TOPOGRAPHY OF CHROMOSOMES

Cytological features of chromosomes

1. Chromosome size--->May have a large variation within an organism's genome.

2. Centromere--->Structure that attaches to spindle fibers during mitosis and meiosis.
(constricted appearance at metaphase)

A. Metacentric--->Centromere in middle.

B. Acrocentric--->Centromere off-center.

C. Telocentric--->Centromere at one end.

D. Acentric--->Chromosome fragment lacking a centromere. Lost since no spindle fiber attachment.

4. Nucleolar Organizer--->Clusters of ribosomal RNA (rRNA) genes (secondary constriction).

5. Euchromatin--->Active DNA (poor staining, lightly packed).

6. Heterochromatin--->Densely staining, highly compacted.

A. Constitutive--->Always inactive, permanent feature of chromosomal region.

B. Facultative--->Present or absent at various times (active or inactive).

8. Polytene Chromosomes--->Unusual feature of some dipterans (flies). (See Fig. 8-4)
Many rounds of replication, 1024 chromatids, characteristic banding pattern.
Balbiani rings or "puffs" are active regions of RNA synthesis. (i.e.) active DNA

CHROMOSOMAL REARRANGEMENTS

A. Deletions--->Loss of a region caused by a chromosomal break.

B. Duplication-->Reciprocal change of a deletion.

C. Inversion--->Region rotated 180°.

D. Translocation--->Exchange of parts of non-homologous chromosomes.

DELETIONS

Homozygous usually fatal, heterozygous often fatal (gene dosage).
Some small deletions viable as heterozygote.
Visualized as a deletion loop during meiosis.

Visualized by a change in matching banding patterns in polytene chromosomes.
Deletions can never revert to WT.

Pseudodominance--->Deletion will "uncover" recessive alleles on the other chromosome, thus the recessive phenotype is expressed.

Small deletions can be mapped due to pseudodominance, useful in correlating linkage and cytological maps. (Fig. 8-8)

Humans--->Usually caused by a new germinal mutation in one parent.
(e.g.) cri du chat syndrome (See Fig. 8-9)
Tip of chromosome 5 deleted

DUPLICATIONS

Can be adjacent to each other or the second copy may be in a novel location on the same or a different chromosome.

3 copies/cell in a diploid.

Duplication heterozygotes can result in unusual pairing structures at meiosis. (See Fig. 8-13)

Duplications can arise from breaking, adding new DNA, then rejoining OR by unequal crossing over (Fig. 8-15).

Usually difficult to detect phenotypically.

At meiosis a loop structure may be detected or extra repeats of certain bands may be seen.

Tandem Duplication---> ABCBCD

Reverse Duplication---> ABCCBD

Hemoglobin (Hb) gene family provides evidence for duplication generated by unequal crossing over.

Human homozygous duplications have never been detected (probably lethal).

Abnormalities due to gene dosage.

X-OVER DIAGRAMS

Follow one chromosome from one end, through

X-over point, to the end of the exchanged DNA.

INVERSIONS

Result in no net change of genetic material.

Generally viable without phenotypic abnormalities unless breakage occurs in an essential gene. Then they are lethal if homozygous.

Paired homologs form an inversion loop at meiosis if heterozygous (See Fig. 8-18).

PARACENTRIC INVERSION--->Centromere outside the inversion. (e.g.) AB_lCDE--->AB_lDCE

X-overs between paracentric inversion and a WT chromosome results in a dicentric and an acentric chromosome.

Acentric fragment is lost.

Dicentric breaks randomly in bridge. (Fig. 8-19)

PERICENTRIC INVERSION--->Inversion spans the centromere. (e.g.) AB_lCDE--->AC_lBDE

X-overs between a pericentric inversion and a WT chromosome result in products with a deletion and a duplication of different parts of the chromosome.

Results in partial sterility. (Fig. 8-20)

Two mechanisms reduce the number of recombinants among the progeny of inversion heterozygotes.
  A. Inviable deletion products.
    B. Inhibition of pairing in the inverted region.

Diagnostic features of inversions

1. Decreased recombinant frequency.

2. Inversion loops.

3. Partial sterility.

4. Inverted arrangements of chromosomal landmarks.

(e.g.) Centromere position
 ___l__________   _{Normal 4:1 ratio}
 ______l_______   _{Inversion 3:2 ratio}

2% of humans carry inversions.

TRANSLOCATIONS--->When 2 nonhomologous chromosomes mutate by exchanging parts.

Viable unless breakpoint is in an essential gene.

RECIPROCAL TRANSLOCATION---> Segment from one chromosome is exchanged with a segment from another nonhomologous chromosome so that two translocation products are generated simultaneously (most common). Fig. 8-26

Diagnostic features of translocations

1. Establishes new linkage groups
(i.e.) gene now on different chromosome

2. May alter size of chromosomes and centromere position.

3. Causes semisterility--->50% of plant gametes and mammalian zygotes are inviable.

HUMANS--->Always in heterozygous state.

