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Answer Key to Problem Set 5

by: AllieStarks

Answer Key to Problem Set 5 2200

GPA 3.76
General Genetics
David Braun

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These are ALL of the correct answers to problem set #5 in General Genetics. I've taken this class and passed with an "A" last semester and saved all of the answer keys to the problem sets.
General Genetics
David Braun
Study Guide
Answer Key, Genetics, answers
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This 7 page Study Guide was uploaded by AllieStarks on Wednesday October 14, 2015. The Study Guide belongs to 2200 at University of Missouri - Columbia taught by David Braun in Fall 2015. Since its upload, it has received 381 views. For similar materials see General Genetics in Biological Sciences at University of Missouri - Columbia.


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Date Created: 10/14/15
BioSci 2200 General Genetics Problem Set 5 Answer key Gene Interactions Complementation and Epistasis Online Tutorials Visit the University of Arizona online learning biology web site httpwwwbiologvarizonaedumendeliangeneticsmendeliangeneticshtml Complete the Dihybrid Cross problems 12 13 Answers Online Gene Interaction that produces a novel phenotype See the example about peppers in the book pg 1101 1 1 Then do question 25 25 In chickens comb shape is determined by alleles at two loci R r and P p A walnut comb is produced when at least one dominant allele R is present at one locus and at least one dominant allele P is present at a second locus genotype R P A rose comb is produced when at least one dominant allele is present at the rst locus and two recessive alleles are present at the second locus genotype R pp A pea comb is produced when two recessive alleles are present at the rst locus and at least one dominant allele is present at the second genotype rr P If two recessive alleles are present at the rst and at the second locus rr pp a single comb is produced a lhl 7 1 Comb shape in chickens a walnut b rose 0 pea 1 single Progeny with what types of combs and in what proportions will result from the following crosses a RR PP X rr pp All walnut Rr Pp b Rr Pp X rr pp 1 walnut Rr Pp 1 rose Rr pp 1 pea rr Pp 1 single rr pp 6 Rr Pp X Rr Pp 916 walnut R P 316 rose R pp 316 pea rr P 116 single rr pp d Rr pp x Rr pp 3 rose R pp 1 single rr pp e Rr pp X rr Pp 1 walnut Rr Pp 1 rose Rr pp IA pea rr Pp 1 single rr pp f Rr pp gtlt rr pp 12 rose Rr pp 12 single rr pp Gene interactions involving Epistasis Before attempting the problems review the lecture notes and the assigned reading pg 110117 Pay special attention to Connecting concepts on pg 115116 and Table 52 3 What is gene interaction What is the difference between an epistatic gene and a hypostatic gene Answer Gene interaction is the determination of a single trait or phenotype by genes at more than one locus the e ect of one gene on a trait depends on the ejfects of a dijferent gene located elsewhere in the genome One type of gene interaction is epistasis The alleles at the epistatic gene mask or repress the ejfects of alleles at another gene locus The gene whose alleles are masked or repressed is called the hypostatic gene 4 What is a recessive epistatic gene Answer Recessive epistasis occurs when the epistatic gene in a homozygous recessive state masks the interacting gene or genes In the example from the text being homozygous recessive at the locus for deposition of color in hair shafts ee completely masked the e ect of the color locus regardless of whether it had the dominant black B or recessive brown bb allele See the Lecture notes and book pg 111117 and then do questions 27 28 29 32 Duplicate Dominant Epistasis 151 27 A variety of opium poppy Papaver somniferum L having lacerate leaves was crossed with a variety that has normal leaves All the F1 had lacerate leaves Two F 1 plants were interbred to produce the F 2 Of the F 2 249 had lacerate leaves and 16 had normal leaves Give genotypes for all the plants in the P F 1 and F 2 generations Explain how lacerate leaves are determined in the opium poppy Answer The F1 progeny tell us that lacerate is dominant over normal leaves In the F 2 24916 does not come close to a 31 ratio Let s see if these numbers fit a dihybrid ratio Dividing 265 total progeny by 16 because dihybrid ratios are based on 16ths we see that 116 of 265 is 1656 Therefore the F 2 progeny are very close to 1516 lacerate 116 normal a modified dihybrid ratio If we symbolize the two genes as A and B then P AABB lacerate x aabb normal F1 Aa Bb gtlt Aa Bb all lacerate F2 916 A B lacerate like F1 316 A bb lacerate 316 aa B lacerate 116 aa bb normal Resulting in 1516 lacerate and 116 normal A dominant allele at either gene A or gene B or both results in lacerate leaves This is an example of duplicate dominant epistasis as both A and B are dominant and epistatic over bb and aa Dominant Epistasis 