18. Brown eyes (B) are dominant to blue eyes (b). The presence of freckles (F) is dominant to the absence of freckles (f). What
is the probability that two people who are heterozygous for both traits will have a child that has blue eyes and no freckles?
1 answer:
Answer:
For bbff we have only 6.3% probability
Step-by-step explanation:
If the parents are heterozygous for both traits, them they are represented by:
BbFf × BbFf
Parent 1: BbFf
Parent 2: BbFf
We have to find the percentage of occurence of bb × ff, which is a child that has blue eyes and no freckles, with no dominant factor.
By distributing the possibilities in a Punnett square, <em>vide</em> picture. We have the following possibilities:
Genotype Count Percent
bBfF 4 25
BBfF 2 12.5
bBFF 2 12.5
bBff 2 12.5
bbfF 2 12.5
BBFF 1 6.3
BBff 1 6.3
bbFF 1 6.3
bbff 1 6.3
For bbff we have only 6.3% probability
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Answer:
a)
The combined resistance of a circuit consisting of two resistors in parallel is given by:
where
R is the combined resistance
are the two resistors
We can re-write the expression as follows:
Or
In order to see if the function is increasing in r1, we calculate the derivative with respect to r1: if the derivative if > 0, then the function is increasing.
The derivative of R with respect to r1 is:
We notice that the derivative is a fraction of two squared terms: therefore, both factors are positive, so the derivative is always positive, and this means that R is an increasing function of r1.
b)
To solve this part, we use again the expression for R written in part a:
We start by noticing that there is a limit on the allowed values for r1: in fact, r1 must be strictly positive,
So the interval of allowed values for r1 is
From part a), we also said that the function is increasing versus r1 over the whole domain. This means that if we consider a certain interval
a ≤ r1 ≤ b
The maximum of the function (R) will occur at the maximum value of r1 in this interval: so, at