Cri du chat syndrome if the offspring is missing the tip of chromosome 5 due to translocation.

Down syndrome if 2 normal chromosomes 21s and additional segment of 21 due to translocation. Results in high occurrence in family tree.

POSITION EFFECT VARIEGATION--->

When a gene is translocated to a region near heterochromatin of another chromosome.

In some cells the heterochromatin will engulf the gene shutting off expression leading to mutant phenotype.

Fig. 8-34

(e.g.) Burkitt's Lymphoma in humans

(See Fig. 8-35)
 
Chapter 9

MONOPLOID NUMBER--->Number of chromosomes in the basic set of an organism.

EUPLOID--->Organisms with multiples of the monoploid number.

POLYPLOID--->Euploid with more than 2 sets of chromosomes.

A. Monoploid = 1x
B. Diploid = 2x
C. Triploid = 3x
D. etc...

n = x for most organisms except some plants.

(e.g.) wheat x=7 6x=42
                    n=21 2n=42

Monoploid organisms--->male bees, wasps, ants.

Males develop parthogenetically from unfertilized eggs.

Meiosis is not possible (no pairing--->sterile).

Plant Engineering

Generate a monoploid plant from a diploid. Then generate a drug resistant diploid from monoploid.

(Figs. 9-1 and 9-3)

Colchicine inhibits mitotic spindle formation. After mutant selection and growth use for one cell division.

(See Fig. 9-2)

AUTOPOLYPLOIDS--->Multiple chromosome sets from within one species.

ALLOPOLYPLOIDS---> Multiple chromosome sets from closely related species.

Triploids (3x)--->Sterile--->Problems during meiotic segregation.
 

(4x) Tetraploid	X	(2x) Diploid
	(3x) Triploid

(e.g.) Seedless watermelons

AUTOTETRAPLOIDS--->Arise spontaneously from accidental doubling(2X-->4X)or use colchicine.

Advantages: larger plant and fruit

Polyploidy in animals

Reproduction is parthogenetic.

(e.g.) leeches, brine shrimp

Common in amphibians, reptiles, salmon, trout.

Oysters (3x)-->No spawning-->palatable all year.

Most human triploids are spontaneously aborted.

If born, none survive.

ANEUPLOIDY--->Number of one or more chromosomes change during formation of an individual.

Caused by nondisjunction during meiosis (Fig. 9-16).

If (n-1) gamete fertilizes zygote--->monosomic for that particular chromosome (2n-1).

If (n+1) gamete--->trisomic (2n+1).

(2n-2)--->nullisomic

(2n+1+1)--->double trisomic

Monosomics (2n-1)--->Deleterious

1. Missing chromosome disturbs overall balance of chromosomes--->disturbs homeostasis.

2. Hemizygous for that chromosome--->deleterious because recessive alleles expressed phenotypically.

Humans--->About 10% of all human conceptions have a major chromosome abnormality. Most are spontaneously aborted.

TURNER SYNDROME--->44 autosomes with

1X chromosome (1/5000 females).

Sterile, normal intelligence.

ALL MONOSOMICS FOR AUTOSOMES ABORT.

KLINEFELTER SYNDROME--->XXY

(1/1000 males) Lanky builds, retarded, sterile

XYY--->Agressive behavior? (1/1000 males) Fertile.

DOWN SYNDROME--->Trisomy 21

(1.5/1000 births) Most common human aneuploid.

More common than translocation form.

No family history.

Older mothers at greater risk.

Mental retardation.

Females may be fertile producing normal and trisomic children.

Males infertile.

PATAU SYNDROME---Trisomy 13

EDWARD SYNDROME--->Trisomy 18

SOMATIC ANEUPLOIDS--->Aneuploid cells that arise spontaneously in somatic tissue
(genetic mosaics).

(e.g.) Sexual mosaics---> XO/XYY

Caused by XY zygote in which the Y chromosome fails to disjoin at an early mitotic division.

The phenotypic sex depends on where the male and female sectors end up in the body.

If nondisjunction occurs later in development a three-way mosaic arises XY/XO/XYY

Others:

XO/XY--->Probably due to loss of chromosome in male zygote.

XX/XY--->Probably due to a double fertilization (fused twins).

ACUTE MYELOID LEUKEMIA--->aneuploid for either chromosome 8, 9, or 21.

Chapter 10

PROKARYOTES-->Eubacteria and archaebacteria

Divide by binary fission--->genetically identical progeny. (Fig. 10-1)

Single, circular chromosome.

No mitosis, meiosis, nucleus.

Haploid.

BACTERIOPHAGE (PHAGE)--->bacterial virus.

Not free living--->parasitize bacteria to replicate.

CONJUGATION--->Transfer of DNA from one cell to another via direct cell to cell contact.

Requires the fertility (F) factor--->small, circular DNA element--->mini chromosome, can replicate.

