If an AaBbccDd individual is crossed to a AaBbCcDD individual, what is the probability that the first offspring will be of AabbccDd genotype?

The probability that the first offspring will be of AabbccDd genotype is 0.03125.

Step-by-step explanation:

The parents genotype are: AaBbccDd and AaBbCcDD.

         A           a

A      AA         Aa

a       Aa         aa

P (Aa) = 2/4 = 1/2

         B           b

B      BB         Bb

b      Bb         bb

P (bb) = 1/4

         c           c

C      Cc         Cc

c       cc         cc

P (cc) = 2/4 = 1/2

         D           d

D      DD        Dd

D      DD        Dd

P (Dd) = 2/4 = 1/2

Then the probability that the first offspring will be of AabbccDd genotype is:

P (AabbccDd) = (1/2) × (1/4) × (1/2) × (1/2) = 1/32

Thus, the probability that the first offspring will be of AabbccDd genotype is 0.03125.

deez nu tz

step-by-step explanation:

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180