Concept explainers
To analyze:
Seven deletion mutations are tested for their ability to form wild-type recombinants with five point mutations as shown in the table below:
Deletion mutation | |||||||
Point mutation | |||||||
a | - | + | - | - | + | + | - |
b | + | + | + | - | + | - | - |
c | + | + | + | + | - | - | - |
d | - | + | + | - | + | - | - |
e | + | - | - | - | + | + | - |
Using the data from the table, construct a genetic map of the order of point mutations and indicate the segment deleted by each deletion mutation.
Introduction:
The symbol ‘
Trending nowThis is a popular solution!
Chapter 6 Solutions
Pearson eText Genetic Analysis: An Integrated Approach -- Instant Access (Pearson+)
- The following is a list of mutational changes. For eachof the specific mutations described, indicate which ofthe terms in the right-hand column applies, either as adescription of the mutation or as a possible cause.More than one term from the right column can applyto each statement in the left column.1. an A–T base pair in the wild-type gene ischanged to a G–C pair2. an A–T base pair is changed to a T–A pair3. the sequence AAGCTTATCG is changed toAAGCTATCG4. the sequence CAGCAGCAGCAGCAGCAGis changed toCAGCAGCAGCAGCAGCAGCAGCAG5. the sequence AACGTTATCG is changed toAATGTTATCG6. the sequence AACGTCACACACACATCGis changed to AACGTCACATCG7. the sequence AAGCTTATCG is changed toAAGCTTTATCGa. transitionb. basesubstitutionc. transversiond. deletione. insertionf. deaminationg. X-rayirradiationh. intercalatori. slippedmispairingarrow_forwardIn each of the illustrations below, a segment of a chromosome has two copies of a transposable element. In panel a, they are oriented in the same direction, whereas in panel b they are in opposite directions. A double strand break occurs in element A and is repaired by homologous recombination using element B as a repair template. For each case, what will the chromosome look like after homologous recombination occurs? Choose one of the five options below, 1-5.arrow_forwardThe D1S80 locus is located on human chromosome 1 and is characterized by a repeating 16 base pair (bp) sequence. Alleles for this locus vary depending on the number of repeats present, thus affecting the size of the locus. The D1S80 locus also contains two conserved sequences, a 32bp sequence at one end and a 113bp sequence at the other end. If the DNA of an individual is targeted for D1S80 amplification, and one of the resulting amplicons is approximately 785bp in size, how many repeats would be present in this D1S80 allele? The amplicon of interest is indicated by a red arrow in the diagram below.arrow_forward
- Recombination signal sequences are conserved heptamer and nonamer sequences that flank the V, J, and D gene segments which undergo recombination to generate the final V region coding exon. Some of these have 12-nucleotide spacers between the heptamer and nonamer, and others have 23-nucleotide spacers. The reason recombination signal sequences come in these two forms is: To ensure the correct assembly of gene segments so that a VH recombines to a DH and not to another VH, for instance To ensure that the heptamer and nonamer are found on the same face of the DNA double helix To ensure that alpha, lambda, and heavy chains recombine within a locus and not between loci To ensure that alpha, lambda, and heavy chain gene segments do not undergo recombination with non-immunoglobulin genes To ensure that the RAG recombinase cuts the DNA between the last nucleotide of the heptamer and the coding sequencearrow_forwardIs homologous recombination an example of mutation? Explain.arrow_forwardA wildtype gene produces the polypeptide sequence: Wildtype: Met-Ser-Pro-Arg-Leu-Glu-Gly Each of the following polypeptide sequences is the result of a single mutation. Identify the most likely type of mutation causing each, be as specific as possible. M1:Met-Ser-Ser-Arg-Leu-Glu-Gly missense mutation M2:Met-Ser-Pro M3:Met-Ser-Pro-Asp-Trp-Arg-Asp-Lys M4:Met-Ser-Pro-Glu-Gly nonsense mutation frameshift insertion in frame deletion M5:Met-Ser-Pro-Arg-Leu-Glu-Gly in frame insertionarrow_forward