Cells with F are F⁺; cells without F are F^-.

F⁺ cells (donor) produce pili which attach to

F^- cells (recipient).

Donor transfers a copy of F to the recipient.

F always remains in donor (i.e.) replication.

F can integrate into bacterial chromosome(Hfr strain).

Hfr (high frequency of recombination) strains can transfer chromosomal genes which recombine with recipient chromosome generating recombinants.

F'--->Integrated F can pop out of chromosome and carry host genes in addition to F. F' can transfer bacterial genes to recipient cell--->recombinants.

(Figs. 10-2, 10-3)

GRADIENT OF TRANSFER--->Less likely for a recipient to receive later genetic markers (genes) due to chromosome breakage.

Strain A--->met^- bio^- thr⁺ leu⁺ thi⁺     Neither strain
                                                             can grow on minimal media
Strain B--->met⁺ bio⁺ thr^- leu^- thi^-

Mix A and B--->growth on minimal (i.e.)prototrophic

Recombination--->transfer only in one direction.

DETERMINING LINKAGE-->Interrupted mating
 
 

Donor

Recipient

Hfr (str^s a⁺ b⁺ c⁺ d⁺) x F^- (str^r a^- b^- c^- d^-)

Time course, remove and plate samples on rich medium containing streptomycin. Drug prevents survival of donor cells.

Test (screen) surviving cells for transferred genes by replica plating. (Fig. 10-6)

ORIGIN--->Fixed point on donor chromosome where gene transfer starts (Hfr transferred last).

Orientation of integrated F determines polarity (direction) of transfer.

First gene transferred depends on orientation and where F is integrated.

Origin--->terminus (chromosome usually breaks before complete transfer)

Several Hfr strains exist depending on position and orientation of F. (Fig. 10-8)

Bacterial Mapping (E. coli)

Matings result in partial diploid (merodiploid).

Transferred genes must recombine with recipient chromosome.

Results in exchange of genetic material--->requires double X-over (Not a reciprocal change)

(Fig. 10-12)

(Fig. 10-11)
 

Determining Gene Order--->(p. 287)

Hfr str^s met⁺ arg⁺ aro⁺ his⁺ x F^- str^r met^- arg^- aro^- his^-

1. plate on minimal + all supplements except methionine (i.e.) select for met⁺

2. screen for other genes by replica plating
 

100% met+  60% arg+  20% aro+  4% his+

Gene Order  met arg aro his

High Resolution Mapping (Fig. 10-13)

Hfr str^s leu⁺ arg⁺ met⁺ x F^- str^r leu^- arg^- met^-

--------------------------> --------------------

1. select for last gene (i.e.) leu⁺

Thus, every transferred fragment has all 3 genes.

2. screen for the other genes by replica plating

1 map unit=1% X-over in the interval

(Fig. 10-13)

5% leu⁺ arg^- met^- (5 map units)
5% leu⁺ arg⁺ met^- (5 map units)
90% leu⁺ arg⁺ met⁺

And, rare quadruple X-overs leu⁺ arg^- met⁺ (determines the middle gene) Product rule (0.05)(0.05)=0.0025=0.25%

BACTERIAL TRANSFORMATION---> Conversion of one genotype to another by introducing chromosomal DNA.

Requires recombination (similar to Hfr x F^- cross).

If two genes are close to each other both can be transferred on the same DNA fragment.

If far apart, requires two DNA fragments (rare).

Product rule (e.g.) (0.01)(0.01)=0.0001=0.01%

BACTERIOPHAGE GENETICS

Nucleic acid (DNA), protein coat

(e.g.) l, T4, T7, P1 (Fig. 10-17)

Phage attach to bacteria and eject DNA into cell.

Note: there is no experimental evidence for the syringe injection model.

Genetic information directs cell to make more phage particles leading to cell lysis. Progeny infect more bacteria leading to plaque formation.

PLAQUE--->Cleared area in a lawn of bacteria caused by cell lysis.

Plaque morphology and host range are heritable characteristics.

PHAGE CROSS (T2)--->circular genome

(See Figs. 10-20 through 10-24)

h^- Can infect E. coli strains 1 and 2.
h⁺ Only infects strain 1.
r^- Rapid lysis (large plaques)
r⁺ Slow lysis (small plaques)

h^- r⁺ x h⁺ r^-

1) Mix phage and double infect E. coli strain 1.
2) Collect phage progeny (phage lysate).
3) Analyze by plating on a lawn of a mixture of

E. coli strains 1 and 2.

A) h^- r⁺ (clear, small)--->parental
B) h⁺ r^- (turbid, large)--->parental
C) h⁺ r⁺ (turbid, small)--->recombinant
D) h^- r^- (clear, large)--->recombinant

1 m.u.=1% recombinants

Experiment Table 10-3 (3 different r genes)

                                                % of each genotype
 

cross	r- h+	r+ h-	r+ h+	r- h-
1) ra-h+ x ra+ h-	34	42	12	12
2) rb-h+ x rb+ h-	32	56	5.9	6.4
3) rc-h+ x rc+ h-	39	59	0.7	0.9

 
Parentals not equivalent in number but recombinants are due to reciprocal exchange.

