Download This outline is designed to provide you with a general summary of

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
and lectures. Don’t expect to pass this course by confining
yourself to this summary- the best way to do well in the course
is to attend class, read the assignments, and ask questions.
Interspersed within this outline are references to readings in
the Hartwell text, where this subject material can be found.
The blue annotations (with a strikethrough symbol) are for
the older third edition, and the red annotations are for the
new fourth edition, or where both editions correspond.
I. Course Description
A. Text
B. Grading
C. Course Schedule
D. Response System/Registration
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: Pre-Mendelian concepts of inheritance
(Chapter 1 2, Hartwell)
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 (18221884) (Chapter 2, Hartwell)
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: principle 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 ratioapplying 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. 669, 672-677 573,
576-579 for detailed discussion of molecular
mechanisms in Drosophila.)
1. Drosophila: X:A ratio. Numerator/denominator
hypothesis.
2. Mammals: TDF/SRY gene on Y chromosome.
(See Hartwell, pp. 195, 200 194)
III. Patterns of Sex-Linked Expression in Humans. (Chapter 4,
Hartwell)
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. One form is
a mutation in the LH receptor expressed in Leydig
cells in the testes. See notes for model.
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. 127-131 122-125)
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 (Chapter 3, Hartwell)
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, 48-49; 60-61;
301-302; 336-339 46-48; 58-59; 290-291; 322-325)
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 (Chapter 5, Hartwell)
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 (nonparental) classes. Hypothesized that the nonparental 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 sporesrare 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 5-21 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
Total number
1
(PD)
AB
AB
ab
ab
108
2
3
4
(TT)
(PD)
(NPD)
AB
AB
aB
Ab
ab
aB
aB
AB
Ab
ab
ab
Ab
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.
C. Mitotic recombination, allele loss, and cancer. The Rb
locus as example (Hartwell pp. 154 148; 703-9 595-600)
VIII. Human Gene Linkage. (Hartwell, pp. 416-417; see
Thompson, Chapters 10-12); Lod scores are not discussed in
Hartwell 4e).
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 (θ ), 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):
Zθ = log 10 (Probability of family for θ = 0.01, 0.02, etc.)
(Probability of family for θ = 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. The Use of Linkage Analysis to Track and Identify Disease
Genes. (Hartwell pp. 359; 415-19 386-387)
A. Example: Secretor locus and Myotonic Muscular
Dystrophy. Knowing whether two genes are linked can be
of predictive value in testing for certain diseases…one gene
can serve as a marker or test for inheritance of the other.
B. Using polymorphism in the DNA itself can be of
predictive value in testing for certain diseases…the DNA
polymorphisms can serve as a marker to test for inheritance
of the linked disease gene.
C. Linkage, in combination with recombinant DNA
analysis, can then lead to the isolation of the disease gene
itself. e.g. Huntington Disease, cystic fibrosis, Duchenne
Muscular Dystrophy, etc.
XI. Aberrant Asci Patterns, Gene Conversion and Recombination
at the Molecular Level.(Hartwell, 191-200 186-193)
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. 191-200 186193 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 14 13)
2. Aberrations due to chromosome number (Hartwell,
Chapter 14 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. 84-85; 472-474 81-82; 337-340; 411-413)
A. Chromosomes are a complex of primarily DNA and protein
(chromatin). euchromatin: lightly staining, genetically active;
heterochromatin, darkly staining, genetically inactive. (Hartwell, 472-474
337-340; 411-413) 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. (pp. 472-474 337-340; 411-413)
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 see pp. 366-67 343-344); 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. 85-88; 110-112 83-85;
107-109)
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. 82-86; 481-482 80-84;
415-416)
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. Epigenetic Modifications of Chromatin Architecture are involved in
Mammalian X inactivation. (Hartwell pp. 659-60; 562-565)
1. X inactivation center (XIC); Xce: which X is active? XIST gene
in XIC involved in inactivation (RNA).
2. At the XIST locus there is transcription from X that will be
inactivated: the XIST RNA transcript appears to bind to
chromosome from which is is transcribed.
3. Methylation of DNA associated with inactivation.
4. Modification of histones leads to changes in chromatin
architecture and formation of facultative heterochromatin.
5. Not all of X chromosome is inactivated (c.f. Turner Syndrome,
Klinefelter Syndrome). Region of Xp expressed from both X
chromosomes (pseudoautosomal regions; also RPS4, near XIST)
D. Genomic Imprinting. (Hartwell, pp. 660-664 565-568)
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.
E. RNA Interference. (Hartwell, pp. 665-68, 670-72, 726-28; 569-72 57475)
1. Early 1990’s- originally characterized in nematodes and
associated with control of particular genes.
2. Nobel Prize awarded in 2006 to Andrew Fire and Craig Mello
for their discovery of a process through which microRNAs
effect control of gene expression- RNA induced Gene
Silencing…
3. 2000-2005 – micro-RNAs identified and characterized in
numerous organisms.
a. RNA interference – trans-acting single stranded microRNAs that regulate eukaryotic gene expression.
b. Number of miRNAs may exceed the number of protein
coding genes.
c. At least 1/3 of all human genes are estimated to be
microRNA targets.
4. Mechanism
a. miRNAs distinguished by 60-120 ribonucleotide-long
segments of reverse sequence complimentarity – will
snap back spontaneously into hairpin stem-loop
structures.
b. Best characterized in role during posttranscriptional
regulation and control at the translational level.
c. Immediately after transcription, pri-miRNAs are
recognized by enzyme which crops out pre-miRNA
stem loops from larger RNA.
d.
Pre-miRNAs undergo active transport from nucleus to
cytoplasm where they are recognized by enzyme
‘Dicer’.
e.
Dicer reduces the pre-miRNA into a short-lived
miRNA:miRNA duplex which is released and picked
up by RISC.
f. If miRNA and its target mRNA contain perfectly
complementary sequences, miRISC cleaves the mRNA.
RNase rapidly degrades cleavage product.
g. If miRNA and its target mRNA have only partial
complementarity, cleavage does not occur. miRISC
remains bound to its target and represses its movement
across ribosomes.
h. Evolutionarily, these systems seem to have evolved to
combat viral infections…But we can now use these
mechanisms to target/repress gene activity artificially
for research in gene function. RNAi can be used to
target and reduce gene expression in specific tissues
and it specific developmental contexts.
i. RNAi may have great value therapeutically and in
treatment of disease.
VI. Departures from Diploidy: Heritable Aberrations in Chromosome Number;
Humans. (Hartwell, Chapter 14 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 14 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 nonhomologous 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 "protooncogene" 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. 702-709 595-600)
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, 216-217 208209).
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 agricultural 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. 210-214 202-207)
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 themselves? 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 15)
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+ à 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. Cannot 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. (Hartwell pp. 475-477 418-420)
1. conserved "consensus" sequence: [(AT)nGm]x e.g. tetrahymena:
TTGGGG
2. redundant sequence: TTGGGGTTGGGGTTGGGGTTGGGG...
C. Telomere replication and telomerase.
D. Telomere replication and cell senescence.
VII. Other Variant Forms of DNA Replication
A. Rolling Circle Replication in certain viruses
B. Single Stranded DNA genomes (Hartwell, pp. 179-180 174)
C. Mitochondrial Replication (Hartwell, p. 584-586 499-501)
VI. RNA Genomes.
A. Retrovirus life cycles, and similarities/differences to transposable elements
(Hartwell, pp. 180-181, 184; 270-271; 315-318; 510-513; 704-706; 811-813
176-179; 260-261; 301-306; 447-453; 596-598; 707-708)
B. Reverse Transcriptase
VII. Applications of Bacterial Enzymes; Hybridization Analyses and Molecular Cloning.
(Hartwell, Chapter 9).
A. Labeling Techniques for Nucleic Acid Hybridizations/In Situ Hybridizations
(Hartwell, pp. 319-326 306-310).
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. 375-383; 439-456 348352; 715-727).
1. Automated DNA sequencing (e.g. ABI bioanalyzer vs. Roche 454)
2. Array Technology
XII. DNA à Protein (Hartwell, Chapter 7; pp. 232-239; 284-289 218-231; 274-275;
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 Structure of the Gene: The Bridge between Transmission
Genetics and Molecular Genetics. (Hartwell Chapter 7; pp. 224-232 216-224)
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 Structure.
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, 258-261
250-252)
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 13 12; pp. 466-472 406-411)
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. 658660 561-565)
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. 612-626 521-535)
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)
D. Eukaryotic Systems: Example: Endocrine Control of Gene Expression
(Hartwell, pp. 646-654; 554-560
1. The Nuclear Receptor Superfamily
2. Regulation of gene expression via steroid hormone signaling: Hormone
Response Elements and Promoter Activation
I. Population Genetics and the Hardy-Weinberg Equation. (Thompson, Chapter 21;
Hartwell, pp. 757-762 655-663)
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.