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The correct answer is " gene order could be determined because the final fraction of exconjugants with a marker decreases as the distance from the origin of transfer increases".
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- Equalizing the Expression of X Chromosome Genes in Males and Females Individuals with an XXY genotype are sterile males. If one X is inactivated early in embryogenesis, the genotype of the individual effectively becomes XY. Why will this individual not develop as a normal male?Linkage Mapping Using Dihybrid Testcrosses in Fruit Flies In fruit flies, the normal honey-gray body-color (bl*) is dominant to black (bl), having normal red eyes (pu+) is dominant to purple (pu), and having normal wings (vg+) is dominant to vestigial (vg). The three genes are located close together on the same chromosome. Your task is to use data from three dihybrid testcrosses to (1) calculate recombination frequencies, (2) convert recombination frequencies into map distances, and (3) build a map of the chromosome interval covered by the three genes, indicating their order and distances between them. BLACK PURPLE DIHYBRID CROSS In the parental generation, you mate a pure-breeding wild-type female (bl+/blt;put/pu+) with a pure-breeding black, purple male (bl/bl;pu/pu) to produce an F1 generation that is all wild-type (bl+/bl;pu*/pu). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (bl+/bl;pu*/pu) with tester males, which are black, purple…Question: This is a normal 3 point test cross, except that instead of regular phenotypes, you are looking at DNA markers on a gel. One parent, according to the gel, is heterozygous at each marker. The other parent is homozygous for each marker. (Again, this means it is a test cross: AaEeHh x AAEEHH --but don't be confused by that, because these are not "dominant" and "recessive" per se; the phenotype is just a band on a gel). For each offspring, figure out its genotype (homozygous or heterozygous for each gene. Then, figure that one parent made only AEH gametes, so you can cross that out if it helps.) Then treat it pretty much the same as a 3 point test cross.
- Determining Gene Order 1. We want to map the distance between genes E (tall height), B (blue flower color), and G (thick stalk). Each geen has a recessive allele (e= short, b= white and g= thin). (The actual configuration of the Trihybrid will be deduces as a part of the process) A. Your testcross genotypes will be (use slash notation). Female: ______ Male:______ Based on these results, are the genes linked or not? _______Question- 1. In onions, male sterility is due to the interaction of a chromosomal allele pair hh and “sterile” (S) cytoplasm. All other combinations (i.e. HH/Hh and “sterile” cytoplasm, HH/Hh or hh “fertile” (F) cytoplasm) result in male-fertile plants. The male-sterile trait is incorporated into inbred lines to produce hybrid F1 seed on a commercial scale.a) How would you perpetuate the male-sterile line? Show the cross.b) Briefly outline the method of producing hybrid seed (heterozygote) for the commercial crop. Show the cross.c) Does it matter whether the cytoplasm is fertile or sterile in the male-fertile inbred? Explain.Required information A single-factor cross is one in which the inheritance of only one character and its associated genotypes are followed. Punnett squares are often used to predict the outcomes of simple genetic crosses. Based on Mendel's laws, the genotypes of the parents can be used to predict the genes in their gametes and the resulting progeny. A Punnett square enables you to predict the types of offspring the parents are expected to produce and in what proportions. Sickle cell anemia is a recessive trait in humans. In a cross between two parents who are heterozygous for the gene, what are the gamete possibilities of the parer Mother's gamete possibilities Father's gamete possibilities of 19 Show All MacBook Air 田
- 3 Other (mutant) Gene: Clot, Please use "cl" abbreviation C A homozygous wt organism has been crossed to a homozygous mutant for your gene and for withered wings (whd). The resulting heterozygote is test-crossed. Predict the test cross results.Calibri 12 - BIU A Section 4: Monohybrid test cross A monohybrid cross is a mating where alleles of just one gene are tracked in the offspring. A test cross is a mating in which two parents (P generation) are crossed to give F1 (first filial generation) offspring, and then two F1 parents are crossed to give F2 (second filial generation) offspring. By tracking offspring through two generations, a test cross can reveal patterns of inheritance for particular genes. 9. In a cross between a black and a white guinea pig, all members of the F1 generation are black. The F2 generation is made of % black and % white guinea pigs. Diagram this cross, and list the genotypes and phenotypes for each generation.Extra Question Chapter 4 1. Consider the following cross concerning 4 different gene loci: AaBbCcDd (x) AabbCcdd From this cross, what is the probability of getting a progeny (offspring) with genotype AABbccdd? b. From this cross, what is the probability of getting a progeny (offspring) with genotype AabbCcDd? c. From this cross, what is the probability of getting a male progeny (offspring) with genotype aaBbCcdd? 2. Your neighbor has twelve children. One is blue eye color and short. Two are brown eye color and short. Two are blue eye color and tall. Seven look just like the parents; brown eye color with tall. What can you discover about the genetics of eye color and height of the children? a. How many traits are you dealing with? Each trait has phenotypes: Specify the phenotypes. b. What is the probability of the height of the children? What is the probability of the eye color of the children? (Refer to monohybrid punnett square slides 17-19) c. What are recessive traits based on the…
- 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…Assignment 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.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