Scalloped (sd) is an x-linked recessive and ebony (e) is an autosomal recessive mutation. what proportion of scalloped, ebony females (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.

1/16

Explanation:

The two genes are independent because they are located on different chromosomes.

True breeding scalloped female wild type for ebony (Xsd Xsd e+e+) mates with a true breeding male mutant only for ebony (Xsd+ Y ee).

The female only produces Xsd e+ gametes. The male produces Xsd+ e or Y e gametes. Therefore, the F1 females will have the genotype Xsd Xsd+ e e+ and the F1 males will have the genotype Xsd Y e e+.

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 Xsd Xsd e e genotype and appear in a 1/16 proportion.

fertilization, pictured in figure 24.23a is the process in which gametes (an egg and sperm) fuse to form a zygote. the egg and sperm each contain one set of chromosomes. to ensure that the offspring has only one complete diploid set of chromosomes, only one sperm must fuse with one egg. in mammals, the egg is protected by a layer of extracellular matrix consisting mainly of glycoproteins called the zona pellucida. when a sperm binds to the zona pellucida, a series of biochemical events, called the acrosomal reactions, take place. in placental mammals, the acrosome contains digestive enzymes that initiate the degradation of the glycoprotein matrix protecting the egg and allowing the sperm plasma membrane to fuse with the egg plasma membrane, as illustrated in figure 24.23b. the fusion of these two membranes creates an opening through which the sperm nucleus is transferred into the ovum. the nuclear membranes of the egg and sperm break down and the two haploid genomes condense to form a diploid genome.

explanation: