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Transcript of Genetics Finals

Fundamental Genetics Lecture 8

Linkage and Chromosome Mapping in EukaryotesJohn Donnie A. Ramos, Ph.D.Dept. of Biological Sciences College of Science University of Santo Tomas

Linked GenesGenes located in the same chromosomes Initiated by Thomas Morgan and Alfred Sturtevant Will not segregate independently Affected by crossing over

Linkage vs Independent Assortment

Linked Genes in DrosophilaRed eyes (bw+) dominant to mutant brown eyes (bw) Thin wing veins (hv+) dominant to mutant heavy wing veins (hv)

X-Linked Genes in DrosophilaCross A Gray body (y+) dominant to mutant yellow body (y) Red eyes (w+) dominant to mutant white eyes (w) Cross B Red eyes (w+) dominant to mutant white eyes (w) Normal wings (m+) dominant to mutant miniature wings (m) Due to linkage and crossing over that occurred during meiosis Distance between linked genes is related degree of crossing over

Chromosome MappingDetermining the distances between genes and the order of sequence in a chromosome Uses the frequency of crossingThe shorter the distance between linked genes, the lower the frequency of crossing-over. The longer the distance, the higher the frequency of crossing over to occur.

Frequencies of crossing over:1. Yellow, white 2. White, Miniature 3. Yellow, miniature 00.5 % 34.0 % 35.4 %

1% of crossing over = 1 map unit (centimorgan, cM)

Distance Affects Crossover

Single Crossover

50% are recombinant gametes 50 % non-crossover gametes If distance between genes is more than 50 map units, ~100 % crossing over will occur.

Multiple Crossover

Occurrence of more than one crossovers between non-sister chromatids. Produces double crossover (DCO) gametes If the probability of crossover between A and B is 20% (0.20) and the probility of crossover between B and C is 30% (0.30), the frequency of DCO is 6 % (0.06) Product Law: (0.2)(0.3)=0.06

Three-Point MappingThe percentage of crossing over could be used to map genes in a chromosome Three criteria needed for successful mapping: Genotypes of organisms producing the crossover gametes must be heterozygous for all gene loci Cross must be constructed so that the genotypes of gametes could be determined based on the phenotypes of the offspring. Large number of offspring must be produced

Traits considered: 1. Body color Gray(y+) dominant to yellow (y) 2. Eye color Red eyes (w+) dominant to white (w) 3. Eye shape Normal (ec+) dominant to echinus (ec)

10,000

Determining Gene SequenceSteps: Determine 3 possible ordersw-y-ec (y at the middle) y-ec- w (ec at the middle) y-w-ec (w at the middle)

Perform a theoretical double cross over Compare the theoretical DCO with actual DCO (least no.) Perform theoretical NCOI and NCOII and compare with dataWhite echinus eyes, gray body Red normal eyes, yellow body Yellow body, normal white eyes Gray body, echinus red eyes Yellow body, echinus red eyes Gray body, white normal eyes

Unknown Gene Sequence

Total=1109

Unknown Gene Sequence

Not all crossovers can be detected

Degree of inaccuracy increases with increasing distance between linked genes

Observed vs Expected DCOObserved DCO = double cross-over that actually occurred Example: (44 + 42)/1109 = 0.078 Expected DCO = theoretical double crossoversProduct of all the SCOI and SCOII Example: (82+79+44+42) / 1109 = 0.223 (200+195+44+42) / 1109 = 0.434 DCO exp= (0.223)(0.434) = 0.097

Coefficient of Coincidence and InterferenceCoefficient of Coincidence (C)The measure of actual DCOs that occurred C = Observed DCO / Expected DCO = 0.078/0.097 = 0.804 or 80.4%

Interference (I)phenomenon when a crossover event in one region of a chromosome inhibits a second event to occur in a nearby region) I = 1-C = 1-0.804 = 0.196 or 19.6% Interpretation: 19.6% of expected DCO did not occur or only 80.4% of expected DCO was observed

Problem 1A stock of corn homozygous for the recessive linked genes colorless (c), shrunken (sh), and waxy (wx) was crossed to a stock of homozygous for the dominant wild type alleles of the genes (+ + +). The F1 plants were then backcrossed to the homozygous recessive stock. The F2 results were as follows: Phenotype +++ c sh wx + sh wx c++ a. b. c. d. e. Number 17,959 17,699 509 524 Phenotype + + wx c sh + + sh + c + wx Number 4,455 4,654 20 12

Determine the distance between the c and sh Determine the distance between the sh and wx Determine the distance between c and wx Give the coefficient of coincidence Compute for the interference

Problem 1: Solutionc sh wx c sh wx

xc sh wx + + +

+ + + + + + c sh wx c sh wx

x

NCO SCOI SCOII DCO

+ + + c sh wx + sh wx c + + + + wx c sh + + sh + c + wx Total

= = = = = = = = =

17,959 17,699 509 524 4,455 4,654 20 12 45,832

35,658 = 77.80 % 1,033 = 02.25 % 9,109 = 19.87 % 32 = 00.07 %

Problem 1: Solution

Distance between c and sh = (509 + 524 + 20 +12) / 45,832 = 0.0232 or 2.32 % Distance between sh and wx = (4466 + 4654 + 20 + 12) / 45832 = 0.1994 or 19.94 % Distance between c and wx = 2.32 + 19.94 = 22.26

