Note that the two parental gamete types (+ + and bl pu) are the most abundant, as expected. Use the data to calculate the recombination frequency and the genetic map distance between the two genes. Record the map distance in your notes.
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- Linkage Mapping Using a Trihybrid Testcross in Fruit Flies Remember the black, purple, and vestigial genes from Exercise 1? Now you're going to have to construct a map of the same region from trihybrid testcross data. The new map should be similar to the one from Exercise 1, but it won't be identical because we're starting with a new, independent data set. BLACK PURPLE VESTIGIAL TRIHYBRID TESTCROSS In the parental generation, you mate a pure-breeding wild-type female (bl+/bl*pu*/pu+vg+/vg+ with a pure-breeding black, purple, vestigial male (bl/bl;pu/pu vg/vg) to produce an F1 generation that is all wild-type (bl/bliput/pu;vg+/vg). Note that the F1 flies are all trihybrid. Next, you mate several F1 trihybrid females (bl/bl:put/pu;vg+/vg) with tester males, which are black, purple, vestigial (bl/bl;pu/pu;vg/vg). The offspring of this trihybrid testcross are: Phenotype Wild-type Black, purple, vestigial Vestigial Black, purple Purple, vestigial Black Purple Black, vestigial bl pu vg bl pu…mapping gene The genes for ruby eyes (rb), tan body (t) and cut wings (ct) are all found on the X-chromosome of Drosophila melanogaster. All of these are recessive traits. They map in the order rb, ct, t with 12.5 map units between rb and ct and 7.5 map units between ct and t. Suppose you cross a cut wing male with a homozygous female that is both tan and has ruby eyes. What will the F1 females look like? Draw map of the section of the X chromosomes that has these 3 genes for the F1 females Assume you testcross your F1 females. What progeny classes would you expect? ii. Give approximate numbers for each class based on a total of 2000 progeny. Assuming the i=1 and there are no double crossovers. Assuming the i=0 and there are the expected number of double crossovers.Crossover frequencies for some genes on a tomato plant: • Normal leaves (M) and tall plant (D) 12% • Normal leaves (M) and normal tomato (O) 33% • Normal leaves (M) and simple florescence (S) 64% • Tall plant (D) and normal tomato (0) 21% • Tall plant (D) and simple fluorescence (S) 52% Draw a chromosome map and state the cross-over frequency between genes for a normal tomato and simple florescence.
- Analyzing Karyotypes 1. Originally, karyotypic analysis relied only on size and centromere placement to identify chromosomes. Because many chromosomes are similar in size and centromere placement, the identification of individual chromosomes was difficult, and chromosomes were placed into eight groups, identified by the letters A to G. Today, each human chromosome can be readily identified. a. What technical advances led to this improvement in chromosome identification? b. List two ways this improvement can be implemented. c. What clinical information does a karyotype provide?Need help. Knowing that the Curly leaf (Cy) is a dominant gene on chromosome 6, as it is useful in tracking other genes using trisomics for chromosome 6. Assume a Cy Cy cy plant used as a pollen parent where disomic pollen does not function crossed to a Cy cy cy female where disomic eggs do function. A) What ratio would be predicted in the progeny of this cross? B) What if the reciprocal cross was made?Below you will find a spreadsheet of the "class" data, use this to complete data table 2. Remember, the class is a total of 24 students, to calculate frequency use the genotype with the highest frequency for each characteristic and divide that number by 24. Don't forget to answer the question on the bottom of page 107! Gender Letters XY XX Sex X, Y 11 13 XX Homozygous Heterozygous Homozygous Recessive Genotype highest frequency Characteristic Dominant Aa Eyebrows Eye Shape Hitchhikers А, а R, r 14 10 11 3 Rr Hh Н, h D, d Е, е F,f G, g В, Ь 15 Earlobes 4 8. 12 dd Ee Tongue Roll Widows Peak 7 10 13 Ff Face Shape 9. 12 3 Gg Eye Color 14 Bb MacBook Air esc 4) FI F2 F4 F5 F6 F8 F9 F10 F1I F12 @ # $ &
- Gene mapping using the Three-point Testcross a) Given the following alleles that control seed traits:W = wrinkled G = green R = roundw = smooth g = yellow r = oval b) Results of a cross with a triple heterozygote revealed the following phenotypes:30 smooth yellow round4 smooth green round958 wrinkled green round2 wrinkled yellow oval18 wrinkled yellow round946 smooth yellow oval16 smooth green oval26 wrinkled green oval c) Determine the order of the genes and the distance between them in centiMorgan (cM).Construct a gene map to show your results. TIP: Based on the phenotypes, determine the alleles in the gametesuh ec CV + + 10.5 SC 9.1 scute bristles echinus eyes 9.2 ct + crossveinless wings Table 1: phenotype wild-type tapdance feet crossveinless wings tapdance & crossveinless cut wings 15.9 vermilion eyes V + + 66.8 Drosophila X chromosome Use the map provided above for problems 1 & 2. Problem 1: 11.2 10.9 garnet eyes M A new gene is being investigated in fruit flies. The recessive allele of this gene (t) causes the flies' feet to grow tiny tapdance shoes, while the dominant allele (t*) permits wild-type feet to develop. Preliminary studies indicate that this new gene is located on the X-chromosome. You decided to perform a two-point testcross to determine its position relative to the well-established crossveinless wings gene (cv). You cross a female heterozygous for both genes with a testcross male fly and obtain the male offspring results shown in table 1, below. Using this information, answer the following questions: # male offspring 13 405 401 11 forked bristles a) is the original…select whcih ic correct When 2 wildtype alleles are on the same parental chromosome, this is known as [Combined or Coupling or dispersed or heterozygous or Repulsion] . In sharp contrast [Combined or Coupling or dispersed or heterozygous or Repulsion] is when 1 wildtype allele and 1 mutant allele are on the same parental chromosome
- Problem Solving: In Drosophila, the gene that controls red eye color (dominant) versus white eye color is on the X chromosomes. What are the expected phenotypic results if a heterozygous female is crossed with a white-eyed male? In cats, S = short hair; s = long hair; XC = black coat; Xc = yellow coat; XCXc = tortoiseshell (calico) coat. If a long-haired yellow male is crossed with a tortoiseshell female homozygous for short hair, what are the expected phenotypic results?Saved Gene linkage Complete the following statements about gene linkage. Not all choices will be used. independent A chromosome contains a long series of different are in a definite assortment that sequence linkage group Their position, or on the chromosome is fixed. locus alleles All of the alleles on a chromosome form a(n) they tend to be inherited together. because gametes Because they reside so close together on the chromosome, there is a reduced chance that not follow the traditional Mendelian genotypic ratios. will occur and therefore they do 30 acerAssignment 3 Linkage and Recombination 1. In corn, the genes an (anther ear), br (brachytic), and f (fine stripe) are linked. Testeross data are as follows: Number Number 355 2 Progeny Progeny +++ 88 an ++ ++f + br + + br f 21 an +f an br + an br f 2 17 399 55 Determine the linkage map and the genotype of the homozygous parents used to obtain the heterozygote for testcross.