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Concept explainers
a.
To determine:
The results observed in the DNA containing multiple copies of 3′ AATCCC 5′ being used as a probe.
Introduction:
The chromosomes are condensed structures that are formed during the early phase of cell division from the loose network of chromatin thread and then regain their original structure after being divided into daughter cells.
b.
To determine:
The sequence of the probe to be used to track the end of the chromosome in a FISH experiment.
Introduction:
The genes are the sequence of nucleotides that are present on the chromosomes and encode for a specific protein that plays a crucial role in the functioning of the different processes in an organism. The gene is located at specific gene loci and can be structural or regulatory in nature.
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Chapter 11 Solutions
Genetics: From Genes to Genomes, 5th edition
- Below are 9 possible primer pairs.● Determine which primer pair is the best choice.● Explain why the other primers are not good choices.● Calculate the Tm for each primer. Underline or highlight the region of DNA for the primer pair you chose as the best.Forward 1: 5’ gaaataattttgtttaactttaag 3’ Tm =Reverse 1: 5’ gtttaagacaaaatagtctgg 3’ Tm =Forward 2: 5’ gtaactcagctttcaggtcg 3’ Tm =Reverse 2: 5’ tctcggaatgttgcaacagc 3’ Tm =Forward 3: 5’ agattagcggatcctacctg 3’ Tm =Reverse 3: 5’ atgtgtaatcccagcagcag 3’ Tm =Forward 4: 5’ cattgattatttgcacggcg 3’ Tm =Reverse 4: 5’ aaaatcttctctcatccgcc 3’ Tm =Forward 5: 5’ tccataagattagcggatcc 3’ Tm =Reverse 5: 5’ tgcaagcttggctgttttgg 3’ Tm =Forward 6: 5’ gatcctacctgacgcttttta 3’ Tm=Reverse 6: 5’ aaataatgaattcgagctcggt 3’ Tm =Forward 7: 5’ataaaaaaatcgagataaccgtt 3’ Tm =Reverse 7: 5’aggtcgactctagaggatc 3’ Tm =Forward 8: 5’ctacctgttccatggccaac 3’ Tm=Reverse 8: 5’ ttcgggcatggcactcttg 3’ Tm=Forward 9: 5’ tccataagattagcggatcc 3’ Tm =Reverse 9: 5’…arrow_forwardA molecular biologist is investigating homologous recombination. One aim of this study is to reconstitute stages of the process in vitro. (a) Draw diagrams to show how the four synthetic oligonucleotides below could base-pair to form a stable model Holliday junction. W 5’ GATCGCATTGTAGCCGTAGGTCCACTGTAA 3’ X 5’ GTCCCATACGTAGCCGTAGGACATGTACCG 3’ Y 5’ CGGTACATGTCCTACGGCTACAATGCGATC 3’ Z 5’ TTACAGTGGACCTACGGCTACGTATGGGAC 3’ (b) What is branch migration? (c) What is the name of the enzyme that resolves a Holliday junction into two separate DNA duplexes? (d) On your diagram, indicate how the Holliday junction can be resolved in two different ways and draw the structures of the products.arrow_forwardSupercoiled DNA is slightly unwound compared to relaxed DNA and this enables it to assume a more compact structure with enhanced physical stability. Describe the enzymes that control the number of supercoils present in the E. coli chromosome. How much would you have to reduce the linking number to increase the number of supercoils by five?arrow_forward
- How would the appearance of your DNA gel change if a 0.1% agarose gel slab was used? What about a 7% agarose gel slab? For separating small fragments of DNA, is it better to use a 7% gel or a 0.1% gel? Why?arrow_forwardIn Noll’s experiment , explain where DNase I cuts the DNA. Why were the bands on the gel in multiples of 200 bp at lower DNase I concentrations?arrow_forwardA 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?arrow_forward
- Using the first and second base key below, predict the DNA sequence given by the SOLID color sequence. For the key G = green, R = red, Y = yellow, and B = blue. Note that the first base of the sequence is already given ("A"). Give the remaining 8 bases for this sequence. A First base A CCT Second base A CGT BGY R GBRY RBG R Y (G) B Y G)(R) GB )( R )( Y ) ( G) Barrow_forwardIn hybridizing microarrays, we use a solution that contains {3X SSC, 0.5% SDS, 5ug/ml salmon sperm DNA, 5% glycerol}. Describe how to make 100ul of this solution using stocks of 20X SSC, 10% SDS, and 10 mg/ml salmon sperm DNA, sheared, in water. Note that Molecular Biology Grade H2O must be used in making up this solution.arrow_forwardYou have run a gel with 5 uL of amplified DNA in addition to 1 uL of 6X loading dye. Comparing the brightness of this band to your marker, you estimated that your amplified DNA is approximately 62 ng. What was the concentration of your amplified DNA that was loaded? You do not need to account for the volume of loading dye in your calculations.arrow_forward
- 1a) When performing classical Sanger or "dideoxy" sequencing, you set up 4 parallel reactions per template to be sequenced from a specific primer, with each of the four reactions containing a different dideoxynucleotide, and then the four reactions were run in a separate, adjacent lanes on a gel. Why couldn't you combine all 4 dideoxynucleotides with the primer and the template and do the whole reaction in one tube, and then run the set of fragments produced by the reaction mixture on a single lane in an acrylamide gel?arrow_forwardA molecular biologist is investigating homologous recombination. One aim of this study is to reconstitute stages of the process in vitro. Draw diagrams to show how the four synthetic oligonucleotides below could base-pair to form a stable model Holliday junction. W 5’ GATCGCATTGTAGCCGTAGGTCCACTGTAA 3’ X 5’ GTCCCATACGTAGCCGTAGGACATGTACCG 3’ Y 5’ CGGTACATGTCCTACGGCTACAATGCGATC 3’ Z 5’ TTACAGTGGACCTACGGCTACGTATGGGAC 3’arrow_forward5) The restriction enzyme EcoRI recognizes the DNA sequence GAATTC and makes a staggered cut as shown in the blue figure below: GA ATTC CTTAAG If EcoRI cuts the generic fragment shown below, 2 fragments with "sticky ends" (single-stranded DNA) will result. "XXX.." denotes any combination of nucleotides that flank either side of the sequence. 6 base pairs 8bp's 5-СTGTAGAATІCGTTACGA- 3' 3-GACATCTTAAGCAATGCT- 5 CTGTAG EcoRI Digest AATTCGTTACGA GCAATGCT GACATCTTAA A. Given the sequence in the section of DNA shown below, draw boxes around the fragments that will be formed when cutting this region of DNA with EcoRI and HAEIII at the same time. 5'- GTGGCCATTCTAGCGCCGAATTCGGCAATTACGATCGTAGGCCATCGATGTAATTC -3' 3'- CACCGGTAAGATCGCGGCTTAAGCCGTTAATGCTAGCATCCGGTAGCTACATTAAG -5' B. Give the sizes in base pairs of each of the fragments that will be formed after making the cuts with these 2 enzymes (don't count the nucleotides in the single-stranded parts).arrow_forward