1) 12+12=24 m.u.
2) 5.9+6.4= 12.3 m.u.
3) 0.7+0.9= 1.6 m.u.

r_a                           r_b                r_c      h                      
            11.7                  10.7         1.6

r_a                                        r_c     h                      r_b      
               22.4                        1.6         12.3

r_a                 r_b                          h      r_c                      
         11.7                  12.3            1.6

r_a                                               h     r_c             r_b      
         24                                       1.6        10.7

Need to do two of the three following crosses to determine which map is correct.

r_a^- r_b⁺ x r_a⁺ r_b^-
r_a^- r_c⁺ x r_a⁺ r_c^-
r_b^- r_c⁺ x r_b⁺ r_c^-

VIRULENT PHAGES--->Always lytic

TEMPERATE PHAGES--->Lytic or lysogenic

LYSOGENY--->Phage infected bacteria not leading to lysis.

Lysogenic strains carry a prophage (phage DNA with lytic functions turned off). Can be integrated or as a separate circle. UV light or chemicals can induce prophage to take the lytic cycle. (See Fig. 10-26)

TRANSDUCTION--->Ability of phages to transfer bacterial genes from one cell to another by packaging bacterial instead of phage DNA by mistake.

GENERALIZED TRANSDUCING PHAGE-->

Can carry any part of bacterial chromosome.

SPECIALIZED TRANSDUCING PHAGE--->

Can carry only specific chromosome region.

GENERALIZED TRANSDUCTION--->

When a cell is infected by a non-lysogenic phage the bacterial chromosome is broken up. Phage can package pieces of bacterial DNA instead of phage DNA. (e.g.) P1 (Fig. 10-30)

Transducing phage ejects bacterial donor DNA into a recipient strain--->merodiploid--->recombination.

Linkage data from P1 transduction

Genes must be close enough for P1 to transduce them in a single piece of DNA.

1) (Donor) P1 met⁺ arg⁺ x met^- arg^- (Recipient)
2) Select for met⁺ (plate on minimal with arginine)
3) Screen for the % of met⁺ that are also arg⁺.

Strains transduced to met⁺ arg⁺ are cotransductants.

The greater the cotransduction frequency the closer the genes are to each other.
 

(Donor) P1 leu+ thr+ azir x leu- thr- azis (Recipient)

 

Experiment	Selected  markers	Unselected  markers
1	leu+	50% azir 2% thr+
2	thr+	3% leu+ 0% azir
3	leu+ thr+	0% azir

Expt. 1 Tells us leu⁺ is closer to azi^r than thr⁺

thr                   leu azi    OR    thr                azi leu

Expt. 2 Tells us thr⁺ closer to leu⁺ than azi^r

thr                   leu azi

Expt. 3 All 3 genes can't be cotransduced.

SPECIALIZED TRANSDUCTION (e.g.) l

Integration by a single X-over between gal and bio.

Useful for moving specific genes. (i.e.) gal or bio

(Fig. 10-31)

Mapping genes on E. coli Chromosome

1) First use Hfr strains to map genes within about 10-15 min.

2) P1 transduction within the interval established by Hfr conjugation.

REMEMBER--->Conjugation, transformation and transduction require a double X-over to replace DNA in the bacterial chromosome.

(Fig. 10-35)

Chapter 11
 

TRANSFORMING PRINCIPLE

Streptococcus pneumonia

Causes pneumonia in humans, lethal in mice.

(Frederick Griffith-1928) (Fig. 11-1)

Strain 1--->(S)--->normal, virulent, smooth
Strain 2--->(R)--->mutant, avirulent, rough

1) Boil S (dead)
2) Mix boiled S with live R--->inject mice
3) Mice die
4) Live S cells recovered from dead mice.

Interpretation--->cell debris from S converted live R to live S.

This process is called transformation.

What is the transforming principle?

(Avery, MacLeod and McCarty-1944)

Separated various classes of compounds found in boiled S and tested each class for the ability to transform R--->S.

Only DNA caused transformation.
DNA encoded the smooth phenotype.
First demonstration that genes are composed of DNA.
Many scientists still thought protein was the hereditary material.

(Hershey & Chase blender experiment-1952)

(Fig. 11-3)

Used phage T2

Phosphorous (P) in DNA not protein.
Sulfur (S) in protein not DNA.

Incorporate ³²P into T2 DNA or ³⁵S into T2 protein.

Infect E. coli.
³²P ejected by phage not ³⁵S.

Conclusion--->DNA is the hereditary material.
 
