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## I need help with Punnett Square

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### I need help with Punnett Square

Cross between a female fly with brown (BW) eye color (keep all other traits as wild-type) and a male fly with ebony (E) body color (keep all other traits as wild-type)—show the F1 offspring.

Result I've got is here

Results of Cross #1
Parents
(Female: BW) x (Male: E)
Offspring
Phenotype Number Proportion Ratio
Female: + 4982 0.4980 1.000
Male: + 5023 0.5020 1.008
Total 10005

then I Designed a male fly with brown eye (BW) color AND ebony body (E) color (homozygous recessive for both traits), then test cross this fly with an F1 wild-type female fly from the above cross.

Results of Cross #2 Parents (Female: +) x (Male: BW;E)
Offspring
Phenotype Number Proportion Ratio
Female: + 1265 0.1252 1.027
Male: + 1294 0.1281 1.050
Female: BW 1279 0.1266 1.038
Male: BW 1254 0.1241 1.018
Female: E 1275 0.1262 1.035
Male: E 1263 0.1250 1.025
Female: BW;E 1232 0.1219 1.000
Male: BW;E 1241 0.1228 1.007
Total 10103

I did until here but, next questions are tough ....

What was the phenotypic ratio for the offspring resulting from this testcross? Based on this phenotypic ratio, determine whether the F1 wild-type female was double homozygous or double heterozygous for the eye color and body color alleles. Explain your answer, including the Punnett square for these results.

Show a Punnett Square that BEST represents the observed F2 ratios for the F2 offspring of the above cross (male fly with brown eye color and ebony body color with an F1 wild-type female fly from the original cross).

Show a Punnett Square for the F2 offspring of the alternate possibility.

My answers is that the F1 female fly was heterozygous for both the eye and body color alleles. The phenotypic ratio for the testcross was 1:1:1:1 The fly been double homozygous the wild type most likely would have had a ratio of around 3:1 for the offspring.

I know how to draw a punnett Sqare if it's simple but this is too complicated...
Yokie83
Garter

Posts: 1
Joined: Sun Sep 29, 2013 9:00 pm

### Re: I need help with Punnett Square

Yokie83 wrote: The fly been double homozygous the wild type most likely would have had a ratio of around 3:1 for the offspring.

No. All of the offspring would have been wild type. The rest of your answer is correct.

F0: cross female (Female: BW) x (Male: E) = female bbEE x male BBee
F1: All offspring: wild type = BbEe
Next: female wild type F1: BbEe x bbee (BW,E phenotype) male
You get: BbEe, bbEe, Bbee, bbee in 1:1:1:1 ratio you have observed.

'Double homozygous wild type' means there are no 'b' or 'e' alleles (BBEE). If you cross that with bbee (BW,E phenotype) male, you get all BbEe (wild type)... However, I am not sure what they mean when they ask for "F2 offspring of the alternate possibility"...
Cat
King Cobra

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Joined: Thu Feb 14, 2008 7:40 pm

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