Genetics Notes week 5
Genetics Notes week 5 Bios 206
Popular in Genetics
verified elite notetaker
Popular in Biological Sciences
verified elite notetaker
This 6 page Class Notes was uploaded by Becca Sehnert on Friday February 12, 2016. The Class Notes belongs to Bios 206 at University of Nebraska Lincoln taught by Dr. Christensen in Fall 2016. Since its upload, it has received 26 views. For similar materials see Genetics in Biological Sciences at University of Nebraska Lincoln.
Reviews for Genetics Notes week 5
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 02/12/16
WEEK 5 NOTES MONDAY Interference affects recovery of multiple exchanges • Expected frequency of double crossovers between 2 genes can be predicted from distance between them • Interference reduces expected # of mult crossovers when a crossover event in one region of chromosome inhibits a second event nearby o Any chromosomes can be involved in exchanging DNA o Not truly random, cuz measurable interference o 2 events happening simultaneously you multiply • Negative interference –more likely to happen in clusters Coefficient of Coincidence • Observed # of DCOs divided by expected number of DCOs o If =1, means random o C is more than 1 –happens more often than expect • Interference = 1-C • Positive interference means fewer doubles than expected • Negative interference means more Parentals Doubles Map order sc—v—s or s—v—sc Sc-v distance = (150+156+10+14) v-s distance = expected frequency of doubles actual frequency of doubles sc s v 314 sc+ s+ v+ 280 sc+ s v 150 sc s+ v+ 156 sc s+ v 46 sc+ s v+ 30 sc s v+ 10 sc+ v+ v 14 Total 1000 5.4 As distance between genes increases, mapping experiments become more inaccurate • When 2 genes are close together, accuracy of mapping is high • As distance between them increases, frequency of double/triple crossovers increases making mapping less accurate A B C D 68 A b C D 234 a B c D 248 a b c D 66 A B C d 56 A b C d 280 a B c d 262 a b c d 66 A plant hetero for 4 genes is test-crossed. Progeny have following phenotypes. Which of these genes is not linked to any of the others? a. a b. b c. c d. d 2 genes are tightly linked to each other. Which ones are they? A B B C A C C D D B CANT CONCLUDE THAT D IS 50 CM AWAY FROM OTHER GENES. ALL YOU KNOW IS THAT IT ASSORTS INDEPENDENTLY OR ISNT LINKED. KNOW IT’S AT LEAST 50 CM, CANT SAY HOW MUCH MORE. “D is 50 cM or more away from other genes or on different chromosome.” Or “It is not linked to A B C” or “It is not less than 50 cM away” Look at problems in book 5.6 Human gene mapping • Lod (log of odds) score analysis uses probability calculations to demonstrate linkage between 2 genes • Done in humans and any other organism where linkage analysis relies on pedigrees o When you cant set up crosses • Limited to extent of pedigree WEDNESDAY Recombination analysis with pedigrees A= ability to smell asparagus metabolites, a= recessive asparagus anosmia B= dominant photic sneeze response, p= recessive non response Parents A_P_ and aapp –> Know mother is AaPp AP Ap ap aP Don’t know much. Independent assortment or linkage Know its not complete linkage Not enough information LOD Score –difficult to do by hand • Pedigree • Dihybrid test cross • Interesting thing is what they got from mom • Grey – and white + are recombinants, rest are non-recombinants • IF assorting independently o If unlinked, prob of getting recombinant is 0.5 and getting not recombinant 0.5 o Probability of getting this family of 8 is 0.5^8 o If genes are linked at 25cM, prob of getting recombinant is 0.25 (definition of map distance) o Prob of getting nonrecombinant is 1-.25 = .75 o Prob of getting this family is .75^6 x .25^2 o Ratio of prob is odds ratio: .75^6 x .25^2 / .5^8 o Take log10 of odds ratio is lod score, function of assumed map distance o Lod score at 25cM is .45 o Lod score .