This outline is designed to provide you with a general summary of the topics that will be discussed in class. It is not designed to substitute for your own in-class notes, or for the detailed discussions of these topics in the assigned readings. Interspersed within this outline are references to readings in the Hartwell text, where this subject material can be found.

  I. Course Description

A. Text

B. Grading

C. Course Schedule

 II.  What is a Gene? A. It must be transmitted between generations and each individual must have a physical copy of this material.

B. It must provide information to its carriers with respect to biological structure and function.

 III.  Historical Perspectives:PreMendelian concepts of inheritance A. Spontaneous generation (till 1861!)

B. Concepts emphasizing continuity between generations

1. Early Greeks (~500 B.C.) a. Pangenesis: somatic tissue contributes to heritable information

b. Heritable contributions of males and females differ.

 2.  Preformationism (1600's)

3. Studies of early embryogenesis and the concept of epigenesis (~1750-1880)

4. Charles Darwin's concepts of inheritance (1859)

a. gemmule theory

b. the observation of continuous variation across the biological spectrum and an emphasis on blending of traits in offspring.

 C.   Gregor Mendel: a contemporary of Charles Darwin (1822-1884) 1. Entered monastery and was educated at Vienna, Brunn.

2. Started experiments in peas in 1854

3. Published experiments in 1866 in a journal distributed to over 100 libraries in Europe; corresponded with major biologists of his day.

4. Assumed administrative responsibilities as abbot (1868); died 1884 in scientific obscurity.

 IV.   Principles of Transmission Genetics: Mendel's experiments in Peas. A. His experimental organism, the common garden pea, very useful for genetic analysis. 1. Chose characteristics that were unambiguous.

2. Relatively quick reproductive cycle; large numbers of progeny.

3. Organism could be either self-pollinated or outcrossed (hybridized).

4. Took a quantitative approach to data collection.

 B. The monohybrid cross: principles of unit inheritance and segregation. 1. If two plants with differing traits were crossed (P generation) the next generation (F1) always gave rise to plants displaying only one parental character. If the F1 plants are now allowed to self-fertilize, the other parental character reappears in the next generation (F2), representing 25% of the offspring.

2. Generated a hypothesis consistent with his results.

a. Traits could be represented as discrete, particulate entities or factors (represented by symbol).

b. Factors appear to be paired, although some traits are not observed in hybrids even though the factor associated with their expression may be present in the hybrid (concept of dominant and recessive traits).

c. Factors will randomly segregate during production of gametes; one factor is received from each parent; restoration of paired factors upon fertilization.

 C. The dihybrid cross: prinicple of independent assortment 1. Expanded his observations, watching the segregation of two traits simultaneously in the same cross.

2. Results confirmed earlier observations.

a. Two traits observed together will partition independently of one another.

b. observed F2 ratios (9:3:3:1) were two independent 3:1 ratios.

 D. The testcross (backcross): Supported his basic hypothesis. 1. Backcross of F1 monohybrid to single recessive parent produced expected 1:1 ratios.

2. Backcross of F1 dihybrid to double recessive parent produced expected 1:1:1:1 ratios.

 E. Basic Terminology 1. Merkmal (Mendel's German term) = factor = gene.

2. alleles: variant genes for any given trait; capital letter (A) represents dominant; small letter (a) represents recessive.

3. homozygote: identical pair at a given locus; AA or aa. (locus = location on a chromosome)

4. heterozygote: a nonidentical pair at a given locus; Aa.

5. genotype: the genetic composition or complement (for any given gene pair or locus)

6. phenotype: the physical appearance observed based on expression of the genotype.

 V. 19th Century Advances in Cell Biology: Indentifying the Physical Basis for Mendel's Laws. A. Nucleus and discovery of cell structure. (Brown, 1833)

B. Fertilization and pronuclear fusion. (Hertwig et al. ~1875)

C. Mitosis (Fleming et al. ~1880)

D. Germplasm theory (Weismann, ~1880)

E. Meiosis (Van Beneden, Boveri et al. 1880s-early 1900s)

F. Rediscovery of Mendel's Laws (deVries, Correns, von Tschermak; 1900)

 VI. The Sutton-Boveri Hypothesis (1903) A. Both chromosomes and Mendelian factors are found in pairs; of both maternal and paternal origin.

B. Chromosomes of each pair separate during meiosis, as do Mendelian factors.

C. On fertilization, the "doubleness" or the chromosome complement is restored, as are Mendelian factors.

 VII. Segregation, Independent Assortment and Probability. A. Probability = number of anticipated events/total number of possibilities;

e.g. the probability of pulling an ace out of a straight deck of cards: P(ace) = 4/52 or 1/13.

B. Product Rule: For independent events, the probability of several events occurring is the product of their individual probabilities; P(A and B) = P(A) * P(B).

e.g. P (ace of spades) = P (ace) * P (spade) = 1/13 * 1/4 = 1/52

C. Sum Rule: For mutually exclusive events, the probability represents the sum of the individual probabilities. P (A or B) = P(A) + P(B).

e.g. P (ace or king) = 1/13 + 1/13 = 2/13

D. For Mendelian patterns of inheritance, the events that occur during the process of meiosis and fertilization (segregation, independent assortment) can be expressed in probabilistic terms.

1. In a monohybrid cross, the fraction of progeny with a given phenotype follows the sum rule: P(AA or Aa) = 1/4 + 1/2 = 3/4; P(A_ or aa) = 3/4 + 1/4.

2. Assuming independent assortment, a dihybrid cross would be expected to yield a 9:3:3:1 phenotypic ratio- applying the product rule to two independent events.

P( A_ and B_) = P(A_) * P(B_) = 9/16 A_B_

P (A_ and bb) = P(A_) * P(bb) = 3/16 A_bb

P (aa and B_) = P(aa) * P(B_) = 3/16 aaB_

P (aa and bb) = P(aa) * P(bb) = 1/16 aabb

 3. In the cross AaBbCc x AaBbcc, what is the probability of producing progeny with the same phenotype as the first parent? (ans. 9/32) I. Documenting Patterns of Transmission for Single Gene Traits in Humans. A. Importance of Familiy Histories for Human Genetic Analysis

B. Pedigree Symbols and Guidelines

C. Pedigree Patterns for Autosomal Dominant Inheritance.

1. Trait appears in every generation (exceptions; new mutation or incomplete dominance).

2. Trait is passed from affected person to 1/2 of progeny, regardless of sex.

3. Unaffected members will not pass on the trait.

 D. Pedigree Patterns for Autosomal Recessive Inheritance. 1.Trait appears to skip generations, often observed in sibs but not parents.

2. On the average, the trait will show up in 25% of the sibs of an affected proband, regarless of sex.

3. The parents may be consanguineous.

 E. Probability and Risk Assessment.

1. From pedigree, trait appears inherited as an autosomal recessive.

2. From phenotype of II-3, parents are obligate heterozygotes.

3. If parents are obligate heterozygotes, III-2 and III-3 have a 2/3 probability of being heterozygotes.

4. If parents of IV-1 and IV-2 are heterozygotes, IV-1 and IV-2 each stand a 1/2 probability of inheriting disease gene allele from their parents.

5. If they are heterozygotes, they have a 1/4 probability of having an affected child.

6. The risk to IV-1 and IV-2 of having an affected child is therefore:

2/3*2/3*1/2*1/2*1/4 = 1/36 II. Evidence for the Chromosomal Theory of Inheritance A. Sex Chromosome Dimorphism: homo- and heterogametic sexes.

