scalloped (sd) is an X-linked recessive and ebony (e) is an autosomal recessive mutation. What proportion of scalloped, ebony fe
males (relative to whole population) is expected in the F2 starting with a true breeding scalloped female which is wild type for ebony mating with a true breeding male mutant only for ebony.
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.
People<span> who </span>tend<span> to </span>gain weight<span> in the </span>hip area have<span> a </span>higher incidence<span> of </span>coronary heart disease<span> wham do </span>people<span> who </span>gain weight<span> in the</span>abdominal area<span>. False.</span>
The right answer is a person has more than two alleles for a certain trait.
Duplication is a genetic mutation characterized by the doubling of genetic material on a chromosome. There are several mechanisms that result in the duplication of either a large chromosomal portion, a gene, or a nucleotide sequence.