#Question id: 20385
#Unit 13. Methods in Biology
#Question id: 4539
#Unit 3. Fundamental Processes
The yeast S. cerevisiae exists in three forms: two haploid cells of different mating types—a and alpha and the diploid formed when an a and an a cell mate and fuse. The a cell and the a cell each encodes cell-type-specific regulators: a cells make the regulatory protein a1, and alpha cells make the proteins alpha1 and alpha2. A fourth regulatory protein, called Mcm1, is also involved in regulating the mating-type-specific genes (and many other genes) and is present in both cell types which shows Combinatorial Control.
The arrangement of regulators at the promoters of a-specific genes and a-specific genes is
|
cell type: |
gene regulatory proteins: |
target genes: |
|
1.a cell (haploid) |
a1 Mcm1 |
aSG alphaSG |
|
2.αlpha cell (haploid) |
alpha 1 alpha 2 Mcm2 |
hSG |
|
3.a/αlpha cell (diploid) |
a1 Mcm2 alpha2 |
#Question id: 4772
#Unit 8. Inheritance Biology
A black Labrador, Three phenotype of black, brown and yellow is control by non allelic gene interaction. Following some genotype
The following conclusions were made
A. The 9:3:4 ratio obtained upon selfing of AaBb here indicates recessive epistasis
B. This is a recessive epistasis where homozygous bb is epistatic to A and a.
C. When Gene A and B incompletely link, It will be produced all phenotype black, brown and yellow upon selfing of AaBb
D. When Gene A and B incompletely link, test cross of AaBb will be produced all phenotype black, brown and yellow in ratio 1:1:2 respectively
Which of the following conclusions are correct?
#Question id: 10413
#Unit 6. System Physiology – Plant
#Question id: 738
#Unit 1. Molecules and their Interaction Relevant to Biology
Which of the following is correct for peptide bonds
A) C and N are shorter than typical C-N bonds.
B) C and N are longer than typical C-N bonds.
C) C and O are longer than typical C=O bonds.
D) C and O are shorter than typical C=O bonds.