B. The white locus in Drosphila: Thomas Hunt Morgan (1910).

1. Characteristic segregation pattern: white female x normal male shows criss-cross inheritance.

2. Calvin Bridges: Correlation of exceptional progeny with nondisjunction during meiosis.

 C. Sex Determination: Nature of the "Primary" Switch; mechanisms vary. (See Hartwell, pp. 605-610 for detailed discussion of molecular mechanisms in Drosophila.) 1. Drosophila: X:A ratio.

2. Mammals: TDF/SRY gene on Y chromosome. (See Hartwell, pp. 653)

 III. Patterns of Sex-Linked Expression in Humans. A. X-linked Recessive Inheritance. 1. Incidence much higher in males than in females.

2. Affected male will give rise to daughters who are carriers. On average, 1/2 of a carrier female's sons will be affected; half of her daughters will also be carriers..

3. Never father to son transmission.

 B. X-linked Dominant Inheritance. 1. Affected males will have no affected sons and no normal daughters.

2. Transmission by females is not distinguishable from an autosomal dominant.

3. Since most females for autosomal dominants are heterozygous and most males will be hemizygous, the trait is often more severe in males.

 C. Holandric Inheritance. 1. Relatively rare in human populations. Y chromosome, relative to X, is gene poor (constitutive heterochromatin).

2. TDF/SRY gene.

 D. Sex-limited and Sex-Influenced Traits. 1. Sex-limited; e.g. precocious puberty.

2. Sex-influenced; e.g. pattern baldness.

 VI. Probability Revisited. (N.B. See Chapter 5 of Thompson; this is not formally discussed in Hartwell) A. Use of the Binomial Expansion. 1. In a three child family, what is the probability of two boys and 1 girl?

P(boy) = p= 1/2; P(girl) = q= 1/2; n = number of events (births) = 3. All potential possiblities are represented by the binomial expansion, with the specific case in question emphasized:

(p + q)3 = 1 or p3 +3p2q + 3pq2 + q3 = 1; 3p2q = 3(1/2)(1/2)(1/2) = 3/8

For any given term in the expansion (e.g. in a three child family, the odds of two boys and a girl) is also represented by the equation:

where n = number of children = 3, P(boy) = p = 1/2; P(girl) = q = 1/2; s = number of boys= 2; t = number of girls = 1. Note that p + q = 1; s + t = n.

n!/s!t! = (3*2*1)/(2*1)(1) = 3; and

psqt = (1/2) (1/2) (1/2) = 1/8;

therefore;

3* 1/8 = 3/8
      1. The binomial expansion is a special case of the more general multinomial distribution, where the model can be expanded to include more than two alternatives:
B. Testing Experimental Data for Conformity to an Experimental Hypothesis: the Chi Square Test. (See Thompson, ch. 5; discussed in Hartwell with regard to linkage analysis, pp. 117-120) 1. Using Mendel's data set as an example of observed vs. experimental data.

2. Concept of Degrees of Freedom; for genetic tests like these, the degree of freedom is one less than the number of classes.

3. The chi square table and assessment of the probability the differences between the observed and expected values are due to chance.

4. The 0.05 criterion for level of significance.

 I. Dominance Relationships A. Complete Dominance 1. Mendel's pea experiments.

2. A. Garrod: (1902-1909)

a. Demonstrated that human disorders (e.g. alcaptonuria) were transmitted in human pedigrees according to Mendelian laws of inheritance (e.g. alcaptonuria shows a autosomal recessive pattern of inheritance.)

b. Postulated that the reason for the disease was a metabolic defect that was genetically inherited- that genes are ultimately responsible for making enzymes and the genetic defect in alcaptonuria was related to the absence of functional enzyme ("inborn error of metabolism"). Like Mendel, this postulate was ahead of its time, and not rigorously tested until the 1030s-40s.

 B. Incomplete Dominance 1. Transmission genetics could demonstrate that some heterozygotes displayed a phenotype different from either parent, although pattern of segregation followed Mendel's laws. e.g. flower color in 4 o'clock plants. Pure breeding red and white individuals produce pink F1s. Pink F1 x Pink F1 produce 1 red: 2 pink: 1 white offspring; red and white breed true; pink phenotype due to heterozygote state.

2. If phenotype, in certain instances, is reflected by level of enzymatic activity, then some phenotypes may show a threshold effect (e.g. alcaptonuria) where any enzyme activity above a certain level confers a wild type phenotype, and some may show a saturation effect, where incremental amounts of the enzyme may lead a gradient of phenotypes (e.g. sugar deposition in some plant seeds).

 C. Multiple Alleles and codominance 1. Although any individual in a population may contain only two allelic variants (Aa), within the population there may be may different variants (A, a, A1, A2, A3 etc.)

2. Allelic variation can be detected through a number of means:

a. electrophoresis (protein level)

b. nucleic acid analysis (DNA level)

 3. examples: beta globin, alpha1-anti-trypsin.

4. codominance: the heterozygote exhibits a phenotype based on the expression of both alleles. e.g. ABO blood group locus.

5. Molecular discussion of beta globin variants and ABO blood group locus: (Hartwell, 277-78; 308-310; 46-7; 56-7)

 II. Deviation from Mendelian Ratios due to Single Gene, Polygenic, or Environmental Interactions. A. lethal genes. e.g. yellow allele in mice leads to 2:1 ratio if two yellow animals are crossed.

B. genetic heterogeneity. e.g. albinism can be caused by a defect at more than one genetic locus.

C. phenocopy. e.g. kwashiorkhor- environmental factors mimic genetic disorder

D. Variable Expressivity and Penetrance.

1. Variable Expression: single gene effects can be variable in severity of expression and pleiotropic (have several different phenotypic effects). e.g. anonychia and neurofibromatosis.

2. Penetrance: the frequency with which a gene manefests itself in the appropriate genotype: some genes are incompletely penetrant (do not express the appropriate phenotype due to gene or environmental modifiers).

 III. Interactions of Genes with other Genes (epistatic interactions). A. 4 distinct phenotypes (comb shape in chickens).

B. complementary gene action (anthocyanin pigment formation in pea flowers)

C. recessive epistasis (albinism in mice)

D. recessive suppression of a recessive phenotype in Drosophila; purple eye color

E. duplicate genes (fruit shape in shepard's purse)

 I. Linkage and Recombination A. Two genes close together on the same chromosome will not assort independently; they will be transmitted to the same gamete >50% of the time since they are attached to a common centromere.

B. In heterozygotes, linked genes may be arrayed in two possible configurations (phasing relationships).

1. cis (or coupling): both dominant loci on same chromosome and both recessive loci on the same chromosome: (AB/ab).

2. trans (or repulsion): A dominant gene at one locus associated on the same chromosome with a recessive gene at a second locus: (Ab/aB).

 C. Linkage can be followed in a testcross. II. Use of testcrosses to demonstrate linkage (e.g. T.H. Morgan). A. e.g. The white and minature genes on the X chromosome of Drosophila. w m/ + +   x   w m/Y  or The white and yellow genes on the X chromosome of Drosophila. y +/+ w  x   y w/Y B. Parental types recovered in greater frequency than on basis of random assortment. 1. Results deviated from expected 1:1:1:1 ratio; non-recombinant (parental) classes recovered in greater frequency than the recombinant (non-parental) classes. Hypothesized that the non-parental gene combinations arose from physical exchange of chromosomal material during meiosis in the heterozygote.

