Question

1. if an offspring has the genotype aa, what possible combinations of parental genotypes could have produced this offspring?

1. in corn, the trait for tall plants (t) is dominant to the trait for dwarf plants (t) and the trait for colored kernels (c) is dominant to the trait for white kernels (c). in a particular cross of corn plants, the probability of an offspring being tall is 1/2 and the probability of a kernel being colored is 3/4. which of the following most probably represents the parental genotype? include your work to show how you derived your solution.

a. ttcc x ttcc
b. ttcc x ttcc
c. ttcc x ttcc
d. ttcc x ttcc
e. ttcc x ttcc

1. in humans, the allele for albinism (lack of pigment) is recessive to the allele for normal skin pigmentation.

a. if two heterozygous parents have children what is the chance that a child will be albino?
b. if the child is normal, what is the chance that it is a carrier (heterozygous) for the albino allele?
c. if normal parents have an albino child, what is the probability that their next child will be normal for pigment?

Explanation:

Question 6

Step 1: Recall Genotype Inheritance

For an offspring with genotype \( Aa \), we consider possible parental genotype combinations. The principle of segregation (Mendel's law) states that alleles separate during gamete formation. Possible parental crosses that can produce \( Aa \) are:

• \( AA \times aa \): All offspring will be \( Aa \) (since \( AA \) produces \( A \) gametes and \( aa \) produces \( a \) gametes, so \( A \times a = Aa \)).
• \( Aa \times aa \): Using Punnett square, gametes from \( Aa \) are \( A \) and \( a \), from \( aa \) are \( a \). Offspring genotypes: \( Aa \) (50%) and \( aa \) (50%).
• \( Aa \times Aa \): Gametes from each \( Aa \) are \( A \) and \( a \). Offspring genotypes: \( AA \) (25%), \( Aa \) (50%), \( aa \) (25%).
• \( AA \times Aa \): Gametes from \( AA \) are \( A \), from \( Aa \) are \( A \) and \( a \). Offspring genotypes: \( AA \) (50%) and \( Aa \) (50%).

Step 2: List Possible Combinations

So the possible parental genotype combinations are:

1. \( AA \) (parent 1) and \( aa \) (parent 2)
2. \( Aa \) (parent 1) and \( aa \) (parent 2)
3. \( Aa \) (parent 1) and \( Aa \) (parent 2)
4. \( AA \) (parent 1) and \( Aa \) (parent 2)

Step 1: Analyze Traits and Probabilities

We have two traits: tall ( \( T \), dominant) vs. dwarf ( \( t \), recessive) and colored ( \( C \), dominant) vs. white ( \( c \), recessive) kernels. The probability of tall ( \( T\_ \)) is \( \frac{1}{2} \) and colored ( \( C\_ \)) is \( \frac{3}{4} \). We analyze each cross:

Option a: \( TtCc \times ttCc \)

• For height ( \( T/t \)): \( Tt \times tt \). Gametes: \( T, t \) (from \( Tt \)) and \( t, t \) (from \( tt \)). Offspring: \( Tt \) (50%, tall) and \( tt \) (50%, dwarf). So \( P(\text{tall}) = \frac{1}{2} \).
• For color ( \( C/c \)): \( Cc \times Cc \). Gametes: \( C, c \) (from each \( Cc \)). Offspring: \( CC \) (25%), \( Cc \) (50%), \( cc \) (25%). So \( P(\text{colored}) = \frac{3}{4} \) ( \( CC + Cc \) ). This matches the given probabilities.

Option b: \( TtCc \times TtCc \)

• Height: \( Tt \times Tt \). Offspring: \( TT \) (25%), \( Tt \) (50%), \( tt \) (25%). \( P(\text{tall}) = \frac{3}{4} \) (not \( \frac{1}{2} \)). Eliminate.

Option c: \( TtCc \times ttcc \)

• Color: \( Cc \times cc \). Offspring: \( Cc \) (50%), \( cc \) (50%). \( P(\text{colored}) = \frac{1}{2} \) (not \( \frac{3}{4} \)). Eliminate.

Option d: \( TTCc \times ttCc \)

• Height: \( TT \times tt \). Offspring: \( Tt \) (100%, tall). \( P(\text{tall}) = 1 \) (not \( \frac{1}{2} \)). Eliminate.

Option e: \( TTCc \times TtCC \)

• Height: \( TT \times Tt \). Offspring: \( TT, Tt \) (100%, tall). \( P(\text{tall}) = 1 \) (not \( \frac{1}{2} \)). Eliminate.

Step 1: Define Alleles

Let \( A \) be normal allele, \( a \) be albinism allele (recessive). Heterozygous parents: \( Aa \times Aa \).

Step 2: Punnett Square

Gametes: \( A, a \) (from each \( Aa \)).
Offspring genotypes:

• \( AA \): 25% (normal, non - carrier)
• \( Aa \): 50% (normal, carrier)
• \( aa \): 25% (albino)

So \( P(\text{albino})=\frac{1}{4} \).

Answer:

Possible parental genotype combinations: \( AA \times aa \), \( Aa \times aa \), \( Aa \times Aa \), \( AA \times Aa \)

Question 7