Following statements are regarding to the protein phosphatase Cdc14 function in budding yeast.

A. In budding yeast, the protein phosphatase Cdc14 throws the switch between two mutually antagonistic states during anaphase.

B. Cdc14 is kept inactive during most of the cell cycle, but is activated during anaphase by a GTPase signaling pathway known as the mitotic exit network.

C. Mitotic exit network is unresponsive to spindle position and becomes active in anaphase.

D. Cdc14 phosphatase activated by the activation of protein kinase cascade.

E. Cdc14 dephosphorylates APC/CCdh1 and Sic1 to promote mitotic cyclin degradation

and mitotic CDK inactivation, respectively. This process leads to exit from mitosis.

Which of the following combination is incorrect?

TLS Online TPP Program

#Question id: 12703

#Unit 8. Inheritance Biology

In the plants, due to mutation in the mitochondrial genome causing cytoplasmic male sterility can be influenced by the nuclear gene (Rf),  which is dominant over the cytoplasmic mitochondrial genome. If a male sterile plant  is pollinated by the fertile male plant with homozygous (Rf)condition, the progeny obtained in F1 will have

TLS Online TPP Program

#Question id: 12704

#Unit 8. Inheritance Biology

Cytoplasmic male sterility (CMS) in plants is caused by mutation in the mitochondrial genome. CMS can be restored by a nuclear gene, restorer of fertility (Rf), which is a dominant character. 

TLS Online TPP Program

#Question id: 12705

#Unit 8. Inheritance Biology

In Neurospora crasssa, the mutant exhibit poky phenotype, when a female of  poky strain is crossed with a normal strain acting as a male , all progeny individuals shows poky phenotype . however, the reciprocal cross resulted in all normal progeny. These results can be explained on the basis of 

TLS Online TPP Program

#Question id: 12706

#Unit 8. Inheritance Biology

When the genotype is determined by the bi-parental contribution, while the phenotypic effect is determined by the maternal nuclear gene. This effect exhibit on the basis of 

TLS Online TPP Program

#Question id: 12707

#Unit 8. Inheritance Biology

AaBb=100% chance of crossing over between gene A and B than chance of  recombination?

TLS Online TPP Program

#Question id: 12708

#Unit 8. Inheritance Biology

800 meiocytes with the genotype AaBb undergoes meiosis with 60% chance of crossing over with A and B. How many recombinant gamete form?