Question

1. to calculate the possible genotypes of the offspring of a given crossing, you must know the types of gametes that the parents can make. due to independent assortment, you can calculate the probability of each genotype in the offspring for each gene separately, then multiply the probability of each event happening at the same time.

for example, the probability of a dihybrid cross (aabb x aabb) producing an offspring with the genotype aabb can be calculated as:

• probability of aa in a cross between aa x aa = ¼
• probability of bb in a cross between bb x bb = ¼
• probability of aa and bb = ¼ x ¼ = 1/16

applying this same reasoning, calculate the probability of the genotypes below resulting from the crosses indicated:
a) aabb x aabb producing aabb =
b) aabb x aabb producing aabb =
c) aabbcc x aabbcc producing aabbcc =
d) aabbcc x aabbcc producing aabbcc =
e) aabbcc x aabbcc producing aabbcc =
f) considering that a = round seed and b = yellow color, what is the probability of aabb x aabb producing offspring that is round and yellow (regardless of the genotype)?

Explanation:

Step1: Calculate probability for each gene

For \(Aa\times Aa\), \(P(Aa)=\frac{1}{2}\) (using Punnett square: \(AA:Aa:aa = 1:2:1\)). For \(Bb\times Bb\), \(P(Bb)=\frac{1}{2}\) (similarly \(BB:Bb:bb=1:2:1\)).

Step2: Multiply probabilities for independent events

Since \(Aa\) and \(Bb\) are independent (Mendel's law of independent assortment), \(P(AaBb)=P(Aa)\times P(Bb)\)
\(P(AaBb)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\)

Step1: Analyze each gene cross

For \(Aa\times aa\), \(P(Aa)=\frac{1}{2}\) (Punnett square: \(Aa:aa = 1:1\)). For \(Bb\times bb\), \(P(Bb)=\frac{1}{2}\) (Punnett square: \(Bb:bb=1:1\))

Step2: Multiply probabilities

Since the gene crosses are independent, \(P(AaBb)=P(Aa)\times P(Bb)\)
\(P(AaBb)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\)

Step1: Calculate probability for each gene

For \(Aa\times Aa\), \(P(aa)=\frac{1}{4}\). For \(Bb\times Bb\), \(P(bb)=\frac{1}{4}\). For \(Cc\times Cc\), \(P(cc)=\frac{1}{4}\)

Step2: Multiply probabilities

Using the multiplication rule for independent events (Mendel's law of independent assortment), \(P(aabbcc)=P(aa)\times P(bb)\times P(cc)\)
\(P(aabbcc)=\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}=\frac{1}{64}\)

Step1: Analyze each gene cross

For \(AA\times aa\), \(P(Aa) = 1\) (all offspring will be \(Aa\)). For \(bb\times BB\), \(P(Bb)=1\) (all offspring will be \(Bb\)). For \(CC\times cc\), \(P(Cc) = 1\) (all offspring will be \(Cc\))

Step2: Multiply probabilities

Since \(P(Aa) = 1\), \(P(Bb)=1\), \(P(Cc)=1\) and events are independent, \(P(AaBbCc)=P(Aa)\times P(Bb)\times P(Cc)\)
\(P(AaBbCc)=1\times1\times1 = 1\)

Answer:

\(\frac{1}{4}\)