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Genetic Recombination
Recombination is crucial to this process because it allows genes to be reassorted into diverse combinations. Genetic recombination is the process of combining genetic components from two different origins into a single unit. In prokaryotes, genetic recombination takes place by the unilateral transfer of deoxyribonucleic acid. It includes transduction, transformation, and conjugation. The genetic exchange occurring between homologous deoxyribonucleic acid sequences (DNA) from two different sources is termed general recombination. For this to happen, an identical sequence of the two recombining molecules is required. The process of genetic exchange which occurs in eukaryotes during sexual reproduction such as meiosis is an example of this type of genetic recombination.
Microbial Genetics
Genes are the functional units of heredity. They transfer characteristic information from parents to the offspring.
Four E. coli strains of genotype a+ b- are labeled 1, 2, 3,
and 4. Four strains of genotype a- b+ are labeled 5, 6, 7,
and 8. The two genotypes are mixed in all possible combinations and (after incubation) are plated to determine the
frequency of a+ b+ recombinants. The following results
are obtained, where M = many recombinants, L = low
numbers of recombinants, and 0 = no recombinants:On the basis of these results, assign a sex type (either
Hfr, F+, or F-) to each strain.
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- Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. After crossing the F1 generation of the cross between mutant strains 1 and 3, you count and determine the phenotypes of 1,000 colonies (here a colony is equivalent to an individual): 563 colonies that can grow on minimal medium alone; 437 colonies that require adenine…The following results are derived from crosses between Neurospora strain xy and strain ++: Tetrad Class 3 4 ху x+ x+ xy ++ ++ ++ +y +y ху +y 25 ++ 3 124 4 (i) Name the ascus type of each class from 1 to 4 as P, NP or T. (ii) Are genes x and y linked? Explain your answer. (iii) If they are linked, determine the map distance between the two genes. If they are unlinked, provide all the information you can about why you draw this conclusion.Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. A. What conclusions can you make about the alleles of mutant strains 1, 2, and 3 and their relationships with each other? B. What phenomenon is occurring in the cross between mutant strains 1 and 3? After crossing the F1 generation of the cross between mutant strains 1…
- 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 listBaker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. A. What conclusions can you make about the alleles of mutant strains 1, 2, and 3 and their relationships with each other? B. What phenomenon is occurring in the cross between mutant strains 1 and 3?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 + + + + +
- Possible equations to use: PID,] = 1- (PBa + 3/4 Pab + 1/2Pbb)" !3! log (1-PID,) log (Pa8 + 3/4 P8b+ 1/2Pbb) n = 8. A breeder is concerned that his best purebred Angus bull may be a carrier for Curly Calf Syndrome. He decides to conduct some test matings by taking this bull and mating him to 15 of the bulls own daughters to determine if he is a possible carrier of this and any other undesirable recessive genes. Show your work. a. How sure would we be that he is not a carrier if all the calves are born normal? b. How many would you need to be 99% sure he was not a carrier?A Neurospora cross was made between a strain that carried the mating-type allele A and the mutant allele arg-1and another strain that carried the mating-type allele aand the wild-type allele for arg-1 (+). Four hundred linear octads were isolated, and they fell into the sevenclasses given in the table below. (For simplicity, they areshown as tetrads.)a. Deduce the linkage arrangement of the mating-typelocus and the arg-1 locus. Include the centromere orcentromeres on any map that you draw. Label all intervalsin map units.b. Diagram the meiotic divisions that led to class 6. LabelclearlyA blood stain from a crime scene and blood samples from four suspects were analyzed by PCR using fluorescent primers associated with three STR loci: D3S1358, vWA, and FGA. The resulting electrophoretograms are shown below. The numbers beneath each peak identify the allele (upper box) and the height of the peak in relative fluorescence units (lower box). Solve, (a) Since everyone has two copies of each chromosome and therefore, two alleles of each gene, what accounts for the appearance ofonly one allele at some loci? (b) Which suspect is a possible source of the blood? (c) Could the suspect be identifi ed using just one of the three STR loci? (d) What can you conclude about the amount of DNA obtained from Suspect 1 compared to Suspect 4?
- Why is the rate of cotransformation for all three genes (M S F) almost the same as the rate of cotransformation for M F alone?In'a Mendelian monohybrid cross, the F2 generation consists of 1/4 red-flowered, 1/2 pink-flowered, 1/4 white-flowered individuals. What does this indicate? For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). 14px A Ix BIUS Paragraph Arial ABC 田 三 x² X2 田用因 Ť {} +]A cross was performed between a yeast strain that requires methionine and lysine for growth (met− lys−)and another yeast strain, which is met+ lys+. One hundred asci were dissected, and colonies were grownfrom the four spores in each ascus. Cells from thesecolonies were tested for their ability to grow on petriplates containing either minimal medium (min), min+ lysine (lys), min + methionine (met), or min + lys+ met. The asci could be divided into two groupsbased on this analysis:Group 1: In 89 asci, cells from two of the four spore colonies couldgrow on all four kinds of media, while the other two spore coloniescould grow only on min + lys + met.Group 2: In 11 asci, cells from one of the four spore colonies couldgrow on all four kinds of petri plates. Cells from a second one ofthe four spore colonies could grow only on min + lys plates andon min + lys + met plates. Cells from a third of the four sporecolonies could only grow on min + met plates and on min + lys+ met. Cells from the…