Once you've amplified your sequence of interest, you can introduce the restriction enzyme EcoRV (described above) to cut your DNA to help identify its genotype. Match the following genotypes with the number and size of bands you would expect to see for each. Homozygous dominant Two bands: one 86bp and on Heterozygous Homozygous recessive (yellow kernels) [Choose ] Two bands: one 51bp and one 35bp Two bands: one 86bp and one 35bp Three bands: one 86bp, one 51bp and one 35bp One band: 86bp One band: 43bp Two bands: one 86bp and one 51bp One band: 51bp

Once you've amplified your sequence of interest, you can introduce the restriction enzyme EcoRV (described above) to cut your DNA to help identify its genotype. Match the following genotypes with the number and size of bands you would expect to see for each. Homozygous dominant Two bands: one 86bp and on Heterozygous Homozygous recessive (yellow kernels) [Choose ] Two bands: one 51bp and one 35bp Two bands: one 86bp and one 35bp Three bands: one 86bp, one 51bp and one 35bp One band: 86bp One band: 43bp Two bands: one 86bp and one 51bp One band: 51bp

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter18: Dna Technologies: Making And Using Genetically Altered Organisms, And Other Applications

Section: Chapter Questions

Problem 1ITD: You learned in the chapter that an STR locus is a locus where alleles differ in the number of copies...

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In an electrophoretic gel across which is applied a powerful electrical alternating pulsed field, the DNA of thehaploid fungus Neurospora crassa (n = 7) moves slowlybut eventually forms seven bands, which represent DNAfractions that are of different sizes and hence havemoved at different speeds. These bands are presumed tobe the seven chromosomes. How would you show whichband corresponds to which chromosome?
Please answer all parts along with the reason. I'll definitely give a like. Thank you in advance! 1A) From the cross Ab/aB x ab/ab, what is the recombination frequency if the progeny numbers are 17 AB/ab, 72 Ab/ab, 68 aB/ab, and 21 ab/ab? 1B)In human gene mapping, a LOD score is calculated to see if a gene causing a rare disease is linked to a known SNP. The LOD score is -4. This means that 1C) A three-point testcross is used to determine the order of three linked genes. The following crossover progeny result: single crossovers, double crossovers, and no crossovers. To determine the order, the no-crossover progeny must be compared to what other class of progeny?
Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior history of the disease. Consider the following pedigree (the darkly colored symbols represent affected individuals): a. Circle the individual(s) in which the mutation most likely occurred. b. Is the person who is the source of the mutation affected by retinoblastoma? Justify your answer. c. Assuming that the mutant allele is fully penetrant, what is the chance that an affected individual will have an affected child?
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The DNA of every individual in the pedigree shown below has been sequenced at the causative locus. All the non-shaded individuals are wild type apart from III.1. III.1 has been proven to have the causative mutation for this autosomal dominant condition, but they exhibit no symptoms. Based on this small pedigree, what is the level of penetrance for the condition? Please give your answer as a WHOLE percentage, give the number only, no percentage symbol. Answer: The level of penetrance for the condition shown in the pedigree below is Blank 1 percent. 1:1 1:2 Il:1 I1:2 I1:3 Il:4 I1:5 I1:6 II:1 I:2 III:3 III:4 III:3 III:6 III:7 III:8 III:9 III:10 III:11 III12 II:13 III:14 IV:1 | IV:2 IV:3 IV:4 IV:5 IV:6 IV:7 IV:8 IV:9 IV:10 IV:11 IV:12 IV:13 IV:14 IV:15 IV:16 IV:17 IV:18 IV:19 V:1 V:2 V:3 V:4 V:5 V:6 V:7 V:8 V:9 V:10 V:11 V:12
Null mutations are valuable genetic resources becausethey allow a researcher to determine what happens to anorganism in the complete absence of a particular protein. However, it is often not a trivial matter to determinewhether a mutation represents the null state of the gene.a. Geneticists sometimes use the following test forthe nullness of an allele in a diploid organism: If theabnormal phenotype seen in a homozygote for theallele is identical to that seen in a heterozygote(where one chromosome carries the allele in question and the homologous chromosome is known tobe completely deleted for the gene) then the alleleis null. What is the underlying rationale for thistest? What limitations might there be in interpreting such a result?b. Can you think of other methods to determinewhether an allele represents the null state of a particular gene?
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An Hfrstrain that is a *b*c*d* e*f* g *h* is mated with an F strain that is a b e d e f gh. The mating is interrupted at 5 minutes interval, and the genotypes of the F recombinants are determined. The results obtained are tabulated in Table 2. Draw the map of the Hfrchromosome and indicate the position of the origin of transfer, the direction of the transfer and the minutes between genes. Table 2:Entry time of Hfr chromosome into recipient cell. Time a d e f h 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
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