2. The frequency of recombination is a function of how close two loci are linked.

3. The crossover frequency can be directly related to physical distances between genes on the chromosome (A.H. Sturtevant, 1911).

a) the farther apart two loci are located from one another, the more likely a crossover will occur between them.

b) 1% recombination in gametes of heterozygote represents 1 map unit ( m.u. or CentiMorgan, cM).

c) for white and miniature, distance is large (~37.6 m.u.); for yellow and white, distance between genes in much smaller (~1.4 m.u.);

d) the closer the loci, the greater the number of progeny that must be examined to detect a crossover event.

 4. Conclusions from Linkage Experiments. a) Genes are linked through all generations, no random mixing.

b) Genes are arrayed in a specific linear order along the length of a chromosome, which is a species characteristic.

c) Homologs carry the same array of gen loci, but not necessarily the same alleles. During meiosis, homologs could exchange parts by crossing over. Chiasmata represent crossover points between homologs (Janssens, 1909).

 III. Three point testcrosses. A. eg. Corn: Mapping the genes for the rececessive traits lazy growth, glossy leaf, and sugary endosperm.

Progeny From a Three Point Testcross in Corn


Phenotype of testcross progeny Genotype of Gamete from Hybrid Parent Number
normal (wildtype)
Lz Gl Su
286
lazy
lz Gl Su
33
glossy
Lz gl Su
59
sugary
Lz Gl su
4
lazy,glossy
lz gl Su
2
lazy, sugary
lz Gl su
44
glossy,sugary
Lz gl su
40
lazy, glossy,sugary
lz gl su
272
B. Review of data set shows deviation from 1:1:1:1:1:1:1:1 ratio that would be expected for independent assortment; instead, two classes are in very high frequency, two classes are in very low frequency and 4 classes are recovered in intermediate frequencies.

C. The parental class represents the most frequent phenotypic classes amongst the progeny. This allows the determination of the orientation of the dominant and recessive alleles in the heterozygous parent.

D. The double recombinant classes represent the least frequent phenotypic classes amongst the progeny. The 4 intermediate classes represent the single recombinant classes in each interval.

E. Comparison of parental and double recombinant classes allows the determination of which gene is in the middle; from the data set given in class, the phasing relationship and order of the genes in the heterozygous parent was:

Lz  Su  Gl/lz  su  gl F. Once gene array in the parental heterozygote is determined, identify the phenotypic classes in the progeny that represent the single and double crossovers which occur within each gene interval: RF = (#SCO + #DCO)/total progeny for each gene interval. eg. for the example given in class, The gene interval between Lz and Su could be calculated as : RF = ([33 + 40] + [4 + 2])/740 = 0.107 = 10.7 m.u.

E. Interference.

1. A crossover in one area may interfere with crossing over in an adjacent area of the chromosome.

Male Progeny from the mating of Drosophila melanogaster females heterozygous for three X-linked Genes with Wildtype Males


Phenotype
Genotype of female gamete
number
wildtype
+ + +
5
crossveinless
cv + +
2207
echinus
+ ec +
217
cut
+ + ct
265
crossveinless, echinus
cv ec +
273
crossveinless,cut
cv + ct
223
echinus,cut
+ ec ct
2125
crossveinless,echinus,cut
cv ec ct
3
2. This may be observed in three point testcrosses, where the number of expected double crossovers that are actually recovered is less than the number that are expected. e.g. for the example taken from a Drosophila cross in class (see above data set), the calculated map distances between the echinus, crossveinless and cut loci were:

ec-------10.3 m.u.--------cv------8.4 m.u.------ct

and out of 5318 progeny, 8 double crossovers were actually observed. The expected number of double crossovers was therefore:

(0.103)(0.084) = (0.0086)(5318) = 46 expected double crossovers.

coefficient of coincidence = observed double crossovers/expected double crossovers = 0.174

Interference = 1- coefficient of coincidence = 0.826

 IV. Ordered tetrads in fungi. A. Some fungi produce meiotic products, haploid spores, in a saclike structure (ascus). All meiotic products from the same division cycle can be recovered; in some species, each member of the tetrad giving rise to these spores are arrayed in the order of their segregation. Likewise, in some species, a mitotic division follows meiosis giving rise to two spore pairs for each tetrad.

B. Other advantages to genetic analysis.

1. Organism is haploid for majority of life cycle.

2, They produce large numbers of spores- rare events are easily detected.

 C. Life cycle of Ascomycetes: e.g. Neurospora crassa, bread mold. 1. Haploid spores generate hyphal structures.

2. Haploid nuclei of opposite mating types can undergo nuclear fusion and meiosis in fruiting bodies (perithecia).

3. The zygote is the only diploid stage; this transient diploid state undergoes meiosis and then mitosis; 8 ascospores are produced.

4. The meiotic products (tetrads) are maintained in a linear order.

 V. Mapping by Ordered Tetrad Analysis. All meiotic products are recovered in the same structure in an ordered array. This permits the mapping of genes relative to the centromere. A. Calculating a Map Distance Between the Gene and Centromere. 1. If no recombination occurs, you will expect to see a 1st division segregation pattern: a 4:4 array of spores relative to phenotype.

2. If a single crossover occurs between the gene and the centromere, you will expect to see a 2nd division segregation pattern: a 2:2:2:2 or a 2:4:2 array (fig 5-23 in Hartwell).

3. The percentage of 2nd division segregation patterns is related to the distances between the gene and centromere.

1/2(asci with 2nd division segregation pattern) X 100 = map distance
         total number of asci

Remember that you a scoring asci here, and that the "1/2" in the above equation corrects for the fact that only 1/2 of the chromosomes arising from that meiosis will be recombinant.

 B. Gene Linkages. Suppose two loci are present in the same cross. In the example given in class, the cross AB x ab yields the following data set:

Hypothetical Data Set from the Neuospora Cross AB x ab



Ascus Type Ascus Type Ascus Type Ascus Type
Spore pair
1
2
3
4
 
(PD)
(TT)
(PD)
(NPD)
1
AB
AB
AB
aB
2
AB
Ab
ab
aB
3
ab
aB
AB
Ab
4
ab
ab
ab
Ab
Total number
108
60
30
2
1. 3 types of tetrads are produced: a) parental ditype (PD): AB AB ab ab

b) nonparental ditype (NPD): aB aB Ab Ab

c) tetratype (TT): AB Ab aB ab

 2. If genes are unlinked, then the A and B loci might be expected to assort independently; therefore PD = NPD (approximately the same numbers).

3. If the two genes are linked, then the only way to generate a NPD would be through a 4 strand double crossover; and PD>> NPD. In the above data set, the A and B loci are linked.

4. Centromere order can be determined by examining frequency of classes showing 1st and second degree segregation patterns in relation to frequency of NPD classes. For example, the type 3 ascus pattern above is informative; both the A and B loci show a second degree segregation pattern, indicating a cross over between both genes and the centromere. If genes were on opposite sides of the centromere, you would need a double crossover to generate this class. Its frequency relative to the other classes indicates that it is a single crossover class, and that A is closer to the centromere than B: centromere--------A----------B.

5. Recombination between centromere and B:

1/2 (60 + 30) X 100 = 22.5 map units
          200

6. Recombination between centromere and A:

1/2 (30) X 100 = 7.5 map units
      200

The deduced difference between the genes is therefore 15 map units.

