Secretors (genotypes ss and ss) secrete their a and b blood group antigens into their saliva and other body fluids, while nonsec
retors (ss) do not. what would be the apparent phenotypic blood group proportions among the offspring of an iaib ss woman and an iaia ss man if typing was done using saliva?
1 answer:
As you can see from the image presented below when parents with AB Ss and AA ss genotypes cross we get three distinct phenotype groups.
Marked in red are the individuals whose A type blood we can determine from the saliva because they have at least one S allele.
Marked in light green are the individuals whose AB type of blood we can determine from the saliva too.
And marked in orange are the individuals that are homozygous recessive for the s allele, and we could not determine their blood type from their saliva.
Marked in red are the individuals whose A type blood we can determine from the saliva because they have at least one S allele.
Marked in light green are the individuals whose AB type of blood we can determine from the saliva too.
And marked in orange are the individuals that are homozygous recessive for the s allele, and we could not determine their blood type from their saliva.
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Answer:
The age difference between the oldest and the youngest is 48.
Explanation:
Given the following data;
The ratio of the ages of Kisii (K) and Esinam (E) is 3:5.
The ratio of the ages of Esinam (E) and Larry (L) is 3:5.
The sum of the ages of all three is 147.
Since the age of Esinam is common in both ratios, we find the lowest common multiple (LCM);
LCM of 3,5 = 15.
Hence, the ratio of the three ages is now;
K:E:L = 9:15:25
Let their ages be denoted by x;
The sum of all three ages is;
9x + 15x + 25x = 147
49x = 147
x = 147/49
x = 3.
To find the age difference between the oldest and the youngest;
The oldest is Esinam = 25x
The youngest is Kisii = 9x
Therefore, 25x - 9x = 16x
Substituting the value of x, we have;
x = 3; 16x = 16(3) = 48.
The age difference between the oldest and the youngest is 48.