Assume that two pigments, red and blue, mix to give the normal purple color of petunia petal. Separate biochemical pathways synthesize the two pigments, as shown in the top two rows of the accompanying diagram. "White" refers to compounds that are not pigments. (Total lack of pigment results in a white petal.) Red pigment forms from a yellow intermediate that is normally at a concentration too low to color petals.
Pathway I ... → white1 →E blue
Pathway II ... → white2→A yellow→B red
C↑
Pathway III ... →white3 →D white4
( the letters should be on top of the arrows, and the numbers are subscripts)
A third pathway, whose compounds do not contribute pigment to petals, normally does not affect the blue and red pathways, but, if one of its intermediates(white) should build up in concentration, it can be converted into the yellow intermediate of the red pathway. In the diagram, the letters A through E represent enzymes; their corresponding genes, all of which are unlinked, may be symbolized by the same letters. Assume that wild-type alleles are dominant and encode enzyme function and that recessive alleles result in a lack of enzyme function. Deduce which combinations of true-breeding parental genotypes could be crossed to produce F2 progeny in the following ratios:
a. 9 purple: 3 green: 4 blue b. 9 purple: 3 red: 3 blue: 1 white c. 13 purple: 3 blue d. 9 purple: 3 red: 3 green: 1 yellow
To deduce the combinations of true-breeding parental genotypes that could produce the specified ratios of F2 progeny, we need to consider the different pathways and their intermediates.
Let's go through each scenario:
a. 9 purple: 3 green: 4 blue
Purple is the result of combining red and blue pigments. From the given pathways, we know that the red pigment can be synthesized through the pathway II. The blue pigment, on the other hand, is a direct product without any intermediates.
To produce purple petals (9), we need at least one parent to have the red pigment pathway (AAbb genotype), and the other parent to have the blue pigment pathway (aabb genotype). This cross would result in F1 progeny with the genotype AaBb, which can produce both red and blue pigments. When these F1 progeny undergo random assortment in meiosis, they can produce the desired 9 purple (AaBb), 3 green (AAbb or aaBb), and 4 blue (aabb) in a 9:3:4 ratio.
b. 9 purple: 3 red: 3 blue: 1 white
To produce the specified ratio, we need to consider the white pathway and its interaction with the red pathway.
In this case, we should start with a parent that can produce both red and blue pigments to account for the 9 purple (AABB or AaBB). The other parent should lack the enzyme for the white pathway (cc).
Crossing these parents would result in F1 progeny with the genotype AaBbCc, which can produce purple (AABB, AaBB, AABb, AaBb), red (AAbb, Aabb), blue (aabb), and white (cc) pigments. When these F1 progeny undergo random assortment in meiosis, they can produce the desired ratio of 9 purple, 3 red, 3 blue, and 1 white.
c. 13 purple: 3 blue
To produce the specified ratio, we only need parents that can produce purple pigments.
The parental genotypes could be AABB or AaBB. Both of these genotypes can produce only purple pigments. When these parents are crossed, all the F1 progeny will have the genotype AaBb, which can only produce purple pigments. Thus, the F2 progeny will also be purple, resulting in the desired ratio of 13 purple to 3 blue.
d. 9 purple: 3 red: 3 green: 1 yellow
To produce the specified ratio, we need to consider the white pathway and its interaction with the red pathway, as well as the yellow intermediate.
We should start with a parent that can produce both red and green pigments to account for the 9 purple (AAbb or AaBb). The other parent should lack the enzyme for the yellow intermediate (BB).
Crossing these parents would result in F1 progeny with the genotype AaBbBB, which can produce purple (AAbb, AaBb), red (AAbb), green (AaBB), and yellow (AaBbBB) pigments. When these F1 progeny undergo random assortment in meiosis, they can produce the desired ratio of 9 purple, 3 red, 3 green, and 1 yellow.
In summary:
a. Cross AAbb x aabb
b. Cross AABB or AaBB x cc
c. Cross AABB or AaBB
d. Cross AAbb or AaBb x BB
These combinations of true-breeding parental genotypes would produce the specified ratios of F2 progeny.