STRUCTURE OF DNA

DNA composed of 4 nucleotides each containing a different nitrogenous base linked to an identical deoxyribose sugar and a phosphate. (Fig. 11-4)

Adenine (A)--->purine
Guanine (G)--->purine
Cytosine (C)--->pyrimidine
Thymine (T)--->pyrimidine

1) Chargaff's rules

A) T+C=G+A
B) T=A; G=C
C) A+T?G+C

3) Jim Watson & Francis Crick solved the DNA structure in 1953. (Nobel prize)

Double helix.
Each helix composed of two antiparallel strands.
(i.e.) 5'--->3' and 3'--->5'

Each strand composed of nucleotides held together by phosphodiester bonds formed between the phosphate from one nucleotide and the deoxyribose sugar from the neighboring nt.

A from one strand pairs with T in the other strand by 2 hydrogen bonds.
G pairs with C with 3 hydrogen bonds.
Purines pair with pyrimidines. (Fig. 11-5, 11-8).

One strand is complementary to the other.
B-form DNA, right hand helix.

Major and minor grooves.

Implications of DNA structure

1) Suggested a way for DNA replication.
2) Suggested a genetic code in DNA would specify the amino acid sequence in proteins (chapter 13).

DNA REPLICATION

(1958)

DNA replication is semiconservative.
(i.e.) Following replication each daughter duplex contains one parental and one newly synthesized strand. (Fig. 11-13)

Parental strands of duplex must be unwound which requires breaking hydrogen bonds.

Each parental strand serves as a template or mold for synthesis of its complementary strand. (Fig. 11-12)

REPLICATION FORK--->Region of unwound DNA where active replication is taking place.

Complementary base pairing provides the basis of fidelity of DNA replication. (i.e.) each template base dictates the complementary base in the new strand.

Replication is a complex process that requires DNA template, nucleotides, DNA polymerase, and several other enzymes and protein factors.

MECHANISM OF DNA REPLICATION

Refer to Fig. 11-21

DNA POLYMERASE--->catalyzes the condensation reaction that incorporates each dNTP into the growing DNA strand one nt at a time.

Arthur Kornberg--->Nobel prize (late 1950's) for identifying DNA polymerase I from E. coli.

Template + dATP, dCTP-----> Replicated DNA
DNA dGTP, dTTP

dNTP--->dNMP incorporation + pyrophosphate (P-P)

E. coli

1) DNA Polymerase I---> 3 enzyme activities

A) polymerase activity (5'--->3')
B) 3'--->5' exonuclease activity that removes mismatched base pairs (bp)
C) 5'--->3' exonuclease activity that degrades dsDNA.
 

3) DNA polymerase III--->Major polymerase for chromosomal replication.

Contains > 20 different protein subunits.

INITIATION--->DNA pol requires a short DNA or RNA primer for initiation. Primer creates a short duplex.

DNA pol adds nts to the primer to initiate replication.

ORIGIN OF REPLICATION--->Fixed point where DNA replication begins.

Factors required to disrupt H-bonds (dnaA protein) and stabilize ssDNA (ssb).

PRIMASE--->Synthesizes RNA primer (» 30 nt) complementary to the DNA resulting in a duplex.

RNA primer extended by pol III.
Can't initiate without the RNA primer (Fig. 11-26).

BIDIRECTIONAL REPLICATION---> Replication proceeds in both directions, thereby synthesizing both strands ending at the terminus. (Fig. 11-23)

Since DNA pol synthesizes 5'--->3'; DNA pol moves along template 3'--->5'.

LEADING STRAND--->synthesized continuously

LAGGING STRAND--->synthesized in short, discontinuous fragments (Okazaki). (Fig. 11-27)

Lagging strand grows in opposite direction of the replication fork.

pol I fills in ssDNA gaps and removes RNA primer with its 5'--->3' exonuclease.

DNA LIGASE--->Seals the nicks after pol I removes primers.

DNA HELICASES--->Disrupts H-bonds holding the two strands together at the replication fork.

TOPOISOMERASES--->Create or relax supercoils in DNA.

EXONUCLEASE EDITING--->pol I and pol III possess 3'--->5' exonuclease that serves a proofreading and editing function by removing mismatched bases incorporated by mistake.

Proofreading greatly enhances the fidelity of DNA replication (See Fig. 11-34).

DNA replication takes » 40 min in E. coli.

EUKARYOTIC DNA REPLICATION

Similar but more complex.

Takes » 1.4 hr (yeast) to » 200 hr in some cells.

Must coordinate replication of several chromosomes.

Replication occurs in S phase of cell cycle.

Multiple points of origin in each chromosome

(» 400 for 17 yeast chromosomes).

Chapter 12

Biochemical reactions are catalyzed by enzymes that have a 3-D structure that is crucial for their function.

Genes specify the structure of proteins, some of which are enzymes \ genes determine phenotypes of the cell and organism.

ONE GENE-ONE ENZYME HYPOTHESIS

Each gene specifies a particular enzyme.
 

