Answer:
in sexual reproduction we get genes from both our parents and we have a combination of them...that leads to genetic diversity...more diversity means more chances to survive different environment conditions.... in asexual reproduction.. offsprings are identical.if environment conditions are rough...all of them won't stand a chance...and also....in sexual reproduction genetic variation can lead to evolutionary advancements
Answer:
The fate of glucose-6-phosphate,glycolytic intermediates and pentose phosphate pathways are described below
Explanation:
Fate of Glucose -6-phosphate
Glucose-6-phosphate undergo dephosphorylation to form glucose when there is an increase demand of glucose in the body.
Glucose-6-phosphate enters into pentose phosphate pathway to synthesize ribose-5-phosphate which is used during denovo pathway of purine nucleotide biosynthesis.
Fate of glycolytic intermediates
Glyceraldehyde-3-phosphate is an important intermediate of glycolysis.The glyceraldehyde-3-phosphate act as a precursor during lipogenesis that deals with the biosynthesis of triacylglycerol.
Fate of pentose phosphate pathway intermediates
Ribose-5-phosphate and NADPH are the important intermediates of pentone phosphate pathway.
Ribose-5-phosphate act as a substrate molecule during the denovo biosynthesis pathway of purine nucleotides.
NADPH act as a reducing agent during fatty acid biosynthesis process.
A baker has created a new strain of yeast which contains no cytochrome c gene and no cytochrome c protein. this will affect what the yeast strain can do to obtain energy. what will this yeast strain do more of compared to a normal strain
Will this new strain of yeast obtain more or less free energy from glucose in its growing medium?
Answer: Less because cytochrome c is key to the electron transport chain.
Explanation:
The Cytochrome c is an essential component of the electron transport chain. Without this there will be no oxidation and reduction of iron atom will take place which could convert the ferrous ions to ferric ions. Thus the entire process of electron transport chain and energy production in the form of ATP will be compromised. So, there will be no production of energy in the anaerobic fermentation by yeast.
Answers:
A. 50-70% - neutrophils
B. 20-40% - Lymphocytes
C. 2-8% - monocytes
D. 1-4% - eosinophils
E. < 1% - basophils
Explanation:
The blood differential test is used to estimate the percentage of each class of white blood cell (WBC) present in the blood and to indicate the presence of abnormal or immature cells.
The Test is Performed by taking of blood sample which is smeared onto a glass slide, then it's stained with a unique dye to indicate the class of white blood cells.
The Five class of white blood cells are
Neutrophils
Lymphocytes (B cells and T cells)
Monocytes
Eosinophils
Basophils
The different class of white blood cells are given as a percentage:
Neutrophils: 40% to 60%
Lymphocytes: 20% to 40%
Monocytes: 2% to 8%
Eosinophils: 1% to 4%
Basophils: 0.5% to 1%
Band (young neutrophil): 0% to 3%
Answer:
The correct answer is c. production of white blood cells, red blood cells, and platelets.
Explanation:
Hematopoiesis is the formation of all types of blood cellular components and components of plasma like white blood cells, red blood cells, and platelets by hematopoietic system which involves the bone marrow, liver, and spleen.
First, the hemopoietic stem cell divides into common myeloid progenitor and common lymphoid progenitor. From myeloid progenitor RBC, platelets, basophils, eosinophils, neutrophils, and macrophages are formed.
Common lymphoid progenitor gives rise to T- lymphocytes, B- lymphocytes, and natural killer cells. Therefore the correct answer is c. production of white blood cells, red blood cells, and platelets.
