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Homework2Solution - broad leaves 6-7 is just about 1/16 of...

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Biol 214/414 Homework #2 Due (in class) 9-4-09 Solution Suppose there are two kinds of geraniums, the first with blue flowers and broad leaves and the second with red flowers and narrow leaves. Furthermore, suppose it has been determined that each of the two traits in question (flower color and leaf size) is controlled by a single gene, each of these two genes has two alleles, and each of the two types of geranium has been bred so that it is homozygous for both these genes. The blue, broad geraniums are crossed with the red, narrow geraniums, and the offspring (F1) from that cross are then crossed with each other. About 6-7% of the offspring from the second cross (F2) are found to have red flowers and broad leaves. Using symbols of your choice that indicate dominance and recessiveness, give the genotypes of the blue, broad and red, narrow plants used as parents in the first cross. The “tipoff” of what is happening here is the 6-7 % of the F2 that have red flowers and
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Unformatted text preview: broad leaves. 6-7% is just about 1/16 of the total F2 offspring, which would indicate that these are the “1” in the classic 9:3:3:1 ratio we expect from a dihybrid cross in which each of the two genes is responsible for one of the traits, each gene has two alleles, simple dominance and recessiveness holds, and we have both parents (the F1 in this case) being heterozygous for both genes (like the pea color/shape cross we went over in class). This then suggests that the red/broad offspring are double recessive for each gene (let’s denote that as rrbb). Then we can conclude that the R allele gives blue color and is dominant over the r allele, while the B allele gives narrow leaves and is dominant over the b allele. Since we know that both original parents were homozygous, we can conclude that the blue/broad parent was RRbb and the red/narrow parent was rrBB. Thus, all the F1 would have been RrBb, and this explains the result noted for the F2 generation....
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