Problem 1: Solutionc2.32 mu 22.26 mu

sh19.94 mu

wx

C = (0.0007) / (0.0232)(0.1984) = 0.1521 or 15.21% I = 1-C = 1-0.1521 = 0.8479 or 84.79 %

Problem 2In a variety of tomato plant, the mutant genes o (oblate fruit), h (hairy fruit), and c (compound inflorescence) are all located in chromosome 2. These genes are recessive to their wild type alleles round fruit, hairless and single inflorescence, respectively. A testcross mating of an F1 heterozygote for all three genes resulted in the following phenotypes: Phenotypes +++ +h+ o++ oh+ Number 73 2 110 306 Phenotypes ++c +hc o+c ohc Number 348 96 2 63

a. Determine the sequence of the 3 genes in chromosome 2 b. Give genotypes of the homozygous parents (P1) used in making the F1 heterozygote. c. Compute for the map distances between the genes d. Give the coefficient of coincidence and interference

Problem 2: Solution+++ ohc

x

ohc ohc

Inference from given data: Sequence of genes is not correct One chromosome contains 2 wild type alleles while the homolog contains the 3rd wild type allele

NCO: DCO:

o h + = 306 + + c = 348 +h+ = 2 o+c = 2

Three possible orders of the genes involved: o h c o c h h o cFind a sequence that will satisfy both NCO and DCO oh+ ++c +ch o++ +oh c++ Satisfies NCO but not DCO Satisfies DCO but not NCO Satisfies both NCO and DCO

Problem 2: Solution+oh c++

x

ohc ohc

Try if the sequence can satisfy the SCOs = 306 = 348 = 2 = 2 = = = = 73 63 110 96

NCO: DCO: SCO I: SCOII:

oh+ ++c +h+ o+c

(same as + o h) (same as c + +) (same as + + h) (same as c o +)

654 = 0.654 or 65.4% 4 = 0.004 or 0.4% 136 = 0.136 or 13.6% 206 = 0.206 or 20.6%

+++ o h c (same as c o h) o + + (same as + o +) + h c (same as c + h)

Total = 1,000

Problem 1: Solution

Distance between c and o = (73 + 63 + 2 + 2) / 1,000 = 0.140 or 14 % / cM Distance between o and h = (110 + 96 + 20 + 12) / 1000 = 0.210 or 21 % / cM Distance between c and wx = 14 + 21 = 35 cM

Problem 1: Solutionc14 cM 35 cM

o21.0 cM

h

C = (0.004) / (0.14)(0.21) = 0.1361 or 13.61% I = 1-C = 1-0.1361 = 0.8639 or 86.39 %

Fundamental Genetics Lecture 10

DNA Replication and SynthesisJohn Donnie A. Ramos, Ph.D.Dept. of Biological Sciences College of Science University of Santo Tomas

The Flow of Biological InformationReplicationDNA

TranscriptionRNA

TranslationProtein

Modes of DNA Replication

Semiconservative Replication

Semiconservative Replication in ProkaryotesMathew Messelson and Franklin Stahl (1958) heavy isotope of N (contains 1 more neutron) compared to 14N has high sedimentation rate in cesium chloride compared to 14N15N 15N

Semiconservative Replication in ProkaryotesExpected results of the Messelson-Stahl experiment

Semiconservative Replication in EukaryotesJ. Herbert Taylor, Philip Woods, and Walter Hughes (1957)

faba (broad bean)

Used root tip cells from Vicia Monitored replication using 3H-Thymidine to label DNA

Used autoradiography to determine the incorporation of 3H-Thymidine Arrested cells at metaphase using colchicine

Replication of E. coli PlasmidShown by John Cairns (1981) using radioisotopes and radiography Replication starts in a single OriC origin of replication (245 bp) Replication is bidirectional Replication fork unwound DNA helix Replicon replicated DNA Ter region region of replication termination

DNA Synthesis in MicroorganismsDNA polymerase I (928 aa) catalyses the synthesis of DNA in vitro (A. Kornberg, 1957) Requirements:Deoxyribonucleoside triphosphates, dNTPs (dATP, dCTP, dGTP, dTTP) DNA template Primer

Chain Elongation5 to 3 direction of DNA synthesis (requires 3 end of the DNA template) Each step incorporates free 3 OH group for further elongation

DNA replication using DNA polymerase is of high fidelity (highly accurate) With exonuclease activity (proofreading ability)

DNA PolymerasesAll 3 types requires a primer Complex proteins (100,000 Da) Functions of DNA polymerases in vivo DNA Pol I proofreading; removes primers and fills gaps DNA Pol II - mainly involved in DNA repair from external damage DNA Pol III main enzyme involved in DNA synthesisa holoenzyme (>600,000 Da) forms replisome when attached to a replication fork.

Replication in Prokaryotes1. 2. 3. 4. 5.Unwinding of DNA helix Initiation of DNA synthesis DNA synthesis proper (elongation) Sealing gaps Proofreading and error correction

Unwinding of DNA HelixTakes place in oriC (245 bp) repeating 9mers