Class Note for ECOL 320 at UA
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Date Created: 02/06/15
Update Time Line of Revolutions in Genetics T H Morgan A H Sturtevant C B Bridges H J Muller Drosophila recombination mapping Hugo DeVries Theodor Boveri Chromosome motationsa etc Carl Correns Walter Sutton Gregor Mendel Rediscovery Chromosome theory 19101916 1866 1900 19023 l I I I l I Complementation groups genes and alleles Object Determine if two mutations are in the same or different genes e g isolate three arginine auxotrophic mutations in haploid yeast strains arg arg2 arg3 arg arg2 arg3 Cross 1 arg x on arg2 look at diploid phenotype arg arg2 The diploid grows on minimal medium because each mutant came With the Wild type allele of the other gene thereby complementing it So arg and arg2 are in different genes and they are not alleles arg arg2 arg3 Cross 1 arg x 0L arg3 look at diploid phenotype arg arg3 The diploid does not grow on minimal medium because each mutant is in the same gene and there is no wild type copy of that gene in the diploid arg and arg3 do not complement each other so they are different mutations of the same gene and they are alleles We have identified two genes which we might call ARGA and ARGB ARGA has three alleles ARGA argAI and argAj ARGB has two alleles ARGB and argBZ Isolate two more arg auxotrophs test growth of diploids on MM arg arg2 arg3 arg4 arg5 arg m2 arg3 arg4 arg5 arg4 and arg5 are in another complemention group so we now have three complementation groups and have identified three genes Each one may encode an enzyme for a different step in the biosynthesis of arginine Can you explain hw 58 I did this in Discussion and 812 as well as another example of physical and genetic maps 812 Recessive genes a b c d e and f are are closely linked but their order is unknown Three deletions in the region are examined One deletion uncovers a d and 6 another uncovers c d and f and the third uncovers b and c What is the order of the genes Uncovers means that in a heterozygote that has one chromosome with a deletion but is otherwise wild type and the other chromosome has recessive alleles one sees the recessive phenotype for the gene that is uncovered W W Deletion 1 uncovers a d and e so these are in a cluster that includes none of the others adeoraedordae Deletion 2 uncovers c d and f so they are also clustered c d or c d or d c But d must be at the end because it is also uncovered by deletion 2 so we have a e d c or a similar order The third uncovers b and c so these are adjacent to each other We also know that b is not in any of the other clusters so c must be at the end of the combined cluster identified by deletions 1 and 2 The order must be as shown below with the three deletions shown in order below it Note that for mapping purposes deletions are shown as lines where the deleted segments are not chromosomes missing certain regions Wt a e d f c b Dreadfcb deleted deletion Intragenic recombination This simply means recombination between two mutations in the same gene It is detected by making a heterozygote for the two mutations and finding wild type recombinant chromosomes We would also have to do a complementation test to know that the mutations are in the same gene Finding a wild type recombinant merely shows that mfg and mfg2 are not two cases of the same mutation in the same base pair Cross with mfgmfg mfg mfg mfg mfg2 mutant mfg andor 2 X gt mfg2 mfg mfg Wt mfg During our review tomorrow can you explain how to do problems 2 and 3 from homework 2 I reviewed these in Discussion You won t be required to do a complete analysis but might be asked to do part of it 2 In maize a strain homozygous for two recessive mutations liguleless lgand glossy gl was crossed to another strain homozygous for a dominant allele Booster B An F1 plant was backcrossed to the liguleless glossy parent strain The progeny phenotypes and numbers are shown below For example lg gl 172 means that 172 plants were liguleless glossy Make a map with these three genes Carry calculations to two decimal places lg gl 172 F1 had lg B 31 on one chromosome B 162 and lg B glt on the other lg B gl 56 This has B and gl together so it is a recombinant for these loci 48 lg B 51 g1 43 lg 6 B gl 5 543 By comparing the double crossover genotypes to the parental genotypes we can tell that gl is in the middle Distance lggl 51 43 6 5 105 105543 2 01933 2 1933 CM Distance gl B 56 48 6 5 115 115543 2 02118 2 2118 cM Map lg gl B 1933 2118 3 In dragons three genes a b C have been mapped on a chromosome as shown below with distances between the loci in centimorgans Suppose a dragon of genotype A A b b C C was mated to one of genotype a a B B C C and the offspring were testcrossed What is the expected frequency of a a b b C C dragons in the progeny from the test cross Give answer to 5 decimal places Offspring had one chromosome with A b C and the other with a B c So the only way to get a a b b c c is by a double crossover Answer 005003 00015 2 Pdouble crossovers faabbcc AABBCC faabbcc 000152 000075 1 What do we need to know about the Chi squared test for the exam You won t have to do one on an exam Should we be able to recognize what is statistically acceptable according to this test Yes 2 I understand what 11 w and x represent individually but what exactly does nlwlx without pwqx represent It calculates the number of orders 3 On slide 4 of the 15th lecture is a time line of scientific discoveries are these all the names dates and events we need to know for the exam See first slide 4 On slide 19 of the 3207 discussion in the very bottom right corner you have the distance from lZ to gl as 247 I thought that this distance should instead be the result of the addition between lZ su 107 and su gl 147 which is 254 Could you please explain why it is actually 247 My error now fixed Could you discuss lproblem 3 from homework sheet 2 Similar problem done earlier 2 problem 10 from Practice problems 4 Similar problem done earlier 3 problem 410 416 428 410 X linked recessive color blindness Normal couple have normal daughter and color blind son What is the probability that the daughter is heterozygous Color blind son got X from mother and it had the recessive allele b so mother is Bb Father is normal so is BY Daughter gets B from father and could get B orb from mother probability 12 either way 416 Draw a pedigree and fill it in Then do the problem as usual the description in the book should help 428 Did Chi square in Discussion and I have nothing to add to that 1 Will we likely face a multinomial equation Maybe but it would have to be a very simple one Also could you review a binomial problem similar to the problem describe on slide 17 of the probability lecture I have nothing to add to what was already said 2 I was digging around for an updated time line what specific names dates should we know and do what extent should we understand their work lives Names and dates and general description of what they did on time line Details of what they did are in lectures 3 Whyhow do tandem repeats act as molecular markers The number of copies of the repeat can increase or decrease You can view a set of repeats as having many alleles that differ in number of repeats The number of alleles is very high because the mutation rate is high so everybody has a unique set of alleles if one looks at the standard set of repeat regions Each repeat region can be amplified by a unique set of PCR primers and repeats of different lengths can be separated by electrophoresis so we can get a profile of of the alleles for an individual 4 On slide 20 of the chromosome variation lecture if you know d was not in 6 could one infer it is in 5 Yes if we assume that it is not in the interband region We used to think all genes were in the bands but last I heard this was no longer clear 5 Can you explain the ps qt section of the equation for homework problem 24 in chapter 4 I understand the rest of the solution listed in the back of the book but not the 128 The answer is to sum up the binomial probabilities of all the possible ways of getting an even number of males and of females Each one is multiplied by 124l24 which is 1 28
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