7. The above calculation will lead to an underestimate of distance, since the above technique is not accurately estimating the number of double crossovers. The number of double crossovers can be estimated, however from the NPD class (presented in class: diagram found in ZAP room notes). From this class, we can estimate the total number of single and double crossovers occurring in the interval between A and B:

DCO = 4 NPD

T = SCO + 1/2 DCO; therefore

SCO = T - 2 NPD

RF = 1/2 SCO + DCO

RF = 1/2[TT-2NPD] + 4NPD
                total asci

for the above data set:

RF = 1/2(60-4) + 8
                 200

     = 36/200 = 0.18 = 18 map units between the A and B loci. (note our previous estimate was 15 map units, an underestimate).

 VII. Mitotic Recombination. A. Although much rarer than meiotic recombination during gametogenesis, recombination events can be detected in somatic tissue. A classic example is the formation of twin spots during Drosophila development (Stern, 1936).

B. Twin spots have proven useful in the analysis of the developmental effects of mutations which may otherwise cause lethality if inherited in homozygous form in the zygote.

 VIII. Human Gene Linkage. (Hartwell, pp. 392-33; Thompson, Chapters 10-12) A. Information from Traditional Linkage Data. It is obviously difficult to generate large numbers of progeny and perform selective matings in humans. Nevertheless, information on linkage can be obtained from pedigree data using statistical techniques.

B. Lod Scores. Lod score analysis is a statistical technique that permits linkage relationships to be derived based on relative probabilities of linkage given the segregation patterns observed in pedigrees. The statistical data handling is beyond the scope of this course; for those who are interested, I can provide you with a more detailed description of the analysis method. Briefly, linkage data can be represented as a likelihood ratio, which represents the probability that two alleles are linked at some given recombination frequency (q ), divided by the probability that the alleles assort independently (RF = 0.5). The data is usually presented as the log of the likelihood ratios (Lod Score):

Zq = log 10 (Probability of family for q = 0.01, 0.02, etc.)
                  (Probability of family for q = 0.50)

Lod scores from individual experiments can be calculated and are additive; it is possible to arrive at linkage probabilities without knowledge of the phasing relationships in the heterozygous parent. A likelihood ratio ratio of 1000:1 (Lod score = 3) is usually considered "proof" of linkage.

 IX. Assigning Genes to Chromosomes Using Somatic Cell Genetics. (see Thompson, chapter 10) A. Production of Hybrid Cells. Usually a rodent tumor cell line is fused with human fibroblast cells. Cell fusion is promoted by either viral infection or chemicals (polyethylene glycol; PEG). 1. HAT Selection System. Used to enrich for hybrid cells and eliminate parental cells and same-parent fusions from the culture. Growth medium contains a chemical (aminopterin) to shut down the major pathway for nucleotide precursor synthesis- forcing the use of a salvage pathway to produce the subunits needed for DNA. The "salvage" molecules hypoxanthine and thymidine are also present in the growth medium and can be incorporated into DNA, provided the cells have two salvage pathway enzymes- HGPRT (hypoxanthine-guanine phosphoribosyltransferase) and TK (thymidine kinase) necessary to convert them into the appropriate DNA precursors. By using parental lines deficient (i.e. mutant) for one or the other of the two enzyme activities, (e.g. human HGPRT-/TK+; rodent /HGPRT+/TK-; hybrid HGPRT+/TK+) one can select for the rare cell fusions that will produce hybrid cells (complementation).

2. The hybrid cells will eventually undergo nuclear fusion to form heterokaryons. This process will lead to a spontaneous loss of human chromosomes from the hybrid cells. The chromosomes will be lost at random, giving rise to colonies which will contain different portions of the human chromosome complement.

3. The presence or absence of a human gene product can then we assayed in the hybrid cells. Only those hybrid cells containing the human gene (and the chromosome on which it resides) would be expected to produce the human product. Concordance and discordance between the presence or absence of the human gene product (enzyme or protein) and the human chromosomes a hybrid cell line carries establishes what chromosome the gene resides on (synteny).

4. Once synteny is established, the cell lines carrying fragments of the chromsome in question can also be assayed.

5. The above technique requires that we be able to identify individual human chromosomes. It also assumes that we can identify a human gene product in the hybrid cell. We can do both. Indeed, the nucleic acid sequence of the gene itself is sufficiently different to distinguish mouse from human genes, allowing DNA gene probes to be used for mapping purposes.

 X. Aberrant Asci Patterns, Gene Conversion and Recombination at the Molecular Level.(Hartwell, 178-186) A. Aberrant asci patterns (e.g. 3:1:1:3; 6:2; 5:3) have led to insights into the molecular nature of recombination. The recovery of products of a meiotic division in a heterozygote at some ratio other than the expected 1a:1A resulting from reciprocal recombination is known as gene conversion. Because we can trap all the meiotic products of the same division cycle, and because the mitotic division following meiosis allows us to analyze the individual DNA molecules that underwent recombination in each tetrad, we have been able to develop models of how recombination occurs. Many of these models have common features which can explain the origin of aberrant asci during fungal meiosis. These models suggest a process of DNA nicking, strand invasion, heteroduplex formation and DNA repair occurring at the recombination site. Aspects of these models have been verified by observing recombination in other model systems (e.g. bacteria). The Holliday Model is one such model developed to explain gene conversion. Features of the model and how it can explain an aberrant ascus pattern are covered in pp. 178-186 of Hartwell. I. Overview of Mutation Section A. Mutation: there is no genetics without variation

B. Gene or "Point" Mutation: allelic variations occuring within a single gene locus, usually resulting in a change or loss of a small number of nucleotides.  (Hartwell, Chapter 7)

C. Chromosomal Mutation: a change in a segment, chromosome or set of chromosomes, which may, or may not include a change within a single gene locus.

1. Aberrations due to chromosome structure (Hartwell, Chapter 13)

2. Aberrations due to chromosome number (Hartwell, Chapter 13)

 D. We will cover this section in the order given below, first discussing chromosome structure and changes that can occur at the chromosome level. We will return to a discussion of mutation at the allele level prior to a discussion of bacterial genetics, and emphasize the important role bacterial genetics had in laying the groundwork for molecular genetics- the study of information transfer at the molecular level. II. Techniques Used to Visualize Chromosome Structure: Karyotype analysis (Hartwell, pp. 81; 422-24) A. Chromosomes are a complex of primarily DNA and protein (chromatin). euchromatin: lightly staining, genetically active; heterochromatin, darkly staining, genetically inactive. (Hartwell, 422-24) for definitions and description).

B. During interphase, chromosome structure is not resolved due to decondensed state of chromatin.

C. DNA becomes highly compacted during process of mitosis; chromosome structure can be visualized.

D. In humans, a cell population may need to be stimulated to start mitosis before analysis can be performed (e.g. white blood cells (lymphocytes) from peripheral blood; or fetal fibroblasts (amniocytes) taken from amniotic fluid).

1. Interphase cells are induced to enter mitosis by stimulation with a mitogen (e.g. phytohemagluttinin).

2. Cell may be stopped in metaphase by disrupting spindle fiber formation (tubulin polymerization) with the chemical colchicine.

3. The cells must be prepared to preserve chromosome structure and specifically disrupted to spread the chromosomes.

 III. Basic Chromosome Morphology A. Size

B. Centromere position (metacentric, submetacentric, acrocentric, telocentric).

C. Other distinguishing features such as nucleolus organizer regions (secondary constrictions, site of ribosomal RNA genes.