Gene A	Gene B	Gene C	Gene D
Enz. A	Enz. B	Enz. C	Enz. D
1------------>	2 -------->	3 --------->	4 ---------> 5

Genes control biochemical reactions by controlling the production of enzymes.

If one step is blocked by a mutation (inactive enzyme) the cell can't make the final product.

If gene C is mutated the cell will survive if supplied with product 4 or 5.

Some enzymes are composed of several different polypeptides.

All proteins are not enzymes (structural, regulatory).

\ One gene-one protein (polypeptide)

PROTEIN STRUCTURE

Proteins are macromolecules composed of amino acids (AAs)
 

R
H2N-C-COOH	--->side chain
H

20 different AAs each with a different R group.

(See Table 12-3)

AAs linked by peptide bonds.

Condensation reaction (removal of water). (Fig 12-3)

POLYPEPTIDE--->Several linked AAs.
Proteins are large polypeptides.

PRIMARY (I°) STRUCTURE--->Linear array of AAs in a polypeptide (See Fig. 12-4)

SECONDARY (II°) STRUCTURE--->The interrelationships of AAs close to each other in the linear sequence.

Polypeptides fold into repeating structures by forming H-bonds between the backbone C=O & N-H groups of different residues.

a-helix (Fig.12-5) b-sheet (Fig. 12-6)

TERTIARY (III°) STRUCTURE--->3-D architecture of a polypeptide generated by H-bonds, electrostatic, VanDerWalls and disulfide bridges between AA R-groups causing the protein to fold up. (Fig 12-8)

QUATERNARY (IV°) STRUCTURE--->Two or more folded polypeptides bound together forming a complex.

GLOBULAR PROTEINS--->Compact due to folding. (e.g.) enzymes

FIBROUS PROTEINS--->Less folding and less compact. (e.g.) hair, muscle

Gene mutations cause altered proteins.

A change in 1 AA can alter or destroy protein function. Some changes don't have an effect.

(e.g.) Hb^A--->wild type

Hb^S--->sickle cell anemia (Fig. 12-13)

Since genes determine the specific primary AA sequence of proteins, they determine II°, III°,IV°.

COLINEARITY OF GENES & PROTEINS

Charles Yanofsky (Early 1960s)

The linear sequence of nts in a gene determines the linear sequence of AAs in a protein.

5'-->3' gene; N-terminus--->C-terminus protein

ENZYME FUNCTION--->Catalyze chemical reactions by breaking and making chemical bonds. (Actually they lower the activation energy)

Substrate molecule(s) fits into a precise notch of the enzyme called the active site.

Lock and key model (See Fig 12-19)
 

AA changes in the active site usually results in inactivity.

Many human diseases are caused by blocks in biochemical pathways. (Fig. 12-20)

(See Table 12-4)

Genetics Explained by Biochemistry

A) Temperature Sensitive Alleles
(Substitution of an AA)

1) Permissive Temperature--->Produces a functional protein.
2) Non-permissive temperature--->Results in a non-functional protein.

(See Fig. 12-21)

B) Genetic Ratios
 

P	AAbb   x   aaBB  (white)     (white)
F1	AaBb (purple)
F2	9 A-B- (purple)
	3 A-bb (white)	9:7 ratio
	3 aaB- (white)	Complementation
	1 aabb (white)
	A allele	B allele

 

white		white		purple
precursor	------------>	precursor	----------->	pigment
1	enzyme A	2	enzyme B

C) Recessive allele--->produces a non-functional enzyme (a).

D) Dominant allele--->produces a functional enzyme (A).

E) Dominant Mutation (negative-dominant)

Non-functional because of one mutant subunit.
 

Prion Diseases (Stan Prusiner-Nobel Prize-1997)

Degenerative central nervous system disorders.

Mad cow disease
Scrapie in goats
Creutzfeldt-Jakob disease

Misfolded prion protein is infectious. Causes other prion proteins to misfold leading to toxicity.

(See Fig. 12-25)

Genetic Fine Structure--->A gene can be subdivided into a linear array of sites that can be mutated and recombined with the smallest site being a single nt (bp).

(i.e.) X-overs can occur between any nt within a gene, not just between genes.

E.coli B--->1) WT phage T4: small, ragged plaques
                   2) rII phage T4: large, round plaques
E.coli K--->1) WT phage T4: small, ragged plaques
                   2) rII phage T4: No phage growth

Experiment (Figs. 12-29 & 12-30)

1) Infect strain B with 2 different rII mutant phages.

2) Both phage chromosomes can be ejected and some recombine generating a WT and a double mutant phage.

3) Only WT recombinants from doubly infected strain B phage lysate can infect strain K and form plaques.

Very small R.F. can be detected because  10⁹ phage/ml in a T4 lysate. Can detect one recombinant in 10⁹ phage (1/billion).

Can't analyze 1 billion Drosophila.

Deletion Mapping

Can map deletions just like point mutations.

If no WT recombinants are generated then the deletions overlap.

