Answer:
1. The parents genotypes could have been BO and AO
2. wire-hair
Explanation:
There are four possible blood types which are type A, B, AB, O. blood group is the classification of blood based on the presence or the absence of inherited antigenic substances on the surface of the red blood cells. They have hereditary basis and also rely on a series of alternative genes sometimes used in solving dispute of parental heritage. With the four possible blood groups, there are six possible genotypes and these are:
Blood type possible genotypes
Type A AA, AO
Type B BB, BO
Type AB AB
Type O OO
Thus, for parents with blood type B and A to give birth to a child with blood type O, it means their genotype could have been both BO and AO for them to be able to produce a child with OO. a cross between these two could give rise to OO.
Question 2
Wire hair is dominant (S) to smooth (s), thus wire hair could be in the homozygous (SS) and heterozygous form (Ss) and the smooth hair can only be expressed in the homozygous recessive form (ss).
thus, in a cross between homozygous wire haired and smooth haired, we will have:
homozygous wire haired homozygous smooth haired
P gen SS x ss
F1 gen. Ss
phenotype: wire haired
Answer:
All prokaryotes have chromosomal DNA localized in a nucleoid, ribosomes, a cell membrane, and a cell wall.
Explanation:
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Answer:
The correct answer is option d, that is, endocytosis.
Explanation:
A procedure in which a component gains entry within the cell without getting passed through the cell membrane is termed as endocytosis. In this process, the cell captivates the substances from the external of the cell by engulfing and combining them with its plasma membrane.
The process is further differentiated into three distinct kinds. These are phagocytosis, pinocytosis, and receptor-mediated endocytosis.
If the bill is counterfeit and the paper is wood-based, the iodine in the pen solution will react with the starch and leave a dark brown or black mark.