 IV. Differential Banding Techniques. A. It is possible to identify every human chromosome on the basis of a distinctive banding pattern revealed by partially disrupting chromosome and staining with specific dyes.

B. Several Methodologies: require treatments with a variety of agents that will partially disrupt chromatin structure (acids, bases, proteolytic enzymes, heating).

C. Most common techniques:

1. G Banding: standard for comparisons (Paris Convention, 1971).

2. Q Banding: fluorescent staining: similar to G Band pattern.

3. R Banding: banding pattern appears reversed from G Band pattern.

4. C Banding: will stain only the heterochromatin surrounding the centromere.
 

 D. Normal Standard of Resolution : ~ 300-500 bands at metaphase; ~1200 bands at prometaphase, when chromatin is less condensed. (The most recent estimates of the number of genes within human genome is on the order of 25,000 (revised since Hartwell was printed in 2004 and still not a certainty); so this level of cytological resolution is obviously limited in its ability to detect structural defects that may exist in chromosome.)

E. Convention for designating sites on chromosomes. Sites are designated by first giving the chromosome number, followed by the chromosome arm (p= "petite" or short arm; q = long arm). Chromosome regions are identified on the basis of "landmark bands" which define chromsome regions; these regions are numbered away from the centromere. The band found within a region is given last; the individual bands are also numbered away from the centromere:

e.g. In the above example, the arrow would be located at 8q22 in the human chromosome complement.

 V. Morphology of Human Sex Chromosomes. A. Mammalian Y Chromosomes. (Hartwell, pp. 79-82; 103-4) 1. Apparently does not contain many genes, other than those involved in human sex determination, or sperm fertility factors.

2. The chromosome contains much chromatin that appears genetically inert (consitutive heterochromatin). There is much population polymorphism in the size of the long arm of the Y.

3. Pairs with the X chromosome during meiosis along a small region that is homologous to the X (pseudoautosomal region).

4. TDF gene serves as a major developmental switch for determining maleness.

 B. Mammalian X Chromosomes. (Hartwell, pp. 79-82; 431-32;) 1. Females have two copies of the X; males have only one; Is there a dosage compensation mechanism that operates to control and normalize levels of X chromosome gene products?

2. Interphase cells of females show a dark body (Barr body) usually associated with the periphery of nucleus; males have no Barr body; individuals with extra X chromosomes show extra Barr bodies, such that the number of Barr bodies observed are 1 less that the total number of X chromosomes.

3. Lyon Hypothesis: In a normal female (46; XX), one X chromosome becomes functionally inactive early in embryogenesis (facultative heterochromatin). The inactivation process is random, and the chromosome that is inactivated remains so in all progeny cells. Since inactivation is a random event, all females are functionally mosaic; some cells contain one inactivated X, some cells the other. e.g. X linked coat color genes in mice; anhydrotic dysplasia in humans.

 C. Genomic Imprinting. (Hartwell, pp. 596-602) 1. In mammals, a small number of genes are passed on from the parent to the zygote in an inactive form; embryos therefore inherit only one active gene.

2. Some of these genes are inactivated during spermatogenesis; some during oogenesis.

3. If one parent passes a wild-type inactive gene and the other parent passes a mutant allele, the resultant individual will show a mutant phenotype. e.g. Praeder-Willi Syndrome (maternal gene inactive); Angelman Syndrome (paternal gene inactive).

4. Why would one gene be specifically inactivated by a parent? One theory suggests that imprinting is related to competition between embryo and mother for resources. According to this theory, genes that are passed in an active form by males would lead to increased growth during embryogenesis; genes that are passed in an active form by females would tend to retard embryonic growth.

 VI. Departures from Diploidy: Heritable Aberrations in Chromosome Number; Humans. (Hartwell, Chapter 13) A. Aneuploidy: a chromosome number that is not an exact multiple of the haploid chromosome number. Aneuploidies usually arise from de novo errors (nondisjunction) during cell division.

B. In humans, aneuploidy involving sex chromosomes are usually the least severe of the possible aneuploidies:

1. Turner Syndrome (45, X)

2. Klinefelter Syndrome (47, XXY).

3. Multiple Y males (47, XYY).

4. Poly X females (47, XXX; 48, XXXX, etc.)

 C. In humans, aneuploidy involving autosomes is usually fatal; all autosomal monoploidy is fatal; most trisomies are fatal with the exception of Patau, Edwards and Down Syndrome. 1. Patau Syndrome (47, +13): seen approximately 1/4000 births; severe mental retardation and physiological defects. Usually fatal in early infancy.

2. Edwards Syndrome (47, +18): seen in approximately 1/7000 births; severe mental retardation and other physiological defects. Usually fatal in early infancy.

3. Down Syndrome (47, +21): seen in approximately 1/800 births; physical signs in infants are lowered muscle tone, protruding tongue, characteristic folds at eye; spots on iris, characteristic fingerprints (dermatoglypics). Children are at risk for increased cardiac defects, severe mental retardation; older individuals will develop early onset Alzheimer disease.

4. There is a pronounced increase in risk for chromosomal aneuplodies in older women.

 VII. Structural Chromosomal Aberrations (Hartwell, Chapter 13) A. Factors Generating Chromosomal Damage (clastogenic agents) 1. ionizing radiation.

2. some viral infections.

3. chemicals.

 B. Breakage Events. 1. Telomeres represents "proper" chromosomal ends; they are specialized structures that with a highly conserved DNA sequence which marks the natural end of a chromosome.

2. If a break is introduced, new "ends" are not properly terminated and will be subject to DNA repair mechanisms. The joining of inappropriate ends can lead to structural abnormalities.

 C. Reciprocal Translocations: a reciprocal exchange between non-homologous chromosomes. 1. If no genes are damaged at the site of breakage, then individuals heterozygous for the reciprocal translocation (i.e. carrying two translocation chromosomes and two normal chromosomes) may show no overt abnormal phenotype.

2. Pairing problems during meiosis, however, will lead to the formation of tetravalents; abnormal segregation from this structure (adjacent segregation) will lead to unbalanced gametes carrying duplications and deficiencies for certain genes. The result is semisterility in the translocation heterozygote.

3. Robertsonian Translocations: a specific type of reciprocal translocation, where breaks are introduced to opposite sides of the centromeres of two non-homolog acrocentric chromosomes. Repair leads to formation of a small chromosome which is usually lost, and the fusion of the two long arms into a single chromosome.

a) Often seen in closely related species. For example, the metacentric chromosome 2 of humans resembles in banding pattern and gene content two acrocentric chromosomes seen in chimpansees, suggesting a robertsonian fusion occurred during primate evolution.

b) Such translocations can also been seen within species; 14:21 translocations are associated with a heritable form of Down Syndrome.

c) Reciprocal Translocations and Cancer:  If the breakpoint occurs within a gene (damage at the site of breakage), or transfers a gene into a region which affects its ability to be expressed, this can lead to somatic mutations in genes controlling normal growth and cell proliferation.

1) formation of hybrid genes with abnormal phenotypes: in >95% of cases of chronic myelogenous leukemia, the tumor cells contain a clear reciprocal translocation within the abl gene locus that leads to the conversion of this "proto-oncogene" into an "oncogene".

