Theodosius dobzhansky discovered that successful species tend to have a wide variety of genes that do not appear to be useful to
1 answer:
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Answer:
1/16
Explanation:
- scalloped (Xsd) is an X-linked recessive trait to Xsd+ (wild type)
- ebony (e) is an autosomal mutation recessive to e+ (wild type).
The two genes are independent because they are located on different chromosomes.
<h3><u>Parental generation:</u></h3>True breeding scalloped female wild type for ebony (<em>Xsd Xsd e+e+</em>) mates with a true breeding male mutant only for ebony (<em>Xsd+ Y ee</em>).
The female only produces <em>Xsd e+</em> gametes. The male produces <em>Xsd+ e</em> or <em>Y e</em> gametes.<u> Therefore, the F1 females will have the genotype </u><u><em>Xsd Xsd+ e e+ </em></u><u>and the F1 males will have the genotype </u><u><em>Xsd Y e e+.</em></u>
If you complete a Punnett Square with the gametes the two F1 individuals can produce, you will get all the F2 proportions. The scalloped, ebony females have a<em> Xsd Xsd e e</em> genotype and appear in a 1/16 proportion.
