In drosophila, the genes crossveinless and stubble are linked, about 7 map units apart on chromosome 3. cv is a recessive mutant allele of crossveinless (cv+ is wild type), while sb is a dominant mutant allele of stubble (sb+ is wild type). a dihybrid female drosophila with genotype cv sb+/cv+ sb is testcrossed. the proportion of phenotypically wild-type individuals in the progeny of the testcross will be

0.035

Explanation:

cv+ is the wild-type dominant allele over cv, therefore:

Sb is a dominant mutant allele over wild-type Sb+, therefore:

It's the cross between the heterozygous female with a homozygous recessive male. Remember that cv and Sb+ are the recessive alleles.

 X

-The male produces only 1 type of gamete: cv Sb+

-The female produces 4 types of gametes:

The genes are linked and separated by 7 map units. A distance of 7 mu means that 7% of the resulting gametes will be recombinant. Because there are 2 possible recombinant gametes, each of them will appear in 3.5% of the cases.

The genotypes and proportions of the offspring resulting from the test cross can be seen in the Punnett Square. The phenotypically wild-type individuals will have the genotype cv+ Sb+ / cv Sb+ (heterozygous for crossveinless and homozygous recessive for Stubble) and a 0.035 proportion.

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what do the different brain waves do?

explanation:

brainwaves are produced by synchronised electrical pulses from masses of neurons communicating with each other. brainwaves are detected using sensors placed on the scalp. brainwave speed is measured in hertz (cycles per second) and they are divided into bands delineating slow, moderate, and fast waves.

the answer is 100% density i've answered this question before and done the quiz myself.