Answer:
a. Type O blood - No A or B antigens on RBCs-Anti-A antibodies in plasma-Anti-B antibodies in plasma
b. Type A blood - A antigen on RBCs- Anti-B antibodies in plasma
c. Type B blood - B antigen on RBCs- Anti-A antibodies in plasma
d. Type AB blood - A antigen on RBCs- B antigen on RBCs- Neither anti-A or anti-B in plasma
Explanation:
ABO blood grouping system represents multiple allelism which was discovered in humans by Karl Landsteiner. The blood group is determined by the presence or absence of A & B antigens and antibodies.
The 4 blood groups which exist in ABO system are O, A, B and AB. Also, allele A and B are co-dominant i.e. if they both will exist on the surface of an RBC then both will be equally expressed which implies that there will be no dominant or recessive allele.
Type O blood group has H antigen on the surface of RBC and has both the antibodies i.e. antibody A and B. The absence of antigens A and B makes it a universal donor.
Type A blood group has A antigen on RBC and has antibody B in the plasma.
Type B blood group has B antigen on RBC and has antibody A in the plasma.
Type AB blood group has both the antigens i.e. antigen A and B on RBC and does not have any antibody in the plasma which makes it a universal acceptor.
Animal Kingdom. One of the six kingdoms, characteristics include: heterotrophic (obtains energy from eating other organisms), multicellular, eukaryotic and cells have a nucleus but no cell wall or chloroplasts.
Answer:
competitive exclusion principle
Answer:
A) cyclic and predictable
Explanation:
The graph shows changes in the water levels (in feet, ft). The water cycle consists of water that evaporates from the surface (land and sea) and subsequently returns in form of rain and/or snow. In general, water levels may be unpredictable in short term but predictable in the long term. The graph shows that this process is cyclic because it is regularly patterned and also occurs in regular intervals (therefore being predictable).
Answer:
The correct answer would be 93 milliliters.
The same can be explained with the help of the unitary method.
The volume of solution required for 1 kg of weight in one hour = 2.5 milliliters (given in the question).
Thus, the volume of the solution required for 37.3 kg of weight in one hour = 
milliliters
It comes out be 93.25 milliliters.
Thus, around 93 milliliters of the solution should be given to the child of weight 37.3 kg in one hour.
