Answer:
characterized by presence or absence of antigens
the blood types are A, B, O, AB
Explanation:
There are two antigens and two antibodies that are mostly responsible for the ABO types. The specific combination of these four components determines an individual's type in most cases. Erythrocytes and serum were related to the presence of antigens on these erythrocytes and antibodies in the serum. these antigens are A and B, and depending upon which antigen the erythrocytes express, blood either belonged to blood group A or blood group B. A third blood group contained erythrocytes that reacted as if they lacked the properties of A and B, and this group was later called "O" blood group. The fourth blood group AB, was added to the ABO blood group system. These erythrocytes expressed both A and B antigens.
Blood group Antigen present on RBC Antibodies in serum Genotype(s)
A antigen A anti-B AA or AO
B antigen B anti-A BB or BO
AB both A and B antigen none AB
O none anti-A and anti-B OO
D. all of the above
as they are all, evergreen, short and contain flammable oils
Answer and Explanation:
<u>Cross:</u> aa Bb dd Ee x AA bb Dd Ee
We can calculate the probability of getting heterozygous individuals in the progeny by using the <u>product rule</u>. Assuming that these four genes <u>assort independently</u> (<em>events that occur independently from each other</em>), we can infer that the F1 will have the next genotypic proportions for each gene:
1) aa x AA
F1) 4/4=1 Aa
2) Bb x bb
F1) ½ bb
½ Bb
3) dd x Dd
F1) ½ dd
½ Dd
4) Ee x Ee
F1) ¼ EE
2/4 = ½ Ee
¼ ee
So, to know what the probability is that the F1 of being heterozygous for all loci, we must multiply the respective individual probabilities of getting a heterozygous genotype, like this:
1 Aa x ½ Bb x ½ Dd x ½ Ee = 1/8 AaBbDdEe
