Answer:
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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
Answer:
The correct answer will be- 21, 29, 29
Explanation:
In a DNA sequence, the nucleotide base pairs on one strand of DNA are complementary to the base pair on other strand of DNA.
According to the Chargaff rule, Adenine binds Thymine and Cytosine binds Guanosine which shows that the amount of A will equal T and the amount of G will equal C.
Therefore, when the amount of C is 21%, then the amount of G will be 21%.
To find amount of AT= 100-GC
AT= 100-42
AT= 58%
So, AT/2= 29% each
Thus, A=T= 29%
G=C=21%
Answer: .374 and .626
Explanation:
5. total number of alleles = 4712 + 2816 = 7528
frequencey of B allele = B allele / total number of alleles --> 2816/7528 = 0.374
6. total number of alleles = 4712 + 2816 = 7528
freq of b allele = b allele/total number of alleles --> 4712/7528 = 0.626
We can confirm our answer as correct because if we add up the answers to 5 and 6 we should get one:
0.374 + 0.626 = 1
Cation-exchange chromatography is used when the molecule of interest is positively charged, the stationary phase is negatively charged and positively charged molecules are loaded to be attracted to it. So, the amino acids with negative charge will elute the first. Glutamate, leucin, arginine is the order of elution because of their pI values ~3, ~6 ~10.