1231 See the example on squash color in the lecture notes and book pgl 13 28 E W Lindstrom crossed two corn plants with green seedlings and obtained the following progeny 3583 green seedlings 853 Virescentwhite seedlings and 260 yellow seedlings E W Lindstrom 1921 Genetics 6291 110 a Give the genotypes for the green Virescentwhite and yellow progeny There are 4696 total progeny Green appears dominant The ratios at first glance don t fit any type of incomplete dominance for a single locus so we hypothesize multiple loci with gene interactions The simplest case is two loci so we look for a fit to a ratio based on 116 of the total 2935 These numbers are close to a 123 ratio of green39virescentwhiteyellow a modified 933 1 ratio Let s define G and g for one locus and Y and y for the other locus 9 G Y 9 green 3 G yy 3 green Total 12 green 3 gg Y 3 virescentwhite 1 gg yy 1 yellow b Provide an explanation for how color is determined in these seedlings The green arises when the G locus is dominant regardless of the alleles at the other Y locus Yellow requires that both loci be recessive and virescentwhite arises when the G locus is homozygous recessive and the Y locus has a dominant allele c Does epistasis occur among the genes that determine color in the maize seedlings If so which gene is epistatic and which is hypostatic As defined above the G locus is the epistatic locus It is an example of dominant epistatis because a dominant allele at this locus masks the e ect of the Y locus The Y locus is hypostatic and its e ect revealed only when the epistatic locus is homozygous recessive Recessive Epistasis 934 See the Lecture notes and book pg 111113 29 A dog breeder liked yellow and brown Labrador retrievers In an attempt to produce yellow and brown puppies he bought a yellow Labrador male and a brown Labrador female and mated them Unfortunately all the puppies produced in this cross were black a Explain this result Labrador retrievers vary in two loci B and E Black dogs have dominant alleles at both loci B E brown dogs have bb E and yellow dogs have B ee or bb ee Because all the puppies were black they must all have inherited a dominant B allele from the yellow parent and a dominant B allele from the brown parent The brown female parent must have been bb EE and the yellow male must have been BB ee The black puppies were all Bb Ee b How might the breeder go about producing yellow and brown Labradors Simply mating yellow with yellow will produce all yellow Labrador puppies Mating two brown Labradors will produce either all brown puppies if at least one of the parents is homozygous EE or 3 brown and 1 yellow if both parents are heterozygous Ee Duplicate recessive epistasis 97 See the lecture notes and worked problem in the book pg 116117 32 Some sweetpea plants have purple owers and other plants have white owers A homozygous variety of pea that has purple owers is crossed with a homozygous variety that has white owers All the F1 have purple owers When these F1 are self fertilized the F2 appear in a ratio of 916 purple to 716 white a Give genotypes for the purple and white owers in these crosses Answer The F2 ratio of 97 is a modified dihybrid ratio indicating two genes interacting Using A and B as generic gene symbols we can start with the F1 heterozygotes F1 AaBb purple selffertilized F2 916A B purple like F1 316 A bb white 316 aa B white 16 aa bb white Now we see that purple requires dominant alleles for both genes so the purple parent must have been AA BB and the white parent must have been aa bb to give all purple F b Draw a hypothetical biochemical pathway to explain the production of purple and white owers in sweet peas White precursor I 9 white intermediate 2 9 purple pigment Enzyme A Enzyme B 4 Complementation See lecture notes and pg 117 of the book 5 What is a complementation test and what is it used for Solution Complementation tests are used to determine whether di erent recessive mutations a ect the same gene or locus are allelic or whether they a ect di erent genes The two mutations are introduced into the same individual by crossing homozygotes for each of the mutants If the progeny show a mutant phenotype then the mutations are allelic in the same locus If the progeny show a wildtype dominant phenotype then the mutations are in di erent genes and are said to complement each other because each of the mutant parents can supply a functional copy or dominant allele of the gene mutated in the other parent Additional Question 1 1 In Drosophila recessive mutants a b c d e f and g all have the same phenotype namely the absence of red pigment in the eyes In pairwise combinations in complementation tests the following were produced where complementation and no complementation a b a What genotypes were crossed to show complementation between mutants g and a b What genotypes were crossed to show no complementation between mutants g and b c Which mutants have defects in the same gene d How