ETO Class Data Female Male Total Phenotype Observed (O) Expected (E) O-E (O-E)2 (O-E)2/E Wild 187 191 378 Wild Type M/F 378 White Eyed 182 177 359 White Eyed M/E 359 Vestigial M/F White Eyed & Vestigial M/F 69 Vestigal White Eyed & Vestigial 71 140 140 62 61 123 123 Totals 500 500 1000
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Need help figuring out the chi-square analysis. The ratio is 3:3:1:1. See attached spreadsheet.
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- Here are the progeny of this cross: (Note that the categories are not in any particular order.)Fly type # of prog. Phenotype symbols Categorywt eyes black body wt wings 97 grn+ blk crv+Green eyes black body curved wings 709 ParentalGreen eyes wt body wt wings 9Green eyes black body wt wings 162wt eyes wt body wt wings 727wt eyes black body curved wings 12wt eyes wt body curved wings 179Green eyes wt body curved wings 105Total = 2000 9.) Write the phenotype symbols in the right-hand column. The first one has been done for you.10.) Next to that, label all fly categories as parental (NCOs), SCOs, and DCOs. One has been donefor you.11.) After each SCO/DCO label, write which gene got “unlinked” in these offspring.12.) Put these three genes into a genetic map in the proper order.13.) Calculate the genetic distance between the genes and label the map with these distances.14.) Calculate the cross-over interference15.) Return to questions #1-6 above. For question 6, you gave your opinion, but…. A true-breeding strain of Virginia tobacco has dominantalleles determining leaf morphology (M), leaf color(C), and leaf size (S). A Carolina strain is homozygousfor the recessive alleles of these three genes. Thesegenes are found on the same chromosome as follows:M C S6 m.u. 17 m.u.An F1 hybrid between the two strains is now backcrossedto the Carolina strain. Assuming no interference:a. What proportion of the backcross progeny willresemble the Virginia strain for all three traits?b. What proportion of the backcross progeny willresemble the Carolina strain for all three traits?c. What proportion of the backcross progeny will havethe leaf morphology and leaf size of the Virginiastrain but the leaf color of the Carolina strain?d. What proportion of the backcross progeny will havethe leaf morphology and leaf color of the Virginiastrain but the leaf size of the Carolina strain?What do the 3r's refers to?
- Mendelian GeneticsF1 Cross: Yellow, Round x Green, Round GgWw x ggWWCharacter: Pea color & shapeUse Punnett square and fork-line method to check the F2.Show and interpret all the possible genotypes and phenotypes of the offsprings. Mendelian GeneticsF1 Cross: Tall, White, Axial x Dwarf, Violet, Terminal DdwwAA x ddWWaaCharacter: Stem height, Flower color & positionUse fork-line method to check the F2.Show and interpret all the possible genotypes and phenotypes of the offsprings. I. For A and B,1. Identify the type of inheritance. Justify your answer.2. Decode the genotypes of the individuals in the pedigree. (Use letter A for representation of alleles.)3. List down all affected individuals.B.A.Mendelian GeneticsF1 Cross: Yellow, Round x Green, RoundGgWw x ggWWCharacter: Pea color & shapeUse Punnett square and fork-line method to check the F2.Show and interpret all the possible genotypes and phenotypes of the offsprings.Mendelian GeneticsF1 Cross: Tall, White, Axial x Dwarf, Violet, TerminalDdwwAA x ddWWaaCharacter: Stem height, Flower color & positionUse fork-line method to check the F2.Show and interpret all the possible genotypes and phenotypes of the offsprings.B.A.I. For A and B,1. Identify the type ofinheritance. Justifyyour answer.2. Decode the genotypesof the individuals inthe pedigree. (Useletter A forrepresentation ofalleles.)3. List down all affectedThe accompanying pedigree is for a rare, but relativelymild, hereditary disorder of the skin.a. How is the disorder inherited? State reasons for youranswer.b. Give genotypes for as many individuals in thepedigree as possible. (Invent your own defined allelesymbols.)c. Consider the four unaffected children of parentsIII-4 and III-5. In all four-child progenies from parentsof these genotypes, what proportion is expected tocontain all unaffected children?(picture added)
- . Consider the genotypes of two lines of chickens: thepure-line mottled Honduran is i/i ; D/D ; M/M ; W/W, andthe pure-line leghorn is I/I ; d/d ; m/m ; w/w, whereI = white feathers, i = colored feathersD = duplex comb, d = simplex combM = bearded, m = beardlessW = white skin, w = yellow skinThese four genes assort independently. Starting withthese two pure lines, what is the fastest and mostconvenient way of generating a pure line that has coloredfeathers, has a simplex comb, is beardless, and has yellowskin? Make sure that you showa. the breeding pedigree.b. the genotype of each animal represented.c. how many eggs to hatch in each cross, and why thisnumber.d. why your scheme is the fastest and the mostconvenientF1 hybrids between two species of cotton, Gossypium barbadenseand Gossypium hirsutum, are very vigorous plants. However, F1crosses produce many seeds that do not germinate and a high percentageof very weak F2 offspring. Suggest two reasons for theseobservations.ABO Blood TypeThe following pedigree shows the incidence of ABO blood types in a family. dentify the genotypes of the following individuals: Individual Genotype II-1 II-2 II-4 II-5 III-2 III-3
- Karyogram and Identification of Chromosomal Aberrations List down five human chromosomal aberrations. Give the corresponding karyotype for each type then briefly describe the traits of affected individuals.The genotype of F1 individuals in a tetrahybrid cross isAaBbCcDd. Assuming independent assortment of these fourgenes, what are the probabilities that F2 offspring will have thefollowing genotypes?(a) aabbccdd(b) AaBbCcDd(c) AABBCCDD (d) AaBBccDd(e) AaBBCCddNotation: D = disease allele at gene d = wild-type allele at gene Let’s provide some information about each MOI. AD: Each of the F1 parents is heterozygous for the D allele. SLR: The observed F2 data are the offspring from an F1 cross where the father is affected (disease) and the mother is an unaffected carrier (wild-type, or WT). That is, the mother’s genotype has one disease allele and one WT allele in it. In Table 1, I provide the observed data counts (in dark blue). Notice that I stratify by gender. Table 1 Phenotype O E O-E (O-E)^2 (O-E)^2 /E Disease, Male 304 Disease, Female 267 WT, Male 285 WT, Female 301 Total 1157 DF p-value In Table 2, provide expected proportions for two different MOIs: AD and SLR. The values in this table are computed using information from the Punnett Square and the specified MOI. Once the proportions are determined, we can fill in the values in Table 3, E(xpected)…