(e.g.)          ^{______________________________} (gene X)
Deletion A ^____________
Deletion B                          ^_________
Deletion C                                      ^_________

WT can be generated with a cross between A and B or A and C but not B and C.

Point mutations can be mapped with respect to previously mapped deletions.

Deletion A ^____________

Point #1               x

Point #2                                                 x

WT can be generated in cross with point #2 but not point #1.

Point #1 lies in the region delineated by the deletion.

  Exam 2 from 1998

1. The following cross is made between two strains of E. coli:

Hfr trp^- met^- glu⁺ str^s x F^- trp⁺ met⁺ glu^- str^r

Previous interrupted mating experiments demonstrated that glu⁺ enters the recipient strain last. Thus, in the cross glu⁺ recombinants are selected on minimal medium containing tryptophan (trp), methionine (met) and streptomycin (str). These recombinants are screened for the presence of trp^- and met^-. The number of colonies found for each genotype are shown below. Based on this information, which of the following linkage maps is correct?
 
 

trp+ met- glu+	0
trp- met- glu+	255
trp- met+ glu+	27
trp+ met+ glu+	18

A. glu         6         trp             9             met

B. glu         9         trp             6             met

C. trp         6         glu             9             met

D. glu         6         met            9             trp

E. glu         9         met             6             trp
 

2. Five Hfr strains A through D are derived from a single F+ strain of E. coli. The following chart shows the order of entry of the first four genes into an F- strain when each is used in an interrupted-conjugation experiment.
 
 
 

A	B	C	D
mal	his	met	pro
trp	gal	xyl	gal
ser	pro	mal	his
ade	met	trp	ade

 

Which of the following is the order of the genes on the circular chromosome of the original F+ strain?

3. Fill in the blanks of the following statement.

A ___________ crossover between a bacterial chromosome and a linear chromosome fragment results in a ___________ chromosome.

A. single; linear
B. double; linear
C. single, circular
D. triple; circular
E. quadruple; linear
 
 

For questions 4 and 5 match the term with the appropriate response.

4. Specialized transducing phage

5. Temperate phage

A. A phage that requires a specific temperature for infection.
B. A phage that is always lytic.
C. A phage that can transduce any region of a bacterial chromosome.
D. A phage that can be lytic or lysogenic.
E. A phage that can only transduce a particular region of a bacterial chromosome.
 
 

6. Which of the series of events outlines an appropriate, although incomplete, order of events that occur during DNA replication?

A. relaxation of positive supercoils-dnaA binding-primer synthesized-replication by DNA polymerase-primer removal-DNA ligase seals nicks

B. relaxation of positive supercoils-ssb binding-dnaA binding- disruption of hydrogen bonds at the replication fork-replication by DNA polymerase-primer removal

C. dnaA binding-ssb binding-primer synthesized-replication by DNA polymerase-primer removal-DNA ligase seals nicks

D. ssb binding-dnaA binding-primer synthesized-replication by DNA polymerase-primer removal-DNA ligase seals nicks

E. disruption of hydrogen bonds at the origin-dnaA binding-ssb binding-primer synthesized- replication by DNA polymerase-primer removal
 
 

7. While on a space expedition you discover a planet with microscopic life forms that you bring back to earth. With a great deal of effort you learn how to propagate the organisms on petri dishes and in liquid culture. After harvesting several grams of the bacterial-like growth you fractionate the cells and purify a substance that has the transforming principle. After further experimentation you determine that this substance has identical properties as DNA on earth except that the four nitrogenous bases have different chemical structures. Since you discovered the molecular structures you named them We-ine (W), Are-ine (A) Penn-ine (P) and State-ine (S). You find that the quantity of A+S=P+W; A=W; P=S; A+W?P+S. Based on what you know about the chemical properties of DNA, which of these bases are likely to basepair with one another?

A. W with A, P with S

B. W with P, A with S

C. W with S, P with A

D. None of the above

E. All of the above

8. Which of the following processes or concepts are not involved in exonuclease editing during DNA replication?

A. DNA pol III uses its 3'--->5' exonuclease activity.

B. DNA pol III uses its 5'--->3' exonuclease activity.

C. Editing results in the removal of mismatched bases.

D. Proofreading enhances the fidelity of DNA replication.

E. All of the above are involved in exonuclease editing.
 
 

9. Fill in the blanks of the following statement.

____________disrupts hydrogen bonds at the replication fork, whereas____________disrupts hydrogen bonds at the origin of replication.

A. DNA helicase; DNA helicase

B. DNA gyrase; DNA helicase

C. DNA helicase; dnaA protein

D. topoisomerase; dnaA protein

E. topoisomerase; topoisomerase
 
 

10. Fill in the blanks of the following statement.

____________relaxes supercoils in front of the replication fork, whereas____________creates supercoils in the newly synthesized daughter duplexes.

