb OR PPbb X ppBB (crossing truebreeding parents -- note that either parent shown will result in the same F1)
F1 Genotype: PpBb
Phenotype: purple, short 
F2  1/16PPBB,  2/16 PPBb, 1/16 PPbb, 2/16 PpBB, 4/16 PpBb, 2/16 Ppbb, 1/16ppBB, 2/16ppBb, 1/16 ppbb Genotypes
9/16 purple, short 3/16 purple tall 3/16 white short and 1/16 white tall Phenotypes

A dihybrid cross results in a 1:2:1:2:4:2:1:2:1 genotypic ratio and a 9:3:3:1 phenotypic ratio

bulletReview the branch line diagram  method of determining F2 phenotypic and genotypic ratios. 

**Keep in mind that the above examples of monohybrid and dihybrid crossed are by definition starting with a standard cross of truebreeding parents, and the resulting F2 ratios will always be the same as shown above no matter what trait or pair of traits you are considering.  However it is possible to look at individual matings of particular organisms which will not be truebreeding, and the resulting ratios will not be the same as above.  The nature of the cross would have to be specified in some way, and the resulting phenotypes and genotypes would be calculated in a similar manner.** 

bulletReview Mendel's 4 postulates and how they correlate to the presence of genes located on homologous chromosomes and their behavior (Figure 3-11).  Especially the concept of segregation and independent assortment. 
bulletBe familiar with what a testcross is, how it is done, and what the results tell you.  Always cross to a homozygous recessive. 

Chapter 4  Extensions of Mendelian Genetics

This whole chapter deals with situations which alter the standard monohybrid and dihybrid phenotype ratios (3:1, 9:3:3:1).   When gene expression does not adhere to a simple dominant/recessive mode, or when more than one pair of genes influences the expression of a single character, the classic ratios are modified.  Nevertheless, the fundamental principles of segregation and independent assortment still hold true in these situations.  

bulletIncomplete dominance  -- the heterozygote has an intermediate phenotype.  phenotype ratio = genotype ratio in a standard monohybrid or dihybrid cross. 
Example : four o'clocks R1R1 = red, R1R2 = pink, R2R2 = white
R1R2 x R1R2 ----> 1/4 R1R1, 2/4 R1R2, 1/4 R2R2 = 1/4 red, 2/4 pink, 1/4 white
bulletCodominance --the heterozygote shows phenotype of both alleles.  phenotype ratio = genotype ratio in a standard monohybrid or dihybrid cross.
Example: MN blood groups.
bulletMultiple alleles -- For any given gene the number of alleles within members of a population is not restricted to two.   Multiple alleles = 3 or more alleles for a given gene.  Any individual diploid organism has at most two alleles of any gene. 
Example: ABO Blood groups.  Three alleles: IA, IB, and IO.  IA, IB are dominant to IO.  IA is codominant to IB.
Table 4.1 shows the various ratios of offspring that can result from different crosses. 
Lethal alleles -- Many gene products are essential to an organism's survival.  When a mutant allele is present in may be lethal for the individual. 
bulletA recessive lethal allele is lethal only if two copies of the lethal allele are present (homozygous).  Sometimes the heterozygote has a unique phenotype (see Figure 4-4).   The other normal allele is sufficient to compensate for the mutant one.   Example: AY allele in mice, p. 85
bulletA dominant lethal allele is lethal even when only one copy of the allele is present.  So organisms homozygous or heterozygous for the lethal allele will die. Example: Huntington's disease
bulletWith lethal alleles, the phenotype ratio is altered   such that in a standard monohybrid cross, the F2 phenotype ratio would be 2:1 instead of 3:1.  The dihybrid ratio would also be affected. 
Be able to distinguish between the different types of inheritance, and be able to apply these types of inheriance to new examples.   
Combinations of two gene pairs involving two modes of inheritance modify the 9:3:3:1 ratio -- see Figure 4-5.  Be able to do this type of cross. 
bulletEpistasis -- an example of gene interaction; occurs when the expression of one gene pair masks or modifies the expression of another gene pair.  The involved genes control the expression of the same general phenotypic trait.  Has the effect of combining one or more of the four phenotypic categories of a standard dihybrid cross in various ways. 
bulletRemember, in epistasis, even though only one characteristic is being followed (flower color for example), the phenotype ratio is expressed in 16 parts. This is a way to see that epistasis is occurring
bulletBe familar with the inheritance of sex-linked traits

