*I am reasking this because the last guy just copied and pasted the information

Question

*I am reasking this because the last guy just copied and pasted the information already given; I want to know the math behind the numbers (380/380/20/20). What is posted is the answer key!*

Hello, I was hoping someone could explain/ show the work for this problem. Our teacher only showed us how to get to map units when given the numbers of offspring for each phenotype, but not the other way around, so I have no idea how they got those numbers. Please help! :)

Question 3 While working with a type of beetle that is normally smooth, large, and white, you discover three mutations that lead to the recessive phenotypes bumpy, small, and grey, respectively. You cross true-breeding smooth white beetles to true-breeding bumpy grey beetles and get all smooth white F1 beetles. Then you cross the F1 beetles to true-breeding bumpy grey beetles and, after analyzing 800 F2s, calculate a map distance of 5 cM between the smoothness and color loci. a) What are the four phenotypic classes you got in the F2, and about how many of each did you get? For the body type e., smooth or bumpy) use B or b to designate the alleles. For the color (i.e., white or grey) use G or g to designate the alleles. For the size (i.e arge or small) use L or l to designate the alleles *In each case, use the uppercase letter for the associated with the dominant phenotype and the lower case letter for the allele associated with the recessive phenotype Genotypic class Phenotypic class How many of this phenotype Bbgg Smooth White 380 bbGg 380 Bumpy Grey bbGb Smooth Grey 20 bbgg Bumpy White 20

Explanation / Answer

When we cross true breeding smooth white beetles to true breeding bumpy grey beetles, F1 progeny will be smooth white beetle. The cross is shown below –

       BBGG                           X                                        bbgg ...............................................................P

(smooth, white)                                                          (bumpy, grey)

                                        BbGg ...........................................................................................................F1

                                  (smooth, white)

When this F1 progeny cross with bumpy grey beetle, the F2 progeny will be –

BbGg                               X                                             bbgg ...............................................................P

(smooth, white)                                                         (bumpy, grey)

The F2 progeny will be –

BG

Bg

bG

bg

bg

BbGg (smooth, white)

Bbgg (smooth, grey)

bbGg (bumpy, white)

bbgg (bumpy, grey)

Here,

B – Smooth is dominant over b – bumpy and G – white is dominant over g – grey.

Here, distance between smoothness and colour loci is 5 cM.

So, total number of recombinant is – 5*100/800 = 40   [Total number of progeny in F2 are 800]

As, there are two recombinant, so number of smooth, grey and bumpy white will be 40/2 = 20 each.

Thus, rest are parental progeny.

Now, the number of parental progeny will be – (800 – 40)/2 = 760/2 = 380 each [As there are two parental progeny, so divided by 2]

Now, number of progenies in detail are –

Genotype Class

Phenotype Class

How many of this phenotype

BbGg

(Parental Progeny)

Smooth, white

380

bbgg

(Parental Progeny)

Bumpy, grey

380

Bbgg

(Recombinant)

Smooth, grey

20

bbGg

(Recombinant)

Bumpy, white

20

BG

Bg

bG

bg

bg

BbGg (smooth, white)

Bbgg (smooth, grey)

bbGg (bumpy, white)

bbgg (bumpy, grey)

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