A. topoisomerase; DNA helicase

B. DNA gyrase; DNA helicase

C. DNA helicase; dnaA protein

D. topoisomerase; dnaA protein

E. topoisomerase; topoisomerase
 
 

For questions 11 and 12 match the type of molecular interaction in proteins with the appropriate level of protein structure.

11. disulfide bridge

12. complex of three polypeptides

A. I° structure
B. II° structure
C. III° structure
D. IV° structure
E. V° structure
 
 

13. Which of the following statements are false?

A. The one gene-one enzyme hypothesis had to be modified because some enzymes are encoded by more than one gene.

B. The nucleotide sequence of a gene dictates the final structure of a protein.

C. The 5’--->3’ nucleotide sequence of a gene is colinear with the N-terminus--->C-terminus of a polypeptide.

D. DNA polymerase I contains 5’--->3’ polymerization, 5’--->3’ exonucleolytic and

3’--->5’ exonucleolytic activities.

E. All of the above statements are true.
 
 

14. Which of the following statements is false?

A. Down syndrome can be caused by aneuploidy.

B. Down syndrome can be caused by translocation.

C. The occurrence of Klinefelter syndrome can be attributed to non-disjunction during gamete formation of the male or female parent.

D. The occurrence of "aggressive man" syndrome (XYY) can be attributed to non-disjunction during gamete formation of the male or female parent.

E. All of the above statements are true.
 
 

15. Seven bands in a salivary gland chromosome of Drosophila are shown in the figure below, along with the extent of 6 deletions (Del 1 to Del 6). Recessive alleles a, b, c, d, e, f, and g are known to be in the region, but there order is unknown. When the deletions are combined with each allele, the results in the table shown below were obtained. Note that a - means that the deletion is missing the corresponding wild type allele, while a + means that the corresponding wild type allele is still present. The gene order (left to right) is...

A. f, a, b, c, g, e, d

B. f, a, c, b, g, e, d

C. a, c, b, g, e, d, f

D. f, a, c, b, g, d, e

E. a, b, c, f, g, e, d
 
 

16. Which of the following statements are characteristic of inversions?

A. Crossovers between a pericentric inversion and a wild type chromosome results in a dicentric and an acentric chromosome.

B. Inhibition of chromosome pairing in the inverted region leads to a reduction in the number of recombinant products among the progeny of inversion heterozygotes.

C. Semisterility

D. Crossovers between a paracentric inversion and a wild type chromosome results in products that contain a duplication and a deletion of different regions of the chromosome.

E. All of the above.

For questions 17 and 18 match the term with the appropriate response.

17. Euchromatin
18. Nucleolar organizer

A. Tips of chromosomes
B. Site of spindle fiber attachment
C. Cluster of rRNA genes
D. Inactive DNA
E. Active DNA
 
 

19. The four gametic products from meiosis are: n+1, n+1, n-1, n-1. What does this indicate?

A. Translocation during meiosis I

B. Polyploidy

C. Colchicine contamination

D. Non-disjunction during meiosis I

E. Non-disjunction during meiosis II
 
 

20. Fill in the blanks of the following statement.

Parthogenetic development is characteristic of_____________and_____________animals.

A. monoploid; diploid

B. haploid; triploid

C. haploid; monoploid

D. triploid; tetraploid

E. triploid; monoploid
 
 

21. To demonstrate linkage of two genes A and B by transformation, one needs to demonstrate....

A. transformation of A=transformation of B.

B. transformation by A and B is less than the product of their individual transformation frequencies.

C. transformation of A and B is greater than the product of their individual transformation frequencies.

D. transformation of A and B is less than the sum of their individual transformation frequencies.

E. transformation of A and B is greater than the sum of their individual transformation frequencies.
 
 

22. In a transduction experiment using phage P1, the donor is x+ y+ z- and the recipient is x- y- z+. The donor y+ allele is initially selected after transduction. The y+ colonies are screened for the presence of the other alleles by replica plating. The results are shown below. What is the contransduction frequency for y+ and x+; y+ and z-?

x+ z- 11

x+ z+ 3

x- z- 14

x- z+ 22

A. 6%; 28%

B. 50%; 72%

C. 72%; 34%

D. 34%; 6%

E. 28%; 50%

23. Which of the following statements is false?

A. Prion diseases are caused by misfolded proteins.

B. Crossovers can occur between any two nucleotides in a chromosome.

C. Enzymes catalyze reactions by lowering the activation energy of the reaction.

D. The Hb^S allele contains a one nucleotide deletion.

E. All of the above statements are true.
 
 

24. Fill in the blanks of the following statement.

DNA contains____________bonds, whereas polypeptides contain____________bonds.

A. hydrogen; phosphodiester

B. phosphodiester; peptide

C. phosphomonoester; hydrogen

D. phosphodiester; dipeptide

E. phosphomonoester; peptide
 
 

25. The following structure observed during meiosis of a heterozygote is characteristic of a/an____________?

A. Pericentric inversion

B. Paracentric inversion

C. Translocation

D. Duplication

E. Deletion