4. A variety of squash is available as true breeding strains with white. yellow
ID: 145320 • Letter: 4
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4. A variety of squash is available as true breeding strains with white. yellow or green gourds. Use this information to answer the following questions. A-C are independent of each othor. A. If the phenotype of gourd color is due to allelic interactions, provide an explanation for the inheritance of gourd color in the answer give the genotypes of the parents, Fl and predict F2 offs following cross of true breeding parents (write this underneath the table). In your pring (fill in table). (4) Parents Yellow X Green Phenotypes | 100% yellow and green Genotypes B. If the phenotype of gourd color is due to allelic interactions, provide an explanation for the inheritance of gourd color in the following cross of true breeding parents (write this underneath the table). In your answer give the genotypes of the parents, Fl and predict what you would get in F2 offspring (fill n table) F1 Offspring | 100% white Parents Phenotypes Yellow X white Genotypes C. If the phenotype of gourd color is due to allelic interactions, provide an explanation for the inheritance of gourd color in the following cross of true breeding parents (write this underneath the table). In your answer give the genotypes of the parents, FI and predict what you would get in F2 offspring (fill in table). F2 Offsprin Parents Phenotypes green X white 100% pale green GenotypesExplanation / Answer
A. Let us assume alleles for yellow phenotype are YY or Yy and for green it is yy. Interactions between two allele produce 100% offsprings of yellow and green gourds in F1.
So, if the yellow phenotype is of YY genotype then the cross would be between YY and yy. If so then we would only get yellow phenotype of Yy or heterozygous genotype but no green or yy genotype. But this is not the case here. We have gotten 100% parental phenotype. To produce green phenotype in F1 both the parents should carry atleast one y allele in their genotype. Then the cross must be occurred between Yy and yy. By this we would get exactly same phenotypes in F1. If F1 offsprings are selfed Yy * yy then we will again get 100% yelloe and green phenotypes or Yy and yy genotypes.
Final outcome: Parents genotype are Yy (yellow) and yy (green). F1 and F2 offsprings are also yellow and green i.e, Yy and yy respectively. This follows simple mendelian inheritance pattern and Y is dominant to y.
B. The cross is between yellow and white parents. Lets assume the genotype for white phenotype is WW. The genotypes of yellow phenotype are Yy or YY. If the cross was between Yy and WW. Then one of the offsprings is of yW which will show pale green phenotype (explained in part C). As here only 100% white F1 offsprings are produced so the cross must be happened between YY and WW. So all the progeny is of YW genotype which produces white phenotypes. So genotype for white is dominant to the genotype of yellow phenotype. In the F2 all YW are selfed.
YW * YW
YY YW WY WW
The phenotypes will be yellow and white
Final outcome: parents genotype and phenotype are yellow YY and white WW
F1: all the offsprings are YW or white
F2: YY (yellow), YW, WY, WW (all white)
So W is dominant to Y.
C. Here the cross is between green amd white parents. The F1 offsprings are all pale green. But neither of the parents are of pale green. So the two colors are mixed in F1 progeny. This is the perfect example of incomplete dominance. If the genotype of green phenotype is yy and genotype of white phenotype is WW then the cross will be between yy * WW. All the F1 progenies would be yW and the phenotype of the heterozygous genotype is pale green. The F2 progeny will have yW * yW = yy, yW, Wy, WW.
Final outcome: parents are green (yy) and white (WW) ; F1 progeny is pale green or yW. F2 progeny is yy (green), yW and Wy (pale green) and WW (white). This shows incomplete dominance pattern.
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