The probable genotype of the IV-3 with the perspective of autosomal inheritance will be A because he does not have condition A. Therefore both parents must be in condition A.
The probable genotype of IV-3 with the perspective of set=linked inheritance will be BbY this is because he has B condition. Therefore the mothers' genotype must be XBXb and the father's genotype must be XBY.
The probability of IV-4 having condition A is 1/36
I think its c because both groups receive the same amount of of sunlight
Anaxyrus, woodhousii. From what I remember
<span>The correct way to classify organisms is as follows :Kingdom, such as Eukaryote, Phylum such as Animalia, Phylum such as Chordata, class, order, genus and family are mostly used for classification purposes in taxonomy and species, such as Oreochromis mossambicus, which is the mozambique tilapia.</span>
The molecule which is the final electron acceptor for electrons from photosystem I is (d) NADP⁺.
Photosystem I is the protein complex involved in the process of photosynthesis. It captures the light energy to mediate the transfer of electrons from a series of electron transporters. It is involved in non-cyclic as well as cyclic photophosphorylation.
NADP⁺ is the Nicotinamide Adenine Dinucleotide Phosphate. It acts as a coenzyme. It is an important constituent in various anabolic reactions like Calvin cycle, lipid and nucleic acid syntheses, etc. The oxidized form of NADP⁺ is NADPH. It is present in organisms of almost all kinds.
To know more about photosystem I, here
brainly.com/question/14427520
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