Viewing 1 reply thread
  • Author
    Posts
    • #17546
      Yokie83
      Participant

      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…

    • #114451
      Cat
      Participant
      quote Yokie83:

      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"…

Viewing 1 reply thread
  • You must be logged in to reply to this topic.