3.27 Complementation tests of the recessive mutantg genes a through f produced the data in the accom-panying matrix. The circles represent missing data.beaim Assuming that all of the missing mutant combina-motions would yield data consistent with the entriesithat are known, complete the table by filling eachoneel circle with a + or – as needed.-rt yd bat21igi a rıba O +b c d e fi612nobsdibliw insiqansbosale yns mon las29niYOuConoe66 BA AAfAA

Complementation test is a kind of genetic test which is used to find that whether mutations are…

3.27 Complementation tests of the recessive mutant genes a through f produced the data in the accom- panying matrix. The circles represent missing data. am Assuming that all of the missing mutant combina- o tions would yield data consistent with the entries pithat are known, complete the table by filling each circle with a + or – as needed. - 21 fonob lo le gNA in Jigi a b a O + Cdei O + O - ienobsdi insiqansn - C -- d ono e 66 BA AA f

3.27 Complementation tests of the recessive mutant genes a through f produced the data in the accom- panying matrix. The circles represent missing data. am Assuming that all of the missing mutant combina- o tions would yield data consistent with the entries pithat are known, complete the table by filling each circle with a + or – as needed. - 21 fonob lo le gNA in Jigi a b a O + Cdei O + O - ienobsdi insiqansn - C -- d ono e 66 BA AA f

Biology 2e

2nd Edition

ISBN:9781947172517

Author:Matthew Douglas, Jung Choi, Mary Ann Clark

Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark

Chapter12: Mendel's Experiments And Heredity

Section: Chapter Questions

Problem 26CTQ: The gene for flower position in pea plants exists as axial or terminal alleles. Given that axial is...

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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
A rare blinding disease that has a relation to dengenerative factors is partially penetrant. In the following pedigrees for two families, the affected symptomatic individuals (black circles and squares) have been diagnosed with this disease due to the mutation in mitochondrial DNA m.14484T>C. If III.4 is homoplasmic for m.14484T>C in hair, blood, urine and other tissues examined. What will occur with IV.7 then?
6.8 Define a map unit and explain why map units best reflect the real distances between two genes only when the genes are relatively close together.
The gene controlling ABO blood type and the gene underlying nail-patella syndrome are said to show linkage. What does that mean in terms of their relative locations in the genome? What does it mean in terms of how the two traits are inherited with respect to each other?
1. a. Assume that two transcription factors are required for expression of the blue pigmentationgenes in pansies. (Without the pigment, theflowers are white.) What phenotypic ratioswould you expect from crossing strains heterozygous for wild-type and recessive amorphicalleles for each of the genes encoding thesetranscription factors?b. Now assume that either transcription factor issufficient to get blue color. What phenotypicratios would you expect from crossing the sametwo heterozygous strains?
In mice, the autosomal locus coding for the β-globinchain of hemoglobin is 1 m.u. from the albino locus. Assume for the moment that the same is true in humans. The disease sickle-cell anemia is the result ofhomozygosity for a particular mutation in theβ-globin gene.a. A son is born to an albino man and a woman withsickle-cell anemia. What kinds of gametes will theson form, and in what proportions?b. A daughter is born to a normal man and a womanwho has both albinism and sickle-cell anemia.What kinds of gametes will the daughter form,and in what proportions?c. If the son in part (a) grows up and marries thedaughter in part (b), what is the probability that achild of theirs will be an albino with sickle-cellanemia?
In an in situ hybridization experiment, a certain clonebound to only the X chromosome in a boy with no diseasesymptoms. However, in a boy with Duchenne musculardystrophy (X-linked recessive disease), it bound to theX chromosome and to an autosome. Explain. Could thisclone be useful in isolating the gene for Duchenne muscular dystrophy?
90. Prader-Willi and Anglelman syndroms have disntinct and non-overlapping phenotypes. Both syndroms are commonly associated with deletions of chromosome 15q12 that appears to be cytogenetically identical. Which of the following factors best explains the different phenotypes? (A) Decreased Penetrance (B) Genomic Impriting (C) Locus Heterogenity (D) Post-Translational Modification (E) SOmatic Mosaicism
A recent estimate of the rate of base substitutions atSNP loci is about 1 × 10−8 per nucleotide pair pergamete.a. Based on this estimate, about how many de novomutations (that is, mutations not found in the genomes of your parents) are present in your own genome?b. Where and when did these de novo mutations inyour genome most likely occur?
1. There are two different variations at STR locus THO1. Where did these variations come from? b) The STR locus D21S11 only has one variation. Why?
Below is a figure (here called Figure 1) from “Prognostic Significance of CpG Island Methylator Phenotype and Microsatellite Instability in Gastric Carcinoma,” by An et al., published in Clinical Cancer Research in 2005. The authors look at five microsatellite loci (BAT 25, BAT 26, D2S123, D5S346, and D17S250) in normal (N) and tumor (T) tissue from patients with Gastric Carcinoma. They amplify the loci by PCR and then instead of using standard agarose gel electrophoresis, they run the PCR products through capillary gel electrophoresis and detect bands as they pass a laser near the positive charge terminal. The x-axis in these plots is the time at which the band passed the laser (aka size of the PCR product) and the intensity of the peaks represents the amount of DNA in that band A. Which patient- 18, 30, or 1- shows the most microsatellite instability? Which patient shows the least? How do you know? B. In which repair pathway is it most likely that you will find the driver mutations…
6. A diploid strain of yeast was made by mating a haploidstrain with a genotype w−, x−, y−, and z− with a haploidstrain of opposite mating type that is wild type for thesefour genes. The diploid strain was phenotypically wildtype. Four different X-ray-induced diploid mutantswith the following phenotypes were produced fromthis diploid yeast strain. Assume a single new mutation is present in each strain.Strain 1 w− x+ y− z+Strain 2 w+ x− y− z−Strain 3 w− x+ y− z−Strain 4 w− x+ y+ z+When these mutant diploid strains of yeast go throughmeiosis, each ascus is found to contain only two viablehaploid spores.a. What kind of mutations were induced by X-rays tomake the listed diploid strains?b. Why did two spores in each ascus die?