TLS Online TPP Program

#Question id: 15147


You have isolated a set of five yeast mutants that form dark red colonies instead of the usual white colonies of wild-type yeast. You cross each of the mutants to a wild-type haploid strain and obtain the results shown below;
you cross each haploid mutant strain to a different haploid mutant of the opposite mating type. From the results shown below deduce as much as you can about which mutations lie in the same gene. Clearly state any remaining ambiguities and suggest some general ways that the ambiguities might be resolved
a) That mutants 1 and 3 form one complementation group and are mutations in the same gene (gene A) 
b) That mutations 2 and 5 form a second complementation group and are mutations in a second gene (gene B).
c) That mutations 3 and 5 form a second complementation group and are mutations in a second gene (gene B).
d) The first ambiguity is whether mutant 4 has a mutation in gene A or B, or whether it represents a unique gene.
Which of the following is the correct prediction about mutants?

#Unit 8. Inheritance Biology
  1. b, c and d
  2. a, b and c
  3. c and d only
  4. a, b and d
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TLS Online TPP Program

#Question id: 4480

#Unit 3. Fundamental Processes

The reason for TBP’s unorthodox recognition mechanism is

TLS Online TPP Program

#Question id: 4481

#Unit 3. Fundamental Processes

Choose correct statements about recognition mechanism of TBP ;

A. TBP causes the minor groove to be widened to an almost flat conformation

B. It also bends the DNA by an angle of ~120o

C. The interaction between TBP and DNA involves maximum number of hydrogen bonds between the protein and the edges of the base.

D. Much of the specificity is imposed by two pairs of phenylalanine side chains that intercalate between the base pairs at either end of the recognition sequence and drive the strong bend in the DNA.

TLS Online TPP Program

#Question id: 4482

#Unit 3. Fundamental Processes

The General Transcription Factors of RNA Polymerase, with their subunits, select correct matching

A.     TBP

      1

B.     TFIA

      11

C.     TAFs

       2

D.     TFIIH

      10

TLS Online TPP Program

#Question id: 4483

#Unit 3. Fundamental Processes

TAF42 and TAF62 from Drosophila form a structure similar to that of the

TLS Online TPP Program

#Question id: 4484

#Unit 3. Fundamental Processes

Match General Transcription Factors with their functions;

    i. TAFs

 A. Appears to regulate the binding of TBP to DNA. It does this using an inhibitory flap that binds to the DNA-binding surface of TBP, another example of molecular mimicry. This flap must be displaced for TBP to bind TATA.

  ii. TFIIB

B. Insert into the RNA-exit channel and active center cleft of Pol II in a manner analogous to the sigma region 3/4 linker in the bacterial case.

  iii. TFIIH

 C.  As an ATP-driven translocator of double-stranded DNA.

   iv. TFIIF

D. This two-subunit (in humans) factor associates with Pol II and is recruited to the promoter together with that enzyme.

TLS Online TPP Program

#Question id: 4485

#Unit 3. Fundamental Processes

In multicellular organisms, regulatory sequences can spread thousands of nucleotides from the promoter—both upstream and downstream—and can be made up of tens of regulator binding sites. Often, these binding sites are grouped in units called_?