2) the movement of the myc gene into the region of the immunglobulin gene cluster produces a change in its expression pattern which is correlated with a cancer known as Burkitt Lymphoma. This activation due to change in location is an example of position-effect variegation.

 D. Deletions or Deficiencies 1. A loss of a small region of the chromosome may be correlated with abnormalities arising from the imbalance; most large deletions are inviable. e.g. (46; 5p-) cri du chat syndrome.

2. A loss of a small region may place an individual at risk if the normal gene copy located on the homolog chromosome mutates. e.g. Hereditary predisposition to retinal cancers for individuals deleted for 13q14, the site of the retinoblastoma (Rb) gene locus. (Hartwell, pp. 634-43)

 E. Inversions 1. Once again, if no genes are damaged at the breakpoint, inversion heterozygotes may only encounter problems relative to chromosome segregation during meiosis.

2. If crossing over occurs within inversion, it will produce chromosomes with duplications and deficiencies.

a. paracentric inversions: centromere not included in the inversion; also produces acentric and dicentric chromosomes.

b. pericentric inversions: centromere is included in the inversion; recombinant products have centromeres but still carry duplications and deletions.

 F. Fragile Sites: X-linked Mental Retardation (Hartwell, 202-203). 1. Fragile sites can be detected by chromosome breakage when cells are cultured under special conditions. The chromosomal marker is a characteristic of Fragile X syndrome, present in 1/1000 male births and one of the most common forms of mental retardation in males (~8% of all males with mental retardation).

2. In Fragile X Syndrome the defect is linked to an unstable replication of a triplet repeat in the FMR-1 gene, located at the fragile site. An amplification of this sequence from a "premutation" leads to the full symptoms of this syndrome.

 G. Monitoring of Chromosomal Defects. 1. Amniocentesis. Usually performed at ~16 weeks, when fibroblasts can be collected safely from amniotic fluid. Requires mitogenic stimulation of cells prior to analysis.

2. Chorionic Villus Sampling. Can be performed at ~9th week, from embryonic cells that are already actively dividing.
 

 VIII. Departures from Diploidy in Plants. A. Tolerances for aneuploidy and polyploidy contribute to speciation and facilitate genetic manipulations. 1. Plants don't necessarily need to propagate via sexual reproduction; they can propagate from somatic tissues (e.g. cuttings).

2. Increases in ploidy may lead to aggricultural benefits (e.g. increases in plant size).

3. Monoploids can be generated from gametes and used in selection experiments.

4. Chromosome number can be doubled by colchicine treatment.

 B. Polyploid species: (number of complete chromosome sets > 2) 1. Autopolyploids: contain multiples of the same chromosome set (same species).

2. Allopolyploids: contain mutiples of different chromsome sets (different species).

3. Hybridizations may lead to pairing problems during meiosis.

a. triploids are often sterile- difficulties in meiotic segregation.

b. a doubling of the different chromosome sets in allopolyploids can relieve problems during meiosis: if there are pairing partners for homologs, sterility often relieved. (e.g. Karpechenko, radish and cabbage polyploids). amphidiploid = contains the diploid chromosome complement of both parent species.

c. polyploidy and evolution: origin of common cultivated wheat, an allohexaploid.

 I. Gene Mutation: there is no genetics without allelic variation (Hartwell, Chapter 7) A. Classes of Mutation 1. Somatic; clonal populations can arise due to mutations occuring during development; e.g. sectors of color in plant leaves, cancer (somatic mutation of proto-oncogenes [normal cellular genes involved in cellular growth regulation] to oncogenes [mutant genes causing abnormal cell proliferation or metastasis]).

 2.Germline; capable of being transmitted to the next generation.

 B. "Point Mutations": mutant phenotypes 1. Heteromorphs (null, loss-of-function, gain-of-function)

2. Lethals

3. Conditional mutants (restrictive vs. permissive conditions)

4. Biochemical mutants

a. prototrophs

b. auxotrophs

 5. Resistance mutants a. antibiotic resistance (e.g. strr)

b. viral resistance (tonr)

 C. Selection Systems and Mutant Detection 1. Advantage of microbes a. haploid

b. fast generation times

c. inexpensive

 2. Eukaryotic systems (e.g. ClB of Muller) D. Nature of Genetic Change; random mutation/selection or physiological adaptation? (Hartwell, pp. 194-196) 1. tonr mutants: resistance to T1 bacteriophage infection.

2. Delbruck/Luria fluctuation test: "Mutations of Bacteria from Virus Sensitivity to Virus Resistance": (1943)

3. Lederberg replica plating experiment: (1952)

4. 1945: Schrodinger's "What is Life?" published. How do living organisms order themsevles? Do molecules govern heredity? What is the physical nature of the molecules that govern heredity? "...likely to involve hitherto unknown 'other laws of physics,' which, however, once they have been revealed will form just as integral a part of this science as the former." The influx of individuals trained in physics provided a basis for the revolution in molecular biology.

 I. Recombination in Bacteria (Hartwell, Chapter 14) A. How are genes in bacteria organized? Are there correlates to eukaryotic chromosomes? Is there linkage? How universal are information storage and retrieval mechanisms in Nature?

B. No genetics without allelic variation. What phenotypes do we study?

 II. Mechanisms for Gene Exchange and Recombination in Bacteria A. Conjugation: transferred through direct cell-cell contact.

B. Transduction: transferred through viral infection.

C. Transformation: Cellular uptake of DNA from the environment.

 III. Conjugation A. Lederberg and Tatum (1946). Using bacteria that were multiply auxotrophic, demonstrated that recombination could occur.

B. Davis (1950). The strains used by Lederberg and Tatum required physical contact for transfer. The U-tube experiment.

C. Hayes (1953). There is a directionality of transfer. In this process, there is a donor/recipient relationship. Streptomycin treatment: will allow a donor cell to transfer DNA to a recipient, but str-treated cells will not be capable of reproducing:

1. Donor (treat c str)-->D* (wash out drug and cross c recipient)--> D* + R= recombinants

2. Recipient (treat c str) --->R* (wash out drug and cross to donor)--> D + R*= no recombinants.

3. Idea of unidirectional transfer: D="males"; R="females"; crosses = "matings"

 D. Attributed "maleness" to a specific genetic entity; the "Fertility" Factor. 1. F+ cells would transfer the genotype of maleness at high frequency.

2. F+ cells would transfer other genotypes (e.g. prototrophy, phage resistance) at low frequency.

3. Cavalli (1950): isolated a peculiar strain of male cell that produced prototrophic recombinants at 1000 times the frequency of normal F+ strains.

a. Termed an Hfr strain: "high frequency of recombination".

b. The high frequency donor character pertains only to a limited portion of donor genome.

c. Unlike F+ cells, Hfr cells do not transfer male genotype to recipient cells.

d. F+ ---> Hfr change attributable to a heritable change in the sex factor's properties. Nature of change?

 E. Oriented Gene Transfer and "Time of Entry" Mapping Experiments. 1. Wollman and Jacob: Interrupted mating experiments. Hfr x R cell matings could be conducted under timed conditions to study the kinetics of gene transfer.

2. Waring blender experiment: Set up matings, allow matings to occur for defined time intervals, disrupt the matings with a Waring blender, then plate to select recombinants.

3. Time of Entry Mapping: By plotting frequency of recombinants recovered vs. time, they could extrapolate the "time of entry", which could be used as an index of distances between genes.

