Genetics - Genetics EXAM II Review 1. What general...

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Genetics EXAM II Review 1. What general observation identifies two interacting genes? A. A mutation in either gene is lethal B. Mutations in only one of the genes produce an observable phenotype C. The combined effect of mutations in both genes is the sum of their individual effects *D. The combined effect of mutations in both genes is not the sum of their individual effects E. Mutations in either gene produce a distinct phenotype Two mutations can be recognized as interacting if their combined phenotypic effect is different than predicted from the sum of their individual effects. 2 . A “-“ indicated no growth on the plate. The answer is ordering the compounds with no growth so you must order them by growing pattern based on which compounds have no growth the most. 3. How can haplosufficiency explain a null (loss of function) allele being recessive? *A. A single functional copy of the gene produces sufficient gene product for normal biological function The definition of haplosufficient: Describes a gene that, in a diploid cell, can promote wild-type function in only one copy (dose). Therefore a single copy of the gene is sufficient! 4. What observations were crucial in allowing Watson and Crick to propose the double helix structure of DNA? *A. Chargraff’s rule of the relative frequency of different nucleotides and Franklin’s X ray diffraction of the DNA molecule. Chargraff discovered 2 rules about DNA nucleotides which helped to understand DNA structure. First %A = %T and %G = %C. And second %A ~ %T and %G ~ %C are valid for each of the two DNA strands. Franklin used x-ray diffraction to photograph DNA and showed that it is parallel with two strands. B. Griffith’s observation of a transforming factor. This experiment was used to figure out the identity of genetic material (DNA). Not relevant to structure C. Hershey and Chase’s demonstration of DNA as a genetic material. This experiment used phages to determine whether protein or DNA was the genetic material. (Not relevant to structure) D. The determination of the genetic code. – irrelevant 5. What would you expect to happen during DNA replication in a cell lacking ligase? A. DNA ahead of the replication fork would become increasingly tightly coiled. opposite *B. Okazaki fragments would not be joined. Ligase is like the glue in DNA replication- joining the 3′ end of the gap-filling DNA to the 5′ end of the downstream Okazaki fragment. Therefore without ligase the okazaki fragments would not be joined.
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6. You have the ability to ‘label’ phage so that their DNA becomes radioactive but their protein remains non radioactive. What experiment would most convincingly determine the identity of the material that is injected into bacteria on infection with your labeled phage. Infect bacterial cells with labeled phage and: (Note: phage ghosts are the remnants of the phage left on the outside of the bacterial cell after the phage genetic material has been injected.) A. Test for radiation in everything.
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This note was uploaded on 12/02/2011 for the course BIOL 3301 taught by Professor Gunaratne during the Spring '05 term at University of Houston.

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Genetics - Genetics EXAM II Review 1. What general...

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