- Consider the following coding sequence transcribed from 5' to 3'5' A T G A A G C G C T C A G T A 3' If a guanine is substituted for nucleotide 11 what type of base substitution has occurred (nucleotide level) and what would be the resulting phenotypic effect?arrow_forwardYou have the following DNA coding sequence of a wild-type allele: 5’-ATG TTC CAG CTA GAT GAT ATG CTG GTA ATT GGG GAA CGC GCG CGG TAA-3’ For each of the following mutations: A. State whether the mutation is missense, nonsense, frameshift, or silent. B. Write the codon change that occurs for the missense, nonsense, and silent mutations (ex. GAA -- GAT). C. For frameshift mutations, write out the entire mutant sequence with each codon clearly indicated (if the frameshift creates a new stop codon, end the sequence at the new stop). Using the wild type DNA sequence above as a guide : Write the amino acid sequence of the mutants. Mutant 1: transition at nucleotide 23 Mutant 2: T --> G transversion at nucleotide 29 Mutant 3: an insertion of “A” after nucleotide 14 Mutant 4: transition at nucleotide 7 Mutant 5: An insertion of GG after nucleotide 40 Mutant 6: transition at nucleotide 15 Mutant 7: a deletion of nucleotide 25arrow_forwardThe genetic alteration responsible for sickle-cell anemia in humans involves: a transition mutation from A to G, substituting glutamic acid for valine in a-globin a transversion mutation from T to A, substituting valine for glutamic acid in b-globin a transition mutation from T to C, substituting valine for glutamic acid in b-globin a transversion mutation from G to C, substituting glutamic acid for valine in a-globin a frameshift mutation of one ATC codon, removing glutamic acid from b-globinarrow_forward
- After isolating six Neurospora mutations, you mate A and a representatives of each in all pairwise combinations and determine which mutations complement. Table 1gives the results, with a (+) indicating complementation, and a (-) representing lack of complementation. With the exception of mutant 1, how many different genes are represented in this experiment, or can you tell from these experiments?arrow_forwardThe map shows a wild-type HOAP gene, whereas Mutant 1 has a DNA deletion from base pair 1125 to 1445 and Mutant 2 contains a point mutation at 1372. Predict the sizes of DNA fragments that would be generated in each of the following digests. Wild-type (BamHI): Wild-type (EcoRI): Wild-type (BamHI + EcoRI): Mutant 1 (BamHI): Mutant 1 (EcoRI): Mutant 1 (BamHI + EcoRI): Mutant 2 (BamHI): Mutant 2 (EcoRI): Mutant 2 (BamHI + EcoRI): BamHI O EcoRI 740 Promoter ECORI 1140 HOAP HOAP HP1/ORC Associated Protein 7200 bp Sacl 2100 BamHI 2420 Psti 2700 Hindlll 3440arrow_forwardThe DNA of a deletion of alpha bacteriophage has a length of 15 micrometers instead of 17 micrometers? How many base pairs are missing from this mutant?arrow_forward
- Human Anatomy & Physiology (11th Edition)BiologyISBN:9780134580999Author:Elaine N. Marieb, Katja N. HoehnPublisher:PEARSONBiology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStaxAnatomy & PhysiologyBiologyISBN:9781259398629Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa StouterPublisher:Mcgraw Hill Education,
- Molecular Biology of the Cell (Sixth Edition)BiologyISBN:9780815344322Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter WalterPublisher:W. W. Norton & CompanyLaboratory Manual For Human Anatomy & PhysiologyBiologyISBN:9781260159363Author:Martin, Terry R., Prentice-craver, CynthiaPublisher:McGraw-Hill Publishing Co.Inquiry Into Life (16th Edition)BiologyISBN:9781260231700Author:Sylvia S. Mader, Michael WindelspechtPublisher:McGraw Hill Education