many genes are present Answer a The F1 is normal so g and a are mutants at di erent genes Since two genes are involved the cross can be written as ggAA x GGaa and the F1 can be written as the dihybrid GgAa b The cross produces a mutant F1 so g and b are alleles at the same gene If we assign the new symbol z to this gene we can write its two alleles as zg and z Then the cross is zgzg x zbe and the F1 is zgzb c Mutants that fail to complement each other are in the same gene Mutants g and b fail to complement each other and therefore mutants g and b have defects in one gene mutants a and d are in a second gene and mutants c and e are in a third gene d As seen in part c the three complementation groups identify three genes But always check to make sure you have accounted for all the mutants when adding up the number of genes as in some examples there may be a mutant that represents a single allele in one gene Additional Question 2 2 In Additional Question 1 all of the mutant alleles evaluated in complementation tests were recessive Can a complementation test be used to determine whether two dominant alleles affect the same gene Can a complementation test be used to determine whether a dominant allele affects the same gene as a recessive allele Explain your answers Answer Complementation tests can be used only to determine whether two recessive mutations a ect the same function They cannot be used to determine whether two dominant alleles a ect the same function or whether a dominant allele a ects the same function as a recessive allele When two mutants are crossed in a complementation test the phenotype of the heterozygous F1 is used to infer whether the two mutations a ect the same gene function If the phenotype of the F1 is normal the two mutants a ect di erent gene functions and complement each other If the phenotype of the F1 is abnormal the two mutants a ect the same gene function and do not complement each other Since a dominant mutation always shows a phenotype in a heterozygote a complementation test with a dominant mutant will always produce a mutant phenotype whether or not the two mutants a ect the same function Therefore F1 complementation tests with dominant mutants are not interpretable Additional Question 3 3 The data in the following table are from complementation crosses between 7 different recessive mutants with similar phenotypes wild type progeny mutant progeny d a What genotypes were crossed to show complementation between mutants c and f b What genotypes were crossed to show no complementation between mutants a and e c How many genes are present d Which mutants have defects in the same gene Answer a Mutant c x mutant f produce a normal F I therefore they complement and are alleles of di erent genes The genotypes can be expressed as ch F x C C which produces an F Cch b Mutant a x mutant e produce a mutant F I therefore they fail to complement and are alleles of the same gene If we assign the symbol z for the gene the cross can be written zaz x zeze which produces an F1 zaze c There are four genes Mutants cd g are alleles of one gene mutants a and e are alleles of a second gene mutant b complements all other mutants and hence represents a third gene Mutant f also fails to complement all others and hence represents a fourth gene d Mutants c d g have defects in the same gene Mutants a and e also have defects in the same gene Additional Question 4 41241 In Dr ph ma g 339 TECEES WE animalanal nale 130 er Predueee a black bed QUIET When h m zyg u and an indEpandE y a l ti g auteeemel allele i1 else pmdm 3 lbw3 CD191 Whal h m ztg usa Flies with genetYPEe we we amtb and efe WE are phenetypieelly identical with respect te Ijeugljfr ml FEES with ge type Ev H7 ha ve a grey deF C0101 T leibl39EEdi g fe bfl e be flies are ereesed with true breedmg We we black iee at What will be the phenetype ef the F1 11525 What Ph tYPE and pmpertiens we uld eeeur in the FE generati n CB What ph typi ra DS W ul ya EXP ECt t in the PregenF of these beekereeeee F1 gtlt true breeeling ehenr ii F1 gtlt true breeeling hlaek Atteteer The F1 flies will be livh efe and be Wild type in eeler grey bi The F2 will shew a 9 if eV grey Ei efe black tJEi e black 1 lid efe black retie Lew 9X16 grey and Tide bleel i e The F1 X true breedmg ebenj can be deneted MJJF efe X hEI efe and will give a 1 EiJrf efe 1 l3 e e progeny ratie lift black and 1 greys The F1 X trueebreeding black can be denoted tJJ fi eXeJ gtlt tiff efe and will give a 1 b4 etf 1 EV e pregeny mtiteur 1152 grey and 1 black some details for solving a and b are given below equot 3e 64gt Erie em in at In see vet 5E Leftilfilim I I 1 Hit a gee I35 e31 3 w a efe t 3 17quot it big e eeta taquot elf aetiti e r 1 we quotwe a e 5r 5 F eight get 1 Quiteye e y a a eie H Li Jih eb Swath 1 equot la I E3 5 i e g e ie tab eet i yet be eeh i b ee b eLU b g T39T E rr W I l a e ie Lab 7 r 7 he 2439 but mer hem


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