TLS Online TPP Program

#Question id: 15153


In this problem we will explore some of the many ways that mutations in two different genes can interact to produce different Mendelian ratios. Consider a hypothetical insect species that has red eyes. Imagine mutations in two different unlinked genes that can, in certain combinations, block the formation of red eye pigment yielding mutants with white eyes. In principle, there are two different possible arrangements for two biochemical steps responsible for the formation of red eye pigment. The two genes might act in series such that a mutation in either gene would block the formation of red pigment. Alternatively, the two genes could act in parallel such that mutations in both genes would be required to block the formation of red pigment.
Further complexity arises from the possibility that mutations in either gene that lead to a block in enzymatic activity could be either dominant or recessive. If the crosses is made between a wild type insect with red eyes and a true breeding white eyed strain with mutations in both genes. Such considerations yield the Pathways in parallel with recessive mutations in both genes, determine the phenotype of the F1 progeny and the expected phenotypic ratio of red to white eyed insects in the F2.

#Unit 8. Inheritance Biology
  1. F1 will have red eyes

    Phenotypic ratio in F2 will be 1 white:15 red

  2. F1 will have red eyes

    Phenotypic ratio in F2 will be 3 white: 13 red

  3. F1 will have red eyes

    Phenotypic ratio in F2 will be 1 white:15 red

  4. F1 will have whit eyes

    Phenotypic ratio in F2 will be 9 white: 7 red

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TLS Online TPP Program

#Question id: 1245

#Unit 4. Cell Communication and Cell Signaling

Many PTB-containing proteins act as docking sites for multiple proteins. If these proteins are involved in the RTK signal transduction pathway, they most likely:

TLS Online TPP Program

#Question id: 1246

#Unit 4. Cell Communication and Cell Signaling

An SH2-containing protein contains a mutation that changes its binding pocket such that tyrosine and phosphotyrosine bind with equal affinity. As a result, MEK activity:

TLS Online TPP Program

#Question id: 1247

#Unit 4. Cell Communication and Cell Signaling

Which of the following is NOT true about the role of adapter proteins in the activation of Ras by receptor tyrosine kinases?

TLS Online TPP Program

#Question id: 1248

#Unit 4. Cell Communication and Cell Signaling

By what mechanism does PI-3 phosphate promote activation of protein kinase B (PKB)?

a. recruiting PKB to the plasma membrane

b. recruiting the activating kinase PDK1 to the plasma membrane

c. releasing inhibition of the catalytic site by the PH domain

d. by the activation of FOXO3A

TLS Online TPP Program

#Question id: 1249

#Unit 4. Cell Communication and Cell Signaling

As in all G protein–coupled signalling pathways, timely termination of the rhodopsin signalling pathway requires that all the activated intermediates be inactivated rapidly, restoring the system to its basal state, ready for signalling again. Which of the following mechanism does not terminate rhodopsin signalling?

TLS Online TPP Program

#Question id: 1250

#Unit 4. Cell Communication and Cell Signaling

Few mutants (column A) and their results (column B) are listed in the table given below:

Column AColumn B
A. Gain-of-function of SkiI. Inhibits the ability of activated type I TGF-β receptors (RI) to phosphorylate R-Smad proteins
B. Loss-of-function of RII receptorII. Transcription activation induced by TGF-β and mediated by Smad complexes is shut down
C. Gain-of-function of smad-7
D. Loss-of-function of R-smad

Match the correct mutation with the observe result?