Answer:
Consider the heterozygous oval, thick cell walled bacteria to have the alleles OoTT and the thin cell walled bacteria to have alleles oott. Results will be 50% oval, thick walled bacteria and 50% round, thick walled bacteria. This will be the F1 progeny.
When the oval, thick walled bacteria from the F1 progeny is cross bred with round, thick walled bacteria then 25 percent of the bacteria will be heterozygous oval, thick walled. 25 percent will be heterozygous oval and heterozygous thick walled. 25 percent will be round and thick walled. 25 percent will be round and thin walled.
Answer AND Explanation:
It causes opposite seasons in the Northern and Southern Hemispheres.
obliquity
A decrease in this value indicates a more circular orbit.
eccentricity
It affects the orientation of the North Star, currently located above Earth’s northern axis. precession
Answer:
abt 24 contraction per min is needed to maintain a relatively stable internal solute conc.
paramecium maintain its volume by preventing itself from shrinking by holding in as much water as it can.
When the water solute concentration is reduced, the number of vacuole contractions will increase. But when the water solute concentrations rise, the number of vacuole contraction will decrease.
When the number of vacuole contractions will increase, the water solute concentration is reduced. But when the water solute concentrations rise, the number of vacuole contraction will decrease. So it is cetris paribus, means when the one is increase the other one will decrease.
Answer:
Option D, 75%
Explanation:
Let the genotype of co dominant checkered hen mates and checkered rooster be CcC and CcC
Where Cc – is the allele for chekered skin
C – is the allele for non chekered skin
Co-dominant allele are those which irrespective of being dominant or recessive are expressed equally.
The punnet square for the cross between these two would be
Cc C
Cc CcCc CcC
C CCc CC
3 out of 4 offsprings have chekered allele i.e CcCc, CcC and CcC. Thus, the probability of chekered offsprings would be 75%
Answer:
-During citric acid cycle or Krebs cycle, radioactive carbon will be first appear in citric acid
Explanation:
During cellular respiration, glucose is first converted into pyruvate molecules by the process of glycolysis. These pyruvate molecules go for oxidative decarboxylation, during which acetyl co-enzyme A (acetyl CoA) is formed along with removal of carbon dioxide.
The acetyl co-enzyme enters to the next step in cellular respiration or citric acid cycle or Krebs cycle. The first step of citric acid cycle is formation of citric acid by joining of acetyl CoA and oxaloacetate.
As citric acid is first molecule formed during citric acid cycle, radioactive carbon would be first appear in citric acid.
