Question

ii. 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 was dark skinned. 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 genetics of blood types.

blood types
the four blood types in the abo blood type system are type a, type b, type ab, and type o. these blood types refer to
different versions of carbohydrate molecules which are present on the surface of red blood cells. these blood types
result from 3 different alleles of a gene as shown in the following table. because blood types are determined by more
than 2 alleles, they are an example of: __________________________. (p. 304)

allele	codes for:
$i^b$	type b carbohydrate molecules
$i$	inactive; doesnt code for either type

people with:	have:
type b blood
type ab blood
type o blood

1. complete the following table to relate genotypes to blood types.

genotype	blood type	genotype	blood type
$ii$		$i^b i$
$i^a i$	a	$i^a i^b$

1. in a person with the $i^a i$ genotype, which allele is dominant, $i^a$ or $i$?

1. codominance occurs when both alleles are dominant. which of the genotypes results in a codominant blood type?

were the babies switched?
now you are ready to evaluate whether the baby girls were switched. the following family trees show the blood types
for each person in both families.

family tree 1: earnest (a) and denise (b) have tonja (o)
family tree 2: michael (ab) and danielle (o) have michelle (a) and michael, jr. (b)

Explanation:

Question 1: Complete the table

Step1: Analyze \( I^A I^A \)

The allele \( I^A \) codes for Type A carbohydrate. So two \( I^A \) alleles will result in Type A blood.

Step2: Analyze \( i i \)

The allele \( i \) is inactive. So two \( i \) alleles result in Type O blood (since no A or B carbohydrates are coded).

Step3: Analyze \( I^B I^B \)

The allele \( I^B \) codes for Type B carbohydrate. Two \( I^B \) alleles result in Type B blood.

Step4: Analyze \( I^B i \)

The \( I^B \) allele is dominant over \( i \) (as \( i \) is inactive), so this genotype results in Type B blood.

Step5: Analyze \( I^A I^B \)

Both \( I^A \) and \( I^B \) are dominant (codominance), so this results in Type AB blood.

Filling the table:

Genotype	Blood Type	Genotype	Blood Type
\( i i \)	O	\( I^B i \)	B
\( I^A i \)	A	\( I^A I^B \)	AB

Question 2: Dominant allele in \( I^A i \)

Step1: Recall Genotype - Phenotype

The genotype \( I^A i \) has a blood type of A (from the table in Q1). The \( I^A \) allele codes for Type A, and \( i \) is inactive. So the phenotype (blood type A) shows that \( I^A \) is expressed, meaning \( I^A \) is dominant over \( i \).

Question 3: Codominant genotype

Step1: Recall Codominance Definition

Codominance is when both alleles are dominant and both are expressed in the phenotype.

Step2: Analyze Genotypes

• For \( I^A I^A \): Only A is expressed.
• For \( I^B I^B \): Only B is expressed.
• For \( I^A i \): Only A is expressed ( \( i \) is inactive).
• For \( I^B i \): Only B is expressed ( \( i \) is inactive).
• For \( I^A I^B \): Both A and B carbohydrates are present (since both \( I^A \) and \( I^B \) are dominant and expressed), so this is codominant.

Answer:

s:
1.

Genotype	Blood Type	Genotype	Blood Type
\( i i \)	O	\( I^B i \)	B
\( I^A i \)	A	\( I^A I^B \)	AB

1. The dominant allele is \( I^A \).

1. The genotype \( I^A I^B \) results in a codominant blood type.