>3 is good evidence of linkage. < -2 evidence of no linkage o This family, we need more data to conclude anything • TAKE AWAY –Lod scores exist • Consider series of hypotheses for dist between 2 genes, 0 to 50 cM • For each, calculate prob of getting pedigree • These probabilities are function of hypothesized map distance • Compare prob at each map dist to prob for 50cM (unlinked) • Odds ratio = prob at ?cM/prod abe 50 cM • Lod = Log10 of odds ratio • Lod , -2 good evidence against linkage • Marker –know where and what it is and can see somehow 5.7 Chromosome mapping is now possible using DNA markers and annotated computer databases • DNA markers represent landmarks along chromosome • Recombination frequency Werner syndrome • Premature aging syndrome • Molecular markers are codominant • Use sequencing to find gene • First time markers used to find unknown syndrome • Mutation allows damage to accumulate o Gene therapy o Defect in DNA repair leads to aging 5.8 Crossing over involves a physical exchange between chromatids • Why would you break your chromosomes!? • Is information changed without breaking strand? • OR Is there an actual breakage and reconnection? FRIDAY 7 Sex Determinism XX/XO mode • 2 X chrom =female • 1 X chrom = male • Half male gametes get an X XX/Xy mode • Females all have X • Male have either X or Y • Ygotes with 2 X chrom (homogametous) results in females • Those with 1 X and 1 Y (heterogametous) result in males • LOOK AT SLIDES ZZ/ZW sex determination, females are heterogametic (ZW) sex and males are homogametic (ZZ) sex 7.3 Y chromosome determines maleness in mammals Kleinfelter syndrome • Persons have male genitalia but have more than 1 X crhom (XXY, a 47, of XXY karyotype) • Audiploidy – Turner syndrome • Have single X and no Y crhom (45, X karyotype, aka XO) and have female genitalia • Developmental delays, short, defects in ovaries, heart and kidneys, webbed neck and skeletal abnormalieies • May have few symptoms Why do XO humans have a phenotype? a. Humans with only 1 X are abnormal b. Humans without a Y are abnormal c. Recessive mutations on the X are now hemizygous d. All of the above e. None of the above Inactivate bar body region of extra X chromosome, if don’t have this, the part that stays active on normal isn’t there. Ribosomal protein gene on bar body not shut down and without it, growth is stunted Extra X chromosomes • Presence of 3 X chrom (47, XXX) results in female differentiation • Frequently, 47, XXX women are perfectly normal (cells can coundt) • Other cases, underdeveloped sex characteristics, sterility XYY • Consistently shared characteristic found in 47,XYY karyotype is males are usually tall Y chromosome in mammals • Y has fewer genes than X • Pseudoautosomal regions (PAR) allow it to pair with X in meiosis • Most of Y is “Male specific region of Y” (MSY) including “sex determining region of Y” (SRY) • Genetically silent, different histones, etc. is why it stains differently than X chroms 7.5 Dosage compensation • Diff # copies of X-linked genes in males and females • Mechanisms to equalize expression of genes on X in males and females • In mammals, done by inactivating one of X chrom in females • Inactive X • Bar body still passed on in gametes A_Oo aaOo (male) aaOoSS 8 Variations in chromosome number and arrangement Euploidy –normal number of sets of chromosomes Aneuploidy –loss or gain of less than entire haploid set of chromosomes Polyploidy –gain of entire haploid set of chromosomes Nondisjunction • Chromosome or chromatids fail to disjoin (move to opposite poles during meiosis I or II) • Aneuploidy is generally result Monosomy and Trisomy • Monosomy –loss of 1 chromosome to produce 2n-1 complement • For any autosome, usually not well-tolerated • Trisomies (2n+1 chromosomes) for sex chromosomes have less severe phenotype than trisomy, usually no problems o Pairing in meiosis 1 irregualr o Down syndrome –human trisomy in #21 Seed capsule phenotypes in trisomies of Jimson weed A completely orange female cat is bred with a black male. One of their kittens is a tortoiseshell male, who is found to have Kleinefelter syndrome (XXY) Who gave the abnormal gamete? Father, XY
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'