Genetics Notes week 4
Genetics Notes week 4 Bios 206
Popular in Genetics
verified elite notetaker
Popular in Biological Sciences
verified elite notetaker
This 5 page Class Notes was uploaded by Becca Sehnert on Monday February 8, 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 13 views. For similar materials see Genetics in Biological Sciences at University of Nebraska Lincoln.
Reviews for Genetics Notes week 4
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: 02/08/16
WEEK 3 NOTES Wednesday after Exam 1 and snow day! CHAPTER 5 5.1 Genes assort independently if they are on different chromosomes but show linkage if they are on same chromosome • How often inherited together? • Independent assortment (Mendel) – • Linkage -2 genes on single pair of homologs, no exchange occurs • Homologs find each other in prophase 1 and swap DNA to ask if it matches o If it doesn’t match, they keep looking o This is why we get crossing over o Probability of crossing over (no set way, random) • On same chromosome –complete linkage with no crossing over • Genes on same chromosome are part of a linkage group o # Groups is haploid number (1n) o If crossing over occurs between 2 linked genes, both parental and recombinant (crossed over) gametes • Exchange between 2 non-sister chromatids o Sister chromatid exchange doesn’t do anything o Parental type –sister chromatid that did not undergo crossing over o Cross over gametes –have some DNA from homolog • Distance matters! o Amt of crossing over between 2 loci on single chromosome is proportional to distance between them 5.2 Crossing over is the basis of chromosome mapping • Usually small number of crossovers per chromosome during meiosis • If one happens to fall between 2 genes, recombinant produced • Probability of that is related to distance between genes • If there is an exchange not between genes we want, there is no way of knowing it happened • GENE MAPPING o % of recombinant progeny in test cross is measure of distance between 2 linked genes o if unlinked, see Mandelian pattern o Recombination frequencies between linked genes are additive, so map distances are additive o 1% recombinants defines 1 map unit (m.u.) = 2 centiMorgan (cM) o With testcross and no recombination, get what would expect o Testcross with recombination, mess up ratios and can measure how far genes are apart § Fig 5-4 • Essence of chromosome mapping o Discovery of gene usually starts with mutant phenotype o “What is this gene, and how does it function?” o Sequence it! § LONG sequence for Dosophila white gene § Genome sizes? (in 3pt courier) 3 poin courier • E. coli -0.9 miles (here to capitol building) • Yeast -2.4 miles (to UNL Dairy Store) o “Where is the gene?” o Measure by finding a cell heterozygous for both and ask “How often does crossover land between genes?” o More often that happens, farther apart 2 genes will be o Consequence of single exchange between 2 nonsiter chromatids occurring in tetrad stage If 2 genes are 60 map units (cM) apart, a dihybrid is test-crossed, what percent of progeny will be recombinants? a. 0% b. 30% c. 50% • Cant have greater than 50% recombinants (according to independent assortment, ½ progeny will be parental, ½ will be recombinants) • There is NO crossing over in male Drosophila (only in females) • 1% = 1 m. u. = 1 cM • Map distance is proportion (# recombinants/total progeny) • # recombinants / # progeny • Double cross overs possible –essentially random o Farther apart = don’t know if anything happened o Can only observe UP TO 50% recombinant FRIDAY LECTURE Two-point cross • Looking at only 2 genes at a time • Measure distance between 2 genes • 2 classes of parentals and 2 classes of recombinants o Find % of recombinants If forest dragons, long or short tails and brown or green scales. If a population of short-tailed green dragons mate with population of long-tailed yellow dragons, offspring all have long tails and green scales. These individuals are then testcrossed with following results. -248 STG parentals -234 LTY -32 STY recombinants -28 LTG WHAT IS MAP DISTANCE BETWEEN GENES? *Know long and yellow are dominant *542 total, 60 recombinants, % recombinant =(60/542)x100=11.07% **SO L and G are 11.07 m.u. or cM apart Could find out how many base pairs apart they are –very difficult 5.3 THREE POINT CROSSES • Additional information on gene order • Map gene relative to 2 other genes to find exact location • Single crossovers can be used to determine distances between pairs of genes • Double crossovers (DCO) can be used to determine the order of 3 genes on chromosome • HOW can you tell which represent what? o Expected frequency of double-crossover gametes is MUCh lower than that of either single-crossover gamete class o Most frequenct reciprocal pair of progeny classes represents noncrossovers o Least frequent reciprocal pair of progeny classes represents double crossovers • Mate male known genotype with unknown female genotype and count (and find %) of progeny o LOOK AT CHART ON SLIDES • To determine gene order, compare noncrossover genotypes to double crossover genotypes • One that is exchanged in middle from DCO - Look at charts In cookies, choc chip (ch), nutes (n), and fluffiness (f) are all recessive traits. Females heterozygous for all 3 loci are test-crossed to produce the following progeny Determine map order, and distances between genes. DCO PARENTALS SCO fluffy 2 chips 802 nuts 96 fluffy,chips 104 flubby, nuts 808 chips, nuts 4 fluffy, chips nuts 80 wild type 104 total 2000 What is F1 progeny? F+ ch n+ / f ch+ n Which are DCO? Rare ones Which are parentals? Most Which is in the middle? Nuts (compare parentals and doubles and look at which one moved of the 3. fluffy 2 f ch+ n+ chips 802 f+ ch n+ nuts 96 f+ ch+ n fluffy,chips 104 f ch n+ fluffy, nuts 808 f ch+ n chips, nuts 4 f+ ch n fluffy, chips nuts 80 f ch n wild type 104 f+ ch+ n+ total 2000 96+104+2+4/2000 = 10.3 cM between f and n 80+104+2+4/2000 = 9.5 cM between n and ch if ignore nut gene, don’t get double recombinants. So undercount, so cant use double cross A heterozygote for 3 genes was test-crossed and progeny are listed. Which gene is in middle? C +++ 317 abc 328 a__ 65 _bc 58 _b_ 82 a_b 91 __c 5 ab_ 3 C –not b because middle ones are parentals. Ones different between doubles is C +++ 17 abc 28 a__ 165 _bc 158 _b_ 3 a_c 0 __c 25 ab_ 23 +++ 252 abc 4 a__ 28 _bc 41 _b_ 39 a_c 33 __c 3 ab_ 234 A
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'