Answer
The three metabolic pathways that make up aerobic respiration are really all parts of one larger pathway because the products of early pathways (like NADH) become <u>utilize</u> in the last one.
Explanation
Aerobic respiration is that type of respiration in which glucose molecule is broken down into CO2 and H2O in the presence of oxygen and 36 or 38 ATP molecules are produced.
Aerobic respiration complete in four main steps:
1. Glycolysis
In this step glucose is broken down into 2 molecules of pyruvate acid along with the production of 2 ATP molecules and 2NADH.
2. Oxidation of pyruvate
In this step pyruvate are oxidized in the presence of co-enzyme A to become Acetyl Co-enzyme A. Again 2NADH are formed in this step.
3. Kreb Cycle
It occus in mitochondria. Here acetyle coenzyme A enter Carbon fixation, reduction and regeneration phase. In this cycle 6 NADH, 2FADH2 and 2ATP are formed.
4. Electron transport chain
All NADH that are produced in above steps get oxidize and help in the production of ATP along with the release of electron and proton that help in the formation of water.
a. hybridization
b. incomplete dominance
c. true-breeding
d. the law of segregation
e. polygenetics
Answer:
c. true-breeding
Explanation:
True-breeding is a type of breeding which involves a cross between two parents that are homozygous for a trait leading to the production of offspring from generation to generation that have genotypes that express the same phenotypic traits. The parent plants with purple flowers in the question, are both homozygous for flower color.
A longshore current moves parallel tho the shore. longshore currents are shallow water currents.
Answer:
No changes occur to the K1 value, its concentration remains higher than the concentration of the inhibitor and enzyme inhibitor complex
Explanation:
Uncompetitive inhibition is an example of a reversible inhibition. Reversible inhibitors bind to enzymes by weak non-covalent bonding. Thus the formation and dissociation of this association is rapid. uncompetitve inhibition lowers the Vmax and Km.
Answer:
Option (A), (B), (D) and (F).
Explanation:
Bacteria are involved in the domains of eubacteria and archaea. Bacteria are different from the other organism and shows different in the chemical and cellular structure of the cell.
The translation elongation factors are different in bacteria and other organisms. The translation factors of bacteria are EF-Tu and EF-Ts whereas the eukaryotic elongation factors are eEF-1 subunit α and eEF-1 subunit βγ. A single RNA pol is present in bacteria and three different RNA pol is present in eukaryotes. Peptidoglycan is present in prokaryotes. Phospholipids contain ester linkage in bacteria and ether linkage in archaea.
Thus, the correct answer is option (A), (B), (D) and (F).