Exam 2

Chapter 5:  Linkage, Crossing over, and Mapping

bulletWhat is linkage? 
bulletWhat is the difference between complete linkage and linkage with recombination?
bulletBe able to distinguish between independent assortment, linkage, and linkage with recombination. 
bulletCrossover occurs between non-sister chromatids of a homologous pair of chromosomes.  When does crossover occur? 
bulletAs distance between genes on the same chromosome increase, the degree of crossing over increases (and the number of recombinant gametes increases)
bulletThree point testcross:
bulletmap distance= % of all detectable exchanges occurring between 2 genes = %SCO + %DCO

Chapter 7: Sex Determination and Sex Chromosomes

bulletKlinefelter and Turner syndrome
bulletBe able to recognize a karyotype of these syndromes and correctly indicate the chromosome arrangement (number of chromosomes, sex chromosomes, deviation from normal)
bulletBoth syndromes are the result of nondisjunction in meiosis
bulletWhat did these syndromes tell scientists about sex determination in humans?
bulletWhat is the SRY?  What would happen if a female had a copy of this gene?   If a male lost this gene?
bulletDosage compensation by X inactivation
bulletWhat are Barr bodies? Where are they located?  How many are present in different individuals?
bulletMosaicism -- the Lyon hypothesis states that females are mosaics for all heterozygous X linked alleles. 

Chapter 8: Variation in Chromosome Number and Arrangement

bulletIs monosomy tolerated in plants?  Animals?
bulletDowns syndrome = trisomy of chromosome 21, caused by nondisjunction
bulletEdwards syndrome = trisomy 18
bulletPatau syndrome = trisomy 13
bulletPolyploidy (triploid, tetraploid, etc)
bulletHow is polyploidy tolerated in plants? Animals?
bulletBe able to distinguish between autoploidy and alloploidy
bulletHow are tetraploids made artificially?  Naturally?
bulletBe able to recognize a karyotype and correctly indicate the chromosome arrangement (number of chromosomes, sex chromosomes, deviation from normal) for the above disorders/conditions
bulletreview the types of chromosomal rearrangements: deletion, inversion, duplication, nonreciprocal translocation, reciprocal translocation. 
bulletWhat is the difference between a paracentric inversion and a pericentric inversion?
bulletReview figures 8-21 and 8-22
bulletReview familial down's syndrome

Chapter 9: Extranuclear Inheritance

bulletreview sections 9.2, 9.3 and 9.5, and class notes
bullet What is extranuclear inheritance?
bulletWhat is maternal effect?  How is this different than maternal  inheritance of mitochondrial DNA? How is it diff. than X-linkage?

Chapter 10: Chemical Structure of DNA

bulletDNA = the genetic material.  Several important characteristics:
bulletreplication
bulletstorage of information
bulletexpression of information
bulletvariation by mutation
bulletKnow the components of a nucleotide
bulletThe four nitrogenous bases in DNA are G, A, T , and C.   A and G are the purine, C and T are the pyrimidines
bulletWhat type of bond links nucleotides together?
bulletReview and understand the features of the Watson and Crick Model of DNA:
bulletright handed double helix
bullet2 strands are antiparallel (5' to 3' and 3' to 5')
bulletthe bases of both chains are perpendicular to the axis of the molecule
bulletnitrogenous bases paired by hydrogen bonding (A-T; G-C)
bulletComplete turn of the helix = 34Ao
bullet20 Ao in diameter (= pyrimmidine + purine distance)
bulletmajor and minor grooves
bulletBe able to appropriately label OR DRAW diagram of DNA with the above features
bulletHow is molecule held together by "weak" hydrogen bonds?
bulletHow many hydrogen bonds between G and C?  A and T?