4. Analysis of different Hfr strains (isolated from F+ populations by replica-plate matings on selective media) led to the conclusion that transfer could occur from different points in the genome in either of two different directions. How can this be explained by a consistent model?

5. Both the bacterial chromosome and the F factor were covalently closed circular molecules. Integration of F factor into the chromosome DNA led to Hfr. Orientation and site of the integration event dictated what chromosomal genes were transferred.

 F. F' factors. 1. Occasionally, F factors will become unintegrated from bacterial chromosome in an Hfr cell. This will create an F factor that can carry a portion of the bacterial genome (F' factor) . If it conjugates with a recipient cell, that cell receives a chromosomal gene.

2. F' cells carry two copies of a portion of the genome (merodiploids). There are also capable of transferring that gene at high frequency to recipient cells.

 G. Mapping by selecting for late markers. 1. Time of entry mapping is not very high resolution and is not very useful for mapping genes relatively close to one another.

2. By selecting for a late marker, closely linked genes can be mapped by examining the recombination frequency.

 IV. Transduction A. Lytic Cycle: upon entry into cell, virus replicates, packages and lysis host.

B.Generalized Transduction and Mapping

1. During lytic infection, host genetic material may be packaged.

2. On infection, host genetic material is transferred into new cell, where it may recombine into genome.

3. Since phage particle can only contain limited amounts of DNA, technique can be used for mapping. "Co-transduction".

 C. Lysogeny and Specialized Transduction 1. Some phage can integrate into the host cell genome and replicate along with host chromosome; will not lyse the host cell.

2. Can also leave host chromosome; if unintegration event is imprecise, host genetic material may be included.

3. These phage can now transfer host genes into new cell on infection.

 V. Transformation. A. Certain bacterial strains incorporate DNA from the environment through their membranes. This DNA can recombine into the genome.

B. To pick up DNA, cells must be in a "competent" physiological state.

C. "Co-transformation" frequencies can be used to map the location of genes relative to one another.

 VI. Bacteriophage Genetics. A. Phage Phenotypes defined by ability to grow on a particular host, or by plaque morphology.

B. Mixed infections between mutant strains can be used to map genes.

 I. Biochemical Nature of the Gene (Hartwell, Chapter 6) A. The "Transforming Principle". Experiments with mutant strains of Streptococcus pneumoniae (pneumococcus), F. Griffith, 1928.

B. Avery, MacLeod and McCarty (1944). Biochemically fractionated pneumococcal extracts and tested chemicals for ability to tranform. DNA = transforming principle.

C. Hershey-Chase Experiment (1952). The material injected into a bacterial cell by a phage programs the phage reproductive cycle in a lytic infection. That material is DNA.

 II. Reluctance to Accept DNA as the Genetic Material. A. DNA structure originally considered a monotonous polymer. 1. base composition

2. phosphate-sugar backbone

3. strand polarity

 B. Equivalence Rule of Chargaff (1950). III. X-Ray Crystallography A. Three groups working on developing 3-D reconstruction: Wilkins and Franklin, Watson and Crick, (at Cambridge, England) and Pauling (CalTech).

B. Franklin produces first high resolution images.

C. Watson and Crick incorporate density data, base composition and crystallography into their model.

1. Watson-Crick Structure. The double helix.

2. Implications of Model. "It has not escaped our notice that the specific pairing mechanism we have proposed immediately suggests a possible copying mechanism for the genetic material"- conclusion of 1953 Nature article proposing DNA structure.

3. B-form DNA relative to A-form, Z-form.

 IV. DNA Replication. A. Models for DNA replication.

B. Proof of the semi-conservative model: The Meselson-Stahl Experiment (1958).

C. Origins of Replication: Autoradiographic evidence for bidirectional replication.

 V. Synthesis at the Replication Fork. A. Properties of DNA Polymerases (III and I) 1. Can synthesize in the 5' to 3' direction only.

2. "leading" and "lagging" strand synthesis: discontinous synthesis on lagging strand (Okazaki fragments).

3. Can not synthesize DNA de novo ("anew"). Requires a template and a primer.

 B. RNA priming and the Primosome.

C. Mechanism for Primer Removal. 5'-3' exonuclase activity of Pol I.

D. Error correcting capabilities of Pol III and Pol I. 3'-5' exonuclease activity and proofreading.

E. Resolution of nicks in phosphate-sugar backbone: role of DNA ligase.

F. Unwinding and relief of torsional stress during DNA replication: role of DNA helicases (unwinding of helix) and topoisomerases (relief of torsional stress).

 VI. Problems with replication of linear molecules. A. Properties of DNA polymerase and replication of chromosome ends.

B. Molecular Properties of telomeres. (pp. 258-259)

1. conserved "consensus" sequence: [(AT)nGm]x e.g. tetrahymena: TTGGGG

2. redundant sequence: TTGGGGTTGGGGTTGGGGTTGGGG...

C. Telomere replication and telomerase.

 VII. Other Variant Forms of DNA Replication A. Rolling Circle Replication in certain viruses

B. Single Stranded DNA genomes (Hartwell, pp. 166-7)

C.  Mitochondrial Replication (Hartwell, p. 528)

 VI. RNA Genomes. A. Retrovirus life cycles, and similarities/differences to transposible elements (Hartwell, pp. 168-9; 252-3; 291-2; 461-4; 636-7; 723-5)

B. Reverse Transcriptase

 VII. Applications of Bacterial Enzymes; Hybridization Analyses and Molecular Cloning. (Hartwell, Chapter 9). A. Labelling Techniques for Nucleic Acid Hybridizations/In Situ Hybridizations (Hartwell, pp. 294-7).

B. Restriction Endonucleases

C. DNA Cloning in Microorganisms

1. Use of bacterial plasmids as vectors

2. Restriction digestions and ligations into plasmid vectors

3. Transformation of recombinant plasmid vectors

 D. DNA Sequencing. 1. dideoxynucleotides.

2. Use of chain terminators in copying unknown template with DNA Polymerase.

3. Sizing chain termination products by electrophoresis.

 E.  High Throughput Genomic Analysis. (Hartwell, pp. 348-365) 1.  Automated DNA sequencing

2.  Array Technology

 XII. DNA----> Protein (Hartwell, Chapter 7; pp. 213-224 ; Thompson, pp. 141-143) A. Use of mutations to examine biochemical pathways.

B. The Beadle/Tatum Experimental Protocol to isolate mutants in biochemical pathways in Neurospora.

C. Protein Structure

1. Amino acids

2. The peptide bond and constraints on rotation

3. The role of R groups

4. Protein Primary Structure

5. Protein Secondary Structure

6. Protein Tertiary Structure

7. Protein Quaternary Structure

 XIII. Analysis of the Fine Stucture of the Gene: The Bridge between Transmission Genetics and Molecular Genetics. (Chapter 7; pp. 206-213) A. Establishing Colinearity between Genes and Proteins (1953-63)

B. Major Questions concerning Gene Structure at the molecular level?

1. What is the fundamental unit of structure? (Are genes divisible?)

2. What is the fundamental unit of change?

3. What is the fundamental unit of function?

 XIV. Seymour Benzer and the Genetic Analysis of Gene Stucture.

A. T4 Experimental System.

1. rII locus phenotype

2. Selection system for mapping mutations

3. Schema for spatial ordering of mutations

 B. Observations. 1. Intragenic recombination

2. Mutational hot spots

3. Nucleotide is the unit of change

4. Complementation Analysis and its use to resolve a functional unit (cistron).

 XV. Charles Yanofsky: mutations directly correlate to amino acid changes in proteins.

XVI. The Central Dogma: DNA----> RNA----> Protein (Hartwell, Chapter 8)

A. Evidence for an RNA intermediate 1. Prokaryotes: On phage infection, a short-lived RNA species could be detected. This RNA was similar in composition to phage, not host DNA, and ended up on host ribosomes, the site of protein synthesis.

