Question

in humans, unattached earlobes and freckles are dominant traits. if two individuals heterozygous for both traits reproduce, what is the chance that their child will have attached earlobes and no freckles? 1/4 3/16 1/16 9/16 3/4

Explanation:

Step1: Determine the genotypes

Let \(E\) represent the allele for unattached earlobes (dominant) and \(e\) for attached earlobes (recessive). Let \(F\) represent the allele for freckles (dominant) and \(f\) for no freckles (recessive). The parents are both heterozygous for both traits, so their genotype is \(EeFf\).

Step2: Use the Punnett square or probability rules

For the earlobe trait: \(Ee\times Ee\). The probability of getting \(ee\) (attached earlobes) is \(\frac{1}{4}\) (from \(P(ee)=\frac{1}{4}\) as \(Ee\times Ee\) gives \(EE:Ee:ee = 1:2:1\)).
For the freckle trait: \(Ff\times Ff\). The probability of getting \(ff\) (no freckles) is \(\frac{1}{4}\) (from \(Ff\times Ff\) gives \(FF:Ff:ff=1:2:1\)).

Step3: Apply the multiplication rule (for independent events)

Since the inheritance of earlobe type and freckle presence are independent events (Mendel's law of independent assortment), the probability of both events (attached earlobes and no freckles) occurring simultaneously is \(P(ee\cap ff)=P(ee)\times P(ff)\).
Substitute the values: \(P(ee\cap ff)=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}\)

Answer:

\(\frac{1}{16}\) (corresponds to the third option in the multiple - choice list)