Question 2 Listen You want to determine the distance between 2 mice genes F and G. You take mice which are FG/fg and cross them to mice which are fg/fg. You get 80 mice which are FG/fg 75 mice which are fg/fg 6 mice which are Fg/fg 4 mice which are fG/fg Which mice are recombinant? a) FG/fg b) Fg/fg c) fg/fg d) All of the above
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Problem 2
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- Below is a partially filled in complementation table. Please answer the following questions. 1 2 3 4 LO 5 6 1 2 I 3 4 5 4 + + 5 6 a. If you were to perform a complementation test with mutants 1 and 2, what would be the result? [Select] b. If you were to perform a complementation test with mutants 5 and 6, what would be the result? [Select] c. If you were to perform a complementation test with mutants 3 and 4, what would be the result? [Select] d. There is one complementation group that you can determine from the information above. If you performed a complementation test with mutant 1 and 6 and there was growth, would you say that the mutants were part of that complementation group? [Select] e. What is the minimum number of complementation groups that could exist based on the information above? [Select] f. What is the maximum number of complementation groups that could exist based on the information above? [Select]An Hfrstrain that is a *b*c*d* e*f* g *h* is mated with an F strain that is a b e d e f gh. The mating is interrupted at 5 minutes interval, and the genotypes of the F recombinants are determined. The results obtained are tabulated in Table 2. Draw the map of the Hfrchromosome and indicate the position of the origin of transfer, the direction of the transfer and the minutes between genes. Table 2:Entry time of Hfr chromosome into recipient cell. Time a d e f h 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +Question 7 Which of the following is inconsistent (not consistent) with inheritance of a maternal effect trait? A) Progeny from the same parents have the same maternal effect phenotype as each other but not their mother. B) Progeny can show a recessive phenotype even though the progeny have dominant genotypes. C) Both parents contribute recessive alleles to an offspring yet it has a dominant phenotype for the maternal effect trait. D) Progeny from the same parents have different traits related to the maternal effect gene in question. CS Scanned with CamScanner
- In Figure 6-19,a. what do the square/triangular pegs and holesrepresent?b. is the suppressor mutation alone wild type inphenotype?You are given a Drosophila female that looks wild-type but is heterozygous for mutations intan body (t), miniature wings (m), and white eyes (w). You test cross this female with a tanbodied, miniature winged, and white-eyed homozygous mutant male, and you obtain thefollowing 1400 progeny: Phenotype : number+ + + : 608t m w : 516+ m w : 2t + + : 6+ m + : 39t + w : 46+ + w : 81t m + : 102 Calculate the distance between each pair t-m, m-w, and t-w only using the number ofrecombinants between them (i.e. ignoring the gene in the middle). Draw a linear map with thedistances between genes.Phenotype chr1.1 chr2.8 chr3.12 chr4.12 chr4.55 chr7.88 chr9.55 chr10.1 chrX.1 chrX.5 Purple AA AT Purple AA AT F F GG GC 80 CC CC TT TT TT CT 80 CC CC AT TT TT CT 8 CC CC AT AT GG CT GC TT CC AT AT GT 10 CT Yellow AA AT TT GC Purple and AA TT Yellow F Purple AA AT AA GG TT GC AT TT TT CT Purple AA AT AA GG TT GC AT TT TT CT Purple AA AT AA Purple AA AT AA 80 CC TT GG AT TT TT TT 646 GG CC GG AT TT TT TT Purple and AA TT AT CC 80 Yellow 80 CC GG AT AT GT TT Purple AA AT AT CC CC GC AT F TT TT TT E Purple AA AT AT CC TT GC AT TT TT CC Purple AA AT E 00 CC TT GG AT TT TT CC Yellow AA AT AT GC TT 00 CC AT AT GG CC 80 TT E AT AT GC CC CC AT AT GG TT Yellow AA AT AT GG CC CC AT AT GG Yellow AA ধ 80 CC CC AT AT GT TT GG TT GG AT AT GG TT Purple and AA AA AT GG Yellow Yellow AA AA F 80 CC GG AT TT TT CC 80 GC CC GG AT AT GT CC Purple AA AA AT GC Purple and AT AA Yellow Purple AA AA F F GG 80 CC GG AA TT GT CC
- Is )Given the experimental set up below, and knowing the genes for stalked eyes (st) and black bodies (b) are on the same chromosome and recessive to wild type (un-stalked eyes and light body), answer the questions below the diagram. 1. st st F1 F2 A = 489 B = 56 C= 497 D= 49 A) ) How many recombinant animals are in the F2 generation? B) woH) How far apart, (as both a % recombination and in map units) are the stalk eyed and black bodied genes on this chromosome?(i) For the chromatogram below, what is the sequence of the template DNA from base 115 to 125? CTGTGTGAAATTGT TA T CCGC T CA CA AT T C CACA CA A CATA CGAGC CGGAAG CA TA A 110 120 130 140 150 160 (ii) An allele of a gene has the following change in it's sequence ATG GTG CÁC CTG ACT CCT GTG GAG AAG TCT compared to the wild type ATG GTG CAC CTG ACT CT GAG GAG AAG TCT With reference to the sequence; there is a codon, resulting in a change from is a mutation in the to which mutation.Complementation tests of distinct recessive mutants, 1 through 8, produce the data in the matrix below. A plus (+) indicates complementation, meaning the phenotype of the combined alleles is wild type, and a minus (-) indicates a failure to complement meaning that a mutant phenotype results. Assume that the missing mutant combinations would yield data consistent with the entries that are shown. How many complementation groups are formed by these eight mutants? (Picture attached) A) 2 B) 3 C) 4 D) 5 E) 6
- 4) You design Drosophila crosses to provide recombination data for gene (A), which is located somewhere on the chromosome shown below: Short aristae Long aristae (appendages on head) III Black body Gray body Mutant phenotypes Cinnabar eyes 48.5 57.5 Red eyes 67.0 Vestigial wings Normal wings Brown eyes 104.5 Red eyes Wild-type phenotypes Gene (A) has a recombination frequency of 14% with the vestigial-wing locus and 23% with the brown- eye locus. Where is gene (A) located on this chromosome?(Problem 65a) In the plant Arabidopsis, the loci for pod length (L, long; I, short) and fruit hairs (H, hairy; h, smooth) are linked 16 m.u. apart on the same chromosome. The following crosses were made: (i) L H/L H × 1 h/l h F1 (ii) L h/L h × 1 H/I H F1 If the F1's from cross i and cross ii are crossed, what proportion of the progeny are expected to be I h/l h? Oa. 16.00% Оb. 8.00% Oc. 4.00% Od. 3.36% Oe. 1.28%Females of wild-type Strain A and males of mutant Strain B, as well as females of mutant Strain B and males of wild-type Strain A, make reciprocal crosses. Explain why reciprocal crosses are needed in genetics experiments involving Drosophila fruit flies.