Regarding "D" the coefficient of linkage disequilibrium, if it is greater than or less than zero, the loci at each locus of a haplotype exhibit nonindependence. Group of answer choices True False
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- The R and S loci are 35 m.u. apart. If a plant of genotype of RS/rs is selfed, what progeny phenotype will be seen and in what proportions?"The genotype of the organism producing the crossovergametes must be heterozygous at all loci underconsideration" Explain this statement ?GsnKivd010j2gIRWLIZOMZZ-VibKYvBbo61ylATAQ/viewform RECOMBINATION". For numbers 7-35, reler to the given data below. Glven the following testcross data for com In whlch the genes for fine stripe (f), bronze gleurone (bz) and knotted leaf (Kn) are involved: + = wild type f fine stripe +=wild type bz = bronze gleurone +=wild type Kn knotted leaf %3D Genotype Ko f Number 451 Ko 134 97 436 bz bz bz Ko 18 119 f 24 Kn f bz 86 Total: Your answer 7-8. What would be the recombination frequency or the frequency of the recombinant type between +/Kn and +/f genes? Oa. 16% O b. 16 map units + + + + +
- Explain why the value of the recombination frequencybetween any two genes is limited to 50%.CH 1-4 X с Maya S x Credib x app.wizer.me/learn/00E0AB wizer.me a Maya X A To-do 25% The table gives the genoty Assign x M Dashb x 50% Figurat X C In cows, brown (B) is codominant with white (W). The heterozygous phenotype is brown and white speckles. A farmer decided to cross a brown cow and a white cow in hopes of making all brown and white speckles. What percentage of the offspring will be brown with white speckles? Figurat X 75% Dashboard Incomp X d Interac X Enter class code Go 100%Considering a pair ofhomologous chromosomescontaining a gene having twodifferent alleles how manydifferent genotypes can theindividual present?
- Two crosses were made in Neurospora involvingthe mating type locus and either the ad or p genes.In both cases, the mating type locus (A or a) was oneof the loci whose segregation was scored. One crosswas ad A × ad+ a (cross i), and the other was p A × p+ a(cross ii). From cross i, 10 parental ditype, 9 nonparental ditype, and 1 tetratype asci were seen.From cross ii, the results were 24 parental ditype,3 nonparental ditype, and 27 tetratype asci.a. What are the linkage relationships between themating type locus and the other two loci?b. Although these two crosses were performed inNeurospora, you cannot use the data given tocalculate centromere-to-gene distances for anyof these genes. Why not?Neurospora of genotype a + c are crossed withNeurospora of genotype + b +. (Here, + is shorthandfor the wild-type allele.) The following tetrads areobtained (note that the genotype of the four sporepairs in an ascus are listed, rather than listing alleight spores):a + c a b c + + c + b c a b + a + ca + c a b c a + c a b c a b + a b c+ b + + + + + b + + + + + + c + + ++ b + + + + a b + a + + + + c + b +137 141 26 25 2 3a. In how many cells has meiosis occurred to yieldthese data?b. Give the best genetic map to explain these results.Indicate all relevant genetic distances, both betweengenes and between each gene and the centromere.c. Diagram a meiosis that could give rise to oneof the three tetrads in the class at the far right inthe listTo determine the recombination frequency of two linked genes, what genotypes of organisms are crossed? O Amy cross between two individuals will yield the same results O A homarygous dominant individual with a homozvgos recesshve O A heterozygous individual with a homozypous dominant individual O A heterozygous individual with a homozygous recessive individual O Two heterorygous individuals
- An individual has the following genotype. Gene loci (A) and (B) are 15 m.u. apart. What are the correct frequencies of some of the gametes that car be made by this individual? Bl a O A. Ab = 7.5%; AB = 42.5% B. ab = 25%; aB = 50% O C. AB = 7.5%; aB = 42.5% O D. aB = 15%; Ab = F0% E. aB = 70%; Ab = 15% Reset Selection OMark for Review What's This?FAlpQLSfiOhfAvlhxzCSiUll_6rt-nU5b0WI73UmWOxkOw8OCwk01ng/formResponse B 1 2 Bb x Bb b 4 The fur in both parents in this cross is * 1 B B Bb x Bb b 3 4 brown black O homozygous dominant homozygous recessive 3. 近What would justify the following ratio appearing after phenotyping the outcome of a crossing trial: 8.9: 2.9: 3.2:1? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a b C d Obviously this represents independent assortment based on crossing dihybrid heterozygotes. Obviously this represents gene linkage based on test crossing a dihybrid heterozygote. Obviously this represents the results of a trihybrid test cross. Obviously this represents independent assortment based on crossing monohybrid heterozygotes.