Chapter 11: Replication of DNA

bulletReplication always occurs 5' to 3' (see Figure 11.8)
bulletDNA Polymerase requires 3' end to initiate synthesis, so requires RNA primer (see Figure 11-11, 11-12
bulletLeading strand synthesis is continuous, 5' to 3'
bulletLagging strand synthesis is discontinuous, 5' to 3'.  Many points of initiation necessary because trying to replicate one way, and 5' to 3' is in the opposite direction.
bulletOkazaki fragments --the short pieces of DNA synthesized from primers on the lagging strand. 
bulletDNA Polymerase = dimer, synthesizes both strands simultaneously
bulletFIGURE 11-13.  Know what all of the proteins illustrated are doing for the replication process and be able to identify them in a figure:
bullethelicase
bulletsingle stranded binding protein
bulletgyrase
bulletprimase
bulletDNA Polymerase (In E. Coli, Pol III is main replication protein; Pol I is responsible for filling in gaps left after RNA primers are removed and for "proofreading" capabilities, Pol II repairs damage to the DNA from UV and mutagenic chemicals)
bulletNot included in figure, but important, is DNA ligase, which joins the Okazaki fragments after RNA primers are removed, and gaps are filled in by Pol I
bulletFigure 11-16, 11-17.  How does Telomerase deal with the problem of gaps at the ends of DNA?

 

Exam 3

Chapter 12: DNA organization in Chromosomes

bulletReview chromatin structure info on pages 292-295 and nucleosome structure
bulletWhat proteins are associated with DNA?  What is the nature if the interaction?
bulletreview Figure 12-9
bulletwhat is meant by chromosome remodeling?
bulletWhat does modification of histones by acetylation, methylation or phosphorylation do?
 

Chapter 13: The Genetic Code and Transcription
Chapter 17: Regulation of Gene Expression in Eukaryotes

bulletreview the characteristics of the genetic code (p. 307) 
bulletreview the in-class handout which summarizes studies done to decipher the genetic code (homopolymer, mixed copolymer, repeating copolymers, triplet binding assay)
bulletknow how to determine amino acid sequence for a given mRNA sequence.
bulletProkaryotic Transcription See Figure 13.9
bulletE. coli. RNA polymerase (a, b, b', and s subunits)
bulletWhich subunit(s) contains the active site and is responsible for the catalytic activity of the enzyme?
bulletWhich subunit is a regulatory subunit involved in recognition of the promoter?
bulletProkaryotic promoter region includes a Pribnow box (-10) and sequences at -35
bulletProkaryotic transcripts are polycistronic
bullettranscription and translation simultaneous
bulletEukaryotic Transcription
bulletEukaryotic transcripts are monocistronic
bullet3 different RNA polymerases:
bulletRNA Pol I -- rRNA
bulletRNA Pol II -- mRNA
bulletRNA Pol III -- tRNA, 5S RNA
bulletTranscription and translation are NOT simultaneous; occur in different parts of the cell
bulletMore complex regulation
bulletProcessing: 5' cap, poly A tail, splicing.  Where does each occur?   When?  Why?
bulletRNA Pol II (2 large subunits, 10-15 smaller)
bulletWhat is the function of CAAT box, TATA box, enhancer?
bulletTFIID, TBP
bulletsee p. 412, which summarizes differences between euk and prok transcription
bulletFigure 17-5, eukaryotic promoter structure
bulletFigure 17-11 Assembly of transcription factors

Chapter 14: Translation

bulletFigure 14-1  Ribosomes consist of small and large subunits, made up of both proteins and rRNA
bulletFigure 14-3  tRNA structure (where is anticodon loop?/where does amino acid bind?)
bulletCharging of tRNAs.  Figure 14-5
bulletaminoacyl tRNA synthetase (what does this enzyme do?)
bulletTranslation: Figures 14-6, 14-7, 14-8
bulletyou should be familiar with the general aspects of translation. 
bulletWhat is the Shine Delgarno sequence and what is its significance?
bulletWhat is the function of peptidyl transferase?

 

The new stuff:

Don't forget to review the diseases for the final exam. 

Don't skim new stuff--a large part of the exam will be on this material. There will one problem from chapter 19 and one from chapter 22. 

