Lab: Using Blood Tests To Identify Criminals And Babies - Anatomy And Physiology Worksheet Page 5

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Part 3. Were the babies switched?
Two couples had babies on the same day in the same hospital.
 Denise and Earnest had a girl, Tonja.
 Danielle and Michael had twins, a boy, Michael, Jr., and a girl, Michelle.
Danielle was convinced that there had been a mix-up and she had the wrong girl, since Michael
Jr. and Tonja were both light-skinned, while Michelle had darker skin. Danielle insisted on
blood type tests for both families to check whether there had been a mix-up. In order to
interpret the results of the blood type tests, you will need to understand the basic biology of
blood types.
Genetics of Blood Types
The ABO blood types result from the alleles of a gene that can code for two different versions
of a protein enzyme or an inactive protein as shown in this table:
Allele Codes for a protein that is
A
I
a version of the enzyme that puts Type A carbohydrate molecules on the surface of
red blood cells
B
I
a version of the enzyme that puts Type B carbohydrate molecules on the surface of
red blood cells
i
inactive; doesn't put either type of carbohydrate molecule on red blood cells
12. Each person has two copies of this gene, one inherited from his/her mother and the other
inherited from his/her father. Complete the following table to relate genotypes to blood types.
Genotype This person's cells make
Blood Type
the version of the enzyme that puts Type A carbohydrate
A
A
I
I
molecules on the surface of red blood cells
i i
the inactive protein
the version of the enzyme that puts Type A carbohydrate
A
I
i
molecules on the surface of red blood cells and the inactive
A
protein.
A
A
13. In a person with the I
i genotype, which allele is dominant, I
or i? Explain your reasoning.
14. Complete the following table to describe each of the three genotypes listed.
Genotype Will this person's cells make the version of the enzyme needed to
Blood
attach this carbohydrate on the surface of his or her red blood cells?
Type
B
B
I
I
Type A __yes __ no;
Type B __yes __ no
B
I
i
Type A __yes __ no;
Type B __yes __ no
A
B
I
I
Type A __yes __ no;
Type B __yes __ no

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