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Concept explainers
a.
To determine:
The nature of the rearranged chromosome in each fly and approximate location of breakpoints that define the chromosomal rearrangement. The way to detect the extra
Introduction:
The fruit flies with scarlet-colored eyes are heterozygous for chromosome 3. They have a normal gene. They also have a scarlet mutation and rearranged chromosome 3.
b.
To determine:
Whether one can conclude that at least part of st gene lies within chromosome 3.
Introduction:
The changes in the sequence of nucleotides are called mutations. The agents that cause mutations are known as mutagens. The st gene is the mutagen that causes recessive scarlet mutation in fruit flies.
c.
To determine:
Whether one or more protein-coding exons of st gene lie outside chromosome 3.
Introduction:
The eukaryotic mRNA is composed of two regions. These are the coding and non-coding regions. The coding regions are termed as exon while the non-coding parts are called introns. The introns interfere in the synthesis of proteins. They are removed from the mRNA by the method of RNA splicing.
d.
To determine:
Whether mapping of st gene reflects a recombination experiment or a complementation test.
Introduction:
The test that is used to detect whether recessive mutations are present within the same gene or in different genes is termed as complementation test. The scientists use a table called the complementation table to study and compare the results of the complementation test.
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Chapter 12 Solutions
Genetics: From Genes to Genomes, 5th edition
- Using figure 1 and the following background information answer the following questions. Identification of the genetic cause of hornlessness in cattle has been the subject of intensive genetic and genomic research, culminating in the nomination of two different candidate neomutations on cattle chromosome 1 that are predicted to have arisen 500-1,000 years ago: a complex allele of Friesian origin (PF), an 80,128 base pair (bp) duplication (1909352–1989480 bp), and a second, simple allele of Celtic origin (PC) corresponding to a duplication of 212 bp (chromosome 1 positions 1705834–1706045) in place of a 10-bp deletion (1706051–1706060)We report the use of genome editing using transcription activator-like effector nucleases (TALENs) to introgress the putative PC POLLED allele into the genome of bovine embryo fibroblasts to try and produce a genotype identical to what is achievable using natural mating, but without the attendant genetic drag and admixture. In our previous studies, we…arrow_forwardAn STR on chromosome 5 is very closely linked to a gene involved in a dominant metabolism disorder (one one copy of the mutation causes the disease). There are four different alleles of the STR (Q, B, S, and L), each with different numbers of GATCTCG repeats. The S allele has 2 repeats, the B allele has 7 repeats, the Q allele has 10 repeats, and the L allele has 5 repeats. Part 1 Lane 1 of the gel shown below indicates the locations of PCR products corresponding to all four STR alleles when run together in one lane. DNA was loaded into the wells at the top of the gel. Move the letter of each allele to its correct position at the left of Lane 1 to indicate where each allele runs on the gel. 0000 Gabriel Andre Lane 1 O Maria Andre Maria Juliana B L Part 2 Andre and Maria are the parents in this family; all others shown on the gel are children. Daughter Juliana is born with the metabolism disorder, but daughter Camila does not have the disorder. Assuming no recombination between the STR…arrow_forwardThe haploid human genome consists of approximately 3.2 × 109 bp (or 6.4 × 109 bp per diploid genome). Assuming that DNA is B-DNA, calculate the total length of a single cell’s DNA.arrow_forward
- A pair of paralogous repeats, A and B, have 96% sequence similarity and therefore can promote non-allelic homologous recombination (NAHR). They exist in four possible arrangements in a genome, illustrated below as arrangements 1 – 4. What is the result of NAHR between repeats A and B in arrangement 1? A.Translocation between chromosomes 1 and 2 resulting in monocentric chromosomes B.Deletion or duplication of the region between A and B C.Translocation between chromosomes 1 and 2 resulting in acentric and dicentric chromosomes D.Inversion of the region between A and Barrow_forwardColchicine is a chemical mutagen that inhibits the spindle formation and prevents anaphase, which retains the cell’s single restitution nucleus (doubled chromosome number). Suppose that an onion (2n=16) is subjected to three consecutive rounds of colchicine treatment, what will be the resulting chromosome number of the treated onion?arrow_forwardArabidopsis thaliana has among the smallest genomes in higher plants, with a haploid genome size of about 100 Mb. If this genome is digested with BbvCl, a restriction enzyme which cuts at the sequence CCTCAGC GGAGTCG 1. approximately how many DNA fragments would be produced? Assume the DNA has a random sequence with equal amounts of each base.arrow_forward
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- A diploid (ie, contains TWO sets of chromosomes) organisms with a 45,000-kb haploid (counts only one set of its chromosomes) genome contains 21% G residues. Calculate the number of A, C, G< and T residues in the DNA of each cell in this organism. Can you help explain why this is the answer, thank you! Answer: Since the haploid genome contains 21% G, it must contain 21% C (Because G=C) and 58% A + T, or 29% A and 29% T. Each cell is a diploid, containing 90,000 kb or 9x10^7 bases. Therefore, A=T = (0.29)(9x10^7) = 2.61 x 10^7 bases and G=C=(0.21)(9x10^7) = 1.89x10^7 bases.arrow_forwardTo detect the CAG repeat expansion with a particular gene where 30 repeats in Normal changes to 250 repeats in a certain disease, how can we diagnose the condition. How To identify Y chromosome microdeletion ( which involves the deletion of AZF locus) using conventional karyotyping? If not then why. How will you diagnose a chromosomal translocation event?arrow_forwardA 5-year-old boy with mental retardation is grossly obese and has facial features of Prader-Willi syndrome (PWS). Karyotyping and fluorescent in situ hybridization studies do not show deletion in the usual site on chromosome 15. Which of the following findings is most likely to confirm PWS in this child? ✓ A Deletion in the short arm of chromosome 15 BY Duplication within chromosome 15 Large trinucleotide repeat expansion in the PWS area of chromosome 15 D) Maternal origin of both chromosomes 15 -Translocation in the short arm of chromosome 15arrow_forward
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