Chapter 19: Recombinant DNA Technology

bulletWhat is recombinant DNA?
bulletBasic procedure in recombinant DNA technology involve a series of steps -- see your notes or p. 458
bulletRestriction enzymes
bulletHow are restrictions enzymes used in recombinant DNA technology?
bulletrestriction enzymes recognize palindrome sequences along DNA
bulletWhat is a sticky end? How do they join together?
bulletWhat is DNA ligase used for?
bulletBe familiar with the plasmid vector pUC18
bullet~500 copies per cell
bulletorigin of replication
bulletampicillin resistance gene
bulletlarge number of restriction sites clustered in one region called polylinker
bulletlac Z gene encodes b galactosidase
bullet+ X-gal = blue colonies
bulletlac Z interrupted = white colonies
bulletHow can this vector be used to distinguish between bacterial cells that have non-recombinant plasmid and those that have recombinant plasmid or no plasmid?
bulletFigure 19-13 overview of steps involved in cloning DNA with a plasmid vector
bulletPolymerase chain reaction.  Figure 19-17   Know the basic procedure and its applications.  What temperatures are used and why?
bulletSouthern blotting.  Figure 19-24.  Know the basic procedure and its applications
bulletSequencing: FIGURE 19-26, 19-27 Know the procedure, how to "read" a gel and its applications. 

 

Chapter 22: Applications and Ethics of Genetic Technology

bulletHow is amniocentesis done, and what kind of analysis can be done with samples obtained?
bulletHow can an individual with thalassemia or sickle cell anemia be identified by recombinant DNA techniques?  What are the steps involved in this procedure? (see Figure 22-9)
bulletWhat are ASOs?  How can they be used for genetic testing?  Be able to analyze results of these types of experiments (see figures 22-10. 22-11 )
bullet What are microarrays?  What can they be used for?
bulletWhat are RFLPs?  How can they be used for gene mapping?  (linkage analysis)
bulletWhat is FISH?  How is it done?

 

There will none of Chapter 18 on the exam. 

Chapter 18: Genetics of Cancer--ONLY WHAT WE GET TO IN CLASS

Much of the material we spoke about on cancer was not from the book, but from the two cancer lab handout introductions, and my own other sources, so definitely refer to your notes. 

bulletNormal cells:
bullethave defined lifespan in culture.  They will grow for a while, undergo "crisis" and then die
bulletcontact inhibited -- cell growth ceases under conditions of decreased nutrients, injury, or cell division, or if space is physically limiting (crowding on plate)
bulletcells adhere to tissue culture plate and to one another
bulletCancer cells have distinguishing characteristics:
bulletimmortal
bulletloss of contact inhibition - cells overgrow, pile on top of one another
bulletanchorage independent -- cancer cells are less adherent to each other and the tissue culture plate
bulletmorphological changes
bulletchromosomal aberrations
bulletAll types of cancer have a common distinguishing feature: LOSS OF GROWTH CONTROL
bulletProto-oncogenes
bulletnormal genes, encode normal proteins involved in accelerating growth of the cell
bulletif mutated, become oncogenes
bulletTumor suppressors
bulletnormal function--restrict cellular proliferation
bulletif mutated or inactivated, cellular proliferation without regulation
bulletReview information on p53 -- what type of protein is it?  What is its normal function? what does it interact with?  How does it restrict cell growth?
bulletReview information on pRB --what type of protein is it?  What is its normal function? what does it interact with?  How does it restrict cell growth?
bulletSomatic mutations vs. Germline mutations
bulletmutations include nucleotide changes, chromosomal rearrangements, chromosome loss
bulletCancer cells -- to key events cause change to cancerous cell
bulletuncontrolled growth
bulletability to metastasize
bulletCell cycle -- G1/S and G2/M transitions regulated (see text, ch. 23, pp. 635-638) by two classes of proteins:
bulletprotein kinases (ex. CDK1)
bulletcyclins (ex. cyclin B)
bulletG2/M transition regulated by cyclinB/CDK1 complex (Figure 23.3)
bulletRetinoblastoma
bulletfamilial vs. sporadic (See Figure 23.4)
bulletsymptoms
bulletTumor suppressors:pRB 
bulletHow is it regulated? (See figure 23.5)  When is it active/inactive?   How does this occur?
bulletTumor suppressor: p53
bulletmutations in p53 gene are most common
bulletnormally low levels increase in response to DNA damage
bullet2 responses: arrest cell cycle (How?  Know the p21 story) and initiate apoptosis