2. Eukaryotes: Pulse-chase experiments indicated that an RNA molecule was transported from the nucleus to the cytoplasm (where ribosomes are located.

 B. DNA----->RNA: Transcription (E. coli) 1. RNA polymerase

2. Initiation:Promoter binding and transcriptional control

3. Elongation: Polycistronic mRNA

4. Termination: Release of RNA polymerase and mRNA

 C. Nature of the Genetic Code: Triplet and non-overlapping code

D. Frameshift mutations (see section on induced mutations: (Hartwell, 241-3)

E. Recombinational studies in T4: intragenic suppressor mutations indicate a triplet code.

F. Yanofsky: point mutations generate a single aa change in protein: suggests non-overlapping code.

G. RNA-----> Protein: Translation (E. coli)

1. The "adaptor" hypothesis: tRNA a. tRNA structure

b. tRNA charging: aminoacyl tRNA synthetases

 2. Ribosomes: a. Ribosome structure

b. site of protein synthesis

 H. Steps in Translation: 1. Initiation: Ribosome binding to mRNA and assembly at start codon.

2. Elongation: Synthesis of a polypeptide chain

a. Peptidyl transferase activity of ribosome

b. A and P sites and translocation

 3. Termination: role of specific protein factors in recognizing termination codons. I. "Cracking the Code": use of in vitro protein synthesizings systems and synthetic RNAs to determine which codons specify which amino acids. 1. Use of the enzyme polynucleotide phosphorylase to synthesize synthetic RNAs (Ochoa). Composition of RNA random and dependent on type of nucleotide added. e.g. ( all U = UUUUUUUUUUU; 3/4 G, 1/4 U = GGUGGGUUGGGG...)

2. Translation of synthetic RNA could be accomplished in vitro (ribosomes would assemble, code would be read in triplet fashion, with no overlaps as regular mRNAs are, but initiation could occur at any nucleotide in RNA).

3. Ordered synthesis of RNA could also be accomplished chemically (Khorana). Composition of RNA could be a known alternating pattern of nucleotides e.g. (UG)n = UGUGUGUGUGUG = (Cys-Val)n .

4. Nirenberg: Development of trinucleotide binding assay. Correct nucleotide will allow the assembly of a ribosome and its cognate charged tRNA into a complex that can be trapped on a filter. (see handout).

 J. Codon "Wobble". Number of distinct tRNA molecules is less than number of codons. Some tRNAs can recognize more than 1 codon due to non Watson-Crick base interactions at the third position in the codon and the complemetary first position in the anticodon.

E. The code is univeral (almost). Some exceptions...see information on mitochondrial genome organization. Note that mitochondria carry transcribe their own genes and can translate the mRNAs produced from them; some differences exist in code UGA = Trp, not STOP.

K. Eukaryotic Gene Expression (e.g. beta globin genes).

1. 3 RNA polymerases; Pol I (large rRNA), Pol II (mRNA), Pol III (tRNA, most small nuclear RNAs)

2. Transcription initiation complexes; nature of the eukaryotic promoter; TATA boxes, binding of generalized transcription factors.

3. Formation of primary transcripts.

4. Capping and polyadenylation.

5. Splicing of Primary Transcripts; SnRNPs.

6. Exon Shuffling hypothesis- if exons represent domains with distinct functional properties, can new genes evolve via duplication and genome reorganization?

 XVII. Chromatin Structure. (Hartwell, Chapter 12; pp. 416-422) A. Histones.

B. Nucleosome organization and the 10 nm core particle.

C. Higher order folding.

1. solenoid formation.

2. evidence for a protein scaffold.

3. degrees of compaction: DNA>nucleosome >solenoid >interphase >metaphase.

 XVIII. Regulation of Gene Expression A. DNA/Protein interactions and the control of transcription. (Hartwell, pp. 594-596)

B. Nuclease sensitivity experiments: "active" vs. "inactive" chromatin.

1. nuclease sensitivity experiments of beta globin gene: more suceptible to nuclease attack when genes are capable of transcription.

2. Exclusion of transcription apparatus from sites in DNA; nucleosome phasing and access to the promoter.

 C. The lac operon. (Hartwell, pp. 554-566) 1. Operon organization: polycistronic mRNA synthesized that encodes 3 distinct proteins involved in lactose metabolism and transport. Synthesis of common mRNA (and therefore all three proteins) controlled at transcriptional level.

2. Negative control, Repressor, and Operon Induction: In the absence of lactose, a repressor interacts with a DNA region (operator) near the lac promoter; precludes RNA polymerase from actively transcribing operon (negative control). Binding of small amounts of lactose analogs to the repressor protein will cause it to lose affinity for operator DNA. Removal of repressor protein from DNA will permit RNA polymerase to actively transcribe operon (induction).

3. Positive Control, Catabolite Activator Protein (CAP) and cAMP levels. RNA polymerase activity is stimulated by its interaction with a CAP protein/cAMP complex. Lac operon is not needed if glucose stores are high. High levels of glucose lead to low levels of cAMP in cell; stimulation of operon transcription (positive control) is therefore blocked when cells have sufficient levels of glucose.

4. Jacob and Monod. Genetic investigations led to operon model (1961).

a. operator mutations

b. repressor mutations

c. tests in merodiploids

d. biochemical isolations of repressor and operator (1967)

 I. Population Genetics and the Hardy-Weinberg Equation. (Thompson, Chapter 21; Hartwell, pp. 677-82) A. Within the gene pool of an organism, suppose that two alleles are present at a given locus, and that the frequencies of these two alleles (A and a) can be represented by the values p and q ( p + q = 1). There will be three genotypes present in the population:
AA    Aa     aa

The frequencies of these genotypes can be calculated if the frequencies of the alleles are known. The frequencies of the three genotypes will follow the binomial expansion, (p + q)2 = p2 + 2pq + q2

                                                                           Sperm
 




 A(p)  a(q)
Eggs

 A (p)  AA (p2)  Aa (pq)
 a (q)  Aa (pq)  aa (q2)



where p2 = the frequency of genotype AA, 2pq = the frequency of genotype Aa and q2 = the frequency of genotype aa.

B. The Hardy-Weinberg equation (i.e. relationship of gene frequencies and genotypes) holds only if certain assumptions are met: 1. mating is random; (no assortative or disassortative mating; or inbreeding)

2. there is no selection;

3. there is no migration;

4. there is a large population size; (i.e. a limited possiblity for chance fluctuatons in allele frequences; genetic drift)

5. there is no mutation.

 C. The Hardy-Weinberg equation can be used to estimate heterozygote frequencies for certain traits or diseases. e.g. In Caucasian populations, the frequency of cystic fibrosis cases is approximately 1/2500 individuals. Therefore:

q2 = 0.0004 (1/2500) and q = 0.02. The frequency of p = 1- q, or 0.98. The frequency of a heterozygote would therefore be 2(0.02)(0.98) = .0392. About 1 person in 25 is estimated to be a carrier in Caucasian populations.