Awoman who is a carrier for colorblindness and has five fingers marries a colorblind man with six fingers (his mother had five fingers and the gene for six fingers is dominant). what percent of the kids would you predict to be colorblind and have five fingers?

50% of the kids (all the boys) will have 6 fingers and 50% will be colorblind

Step-by-step explanation:

Women have two X chromosome: XX

Men have one X and one Y chromosome:

Colour blindness is a hereditary condition, passed down from your parents. The gene which is responsible for the condition is carried on the X chromosome, reason why many more men are affected than women

c: colorblindness gene

Woman carrier of colorblindness: XcX (or XXc)

Woman colorblindness: XcXc

Woman not carrier of colorblindness gene or not colorblindness: XX

Man colorblindness: XcY

Man not colorblindness: XY

Dominant gene show their effect even if the individual only has one copy of the gene (e.g. gene for brown eyes is dominant, therefore you only need one copy of the 'brown eye' gene to have brown eyes (although, with two copies you will still have brown eyes))

6: 6 fingers gene (dominant gene)

Woman 6 fingers: X6X (or XX6), X6X6

Woman 5 fingers: XX

Man 6 fingers: X6Y, XY6, X6Y6

Man 5 fingers: XY

In this particular case the woman is a carrier for colorblindness ( XcX or XXc) and have 5 fingers (XX)

The man is colorblind (XcY) and have 6 fingers (X6Y, XY6, X6Y6)

Now we have to look which of the 6 fingers option for the man is the correct one for this particular exercise:

If the mother of the man, have 5 fingers (and as 6 fingers is a dominant gene), she must be XX, so he got the 6 finger gene from the father, then he have the Y chromosome with the 6 finger gene and the X chromosome without 6 finger gene. The right option is XY6

So let's cruz the woman XcX with the man XcY6

The results are XcXc, XXc, XcY6 and XY6

XcXc : girl colorblindness

XXc: girl carrier of colorblindness gene

XcY6: boy colorblindness and 6 fingers

XY6: boy with 6 fingers

50% of the kids (all the boys) will have 6 fingers and 50% will be colorblind

john earns $14.00 for two hours of work.

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2