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use the completed punnett square in part b to answer the questions below about the f2 generation. drag the probabilities on the left to the blanks on the right to answer the questions. terms can be used once, more than once, or not at all.

Respuesta :

Using completed punnett square,we get

p(A) = 3/4 or 75%

1 out of 3 or 33℅

63/64

27/64

you can do it by the theoretical definition of probability, using a formula:

p(A) = n(A) / n(Ω)

where,

p(A) stands for "probability of happening the event A",

n(A) stands for "number of elements or of individual events that satisfy the definition of A",

n(Ω) stands for "total number of possibilities; number of elements in the sample".

Being event A "choosing a seed of any yellow specimen",

n(Ω) = 4 (4 genetic possibilities)

n(A) = 3 (3 out of all of the elements in Ω satisfy the event, or, if being chosen, would mean that the event has happened, as 3 out of 4 seeds are yellow, according to genetics):

p(A) = 3/4 or 75%

Answer 2.

1 out of 3 or 33℅

Only YY will breed true for yellow. Remaining two seeds have Yy genotype which is heterozygous.

Answer 3.

63/64

Simply add up all the possibilities of drawing at least 1 yellow seed

YYY=.75x.75x.75=27/64

YYG=.75x.75x.25=9/64

YGY=.75x.75x.25=9/64

GYY=.75x.75x.25=9/64

YGG=.75x.25x.25=3/64

*GYG=.25x.75x.25=3/64

*GGY=.25x.25x.75=3/64

Note: GGG will yield green seeds with a probability .25x.25x.25 = 1/64

Answer 4.

From solution of answer 3, sum the probabilities of YYG, YGY and GYY i.e. 9+9+9 = 27/64

To know more about completed punnett square here

https://brainly.com/question/11515153

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