ch14 - Chapter 14 14.1 In what ways does our present...

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Chapter 14 14.1 In what ways does our present concept of the gene differ from the pre-1940 or classical concept of the gene? ANS: Prior to 1940, the gene was considered a “bead-on-a-string,” not subdivisible by recombination or mutation. Today, the gene is considered to be the unit of genetic material that codes for one polypeptide. The unit of structure, not subdivisible by recombination or mutation, is known to be the single nucleotide pair. FEEDBACK: 14.1 DIFFICULTY: easy 14.2. What was the first evidence indicating that the unit of function and the unit of structure of genetic material were not the same? ANS: The recombination observed between lz s and lz g , two functionally allelic mutations at the lozenge locus of Drosophila . FEEDBACK: 14.1 DIFFICULTY: easy 14.3 What is the currently accepted operational definition of the gene? ANS: The cis-trans test, which defines the unit of genetic material specifying the amino acid sequence of one polypeptide. FEEDBACK: 14.1 DIFFICULTY: easy 14.4. Based on our current concept of the gene, (a) what is the smallest unit of genetic material that can be changed by mutation, and (b) what is the smallest region of genetic material in which recombination can occur? ANS: (a) Nucleotide. (b) Between adjacent nucleotide base-pairs. FEEDBACK: 14.1 DIFFICULTY: easy 14.5 Four mutant strains of Neurospora require one or more of the supplemented
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metabolites A through D to grow. From the data below (where “+” = growth; “0” = no growth), draw the metabolic pathway for the synthesis of these metabolites and show where the mutant strains (1, 2, 3, and 4) are blocked. Metabolite Mutant A B C D 1 0 0 0 + 2 + + + + 3 0 0 + + 4 + 0 + + ANS: FEEDBACK: 14.3 DIFFICULTY: medium 14.6. Seven mutants of Neurospora are unable to grow on minimal medium unless it is supplemented with one or more of the metabolites A through G. On the basis of the data given below (where “+” indicates growth and “0” indicates no growth), draw a biochemical pathway for the synthesis of these seven substances and show where the mutants are blocked in the pathway. Growth in the Presence of Metabolite(s) Mutant A B C D E F G B + C D + G B + C + E 1 + 0 0 0 0 0 0 0 + + 2 0 + 0 0 0 0 0 + 0 + 3 0 0 + 0 0 0 0 + 0 + 4 0 0 0 + + 0 0 0 + + 5 0 0 0 0 + 0 0 0 0 + 6 + 0 0 0 0 + 0 0 + + 7 0 0 0 0 0 0 + + + + ANS: FEEDBACK: 14.3 DIFFICULTY: hard
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14.7 Of what value are conditional lethal mutations for genetic fine structure analysis? ANS: They provide powerful selective sieves for identifying rare recombinants. This is accomplished by using the restrictive environmental conditions to select wild-type recombinant progeny from crosses between pairs of conditional lethal mutants. FEEDBACK: 14.3 DIFFICULTY: easy 14.8. Eight independently isolated mutants of E. coli , all of which are unable to grow in the absence of histidine (his - ), were examined in all possible cis and trans heterozygotes (partial diploids). All of the cis heterozygotes were able to grow in the absence of histidine. The trans heterozygotes yielded two different responses: some of them grew in
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This note was uploaded on 11/26/2011 for the course BIOLOGY 2c03 taught by Professor Dej during the Spring '11 term at McMaster University.

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ch14 - Chapter 14 14.1 In what ways does our present...

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