Question

How does the map unit correlate with recombination frequency?

	a.	There is no correlation between recombination frequency and map units.
	b.	The number of map units is inversely proportional to the recombination frequency between genes.
	c.	1 map unit is equal to 10% recombination.
	d.	1 map unit is equal to 1% recombination.

A white-eyed Drosophila female would have the genotype:

	a.	XWXW
	b.	XWXw
	c.	XwXw
	d.	XWY
	e.	XwY

Cinnabar eyes is a sex-linked recessive trait in Drosophila. If a cinnabar female is mated with a normal male, what percentage of the F1 males will have cinnabar eyes?

	a.	0%
	b.	50%
	c.	100%
	d.	75%
	e.	25%

Gene S controls the sharpness of spines in a species of cactus. SS and Ss individuals have sharp spines and ss individual have dull spines. A second gene determines whether the cactus has spines at all. NN and Nn individuals have spines and nn individuals have no spines. The relationship between the alleles at these two loci is best described as ________________.

	a.	pleiotropy
	b.	codominance
	c.	incomplete dominance
	d.	complete dominance
	e.	gene-by-gene interaction

How does crossing over break up linkage between alleles?

	a.	It allows male and female gametes to combine randomly at fertilization.
	b.	It involves a physical exchange of segments between homologous chromosomes.
	c.	It involves a physical exchange of segments between sister chromatids.
	d.	It causes alleles on nonhomologous chromosomes to be inherited independently.

Answer 1

Q1) A map unit is related to the recombination frequency. if the genes have the distance of 1 map unit, the recombination frequency is 1%.

Therefore option D is the correct.

Q2) Since the eye colour in Drosophila is a X-linked trait and the Drosphila female has two X chromosomes. Moreover, white eye colour is recessive to Red eye colour. therefore the genotype of White eye colour Drosophila Female is X^wX^w

Therefore Option C is the correct answer.

Q3) Let the genotype of Cinnabar female = X^cX^c

and the genotype of Normal male = XY

Now X^cX^c x XY

Gametes ----->    X^c X Y

F1 -------------> XX^c X^cY

(Normal female) (Cinnabar male)

Therefore the percentage of F1 Cinnabar males = 1/2 x100 = 50%.
Therefore the option B is correct.

Q4) Since SS and Ss has the same phenotype, it means that S is completely dominant over s.

Also NN and Nn individuals have the same phenotype, it means that N is completely dominant over n.

Therefore the option d is correct.

Q5) Crossing over involves the exchange of chromosome segments between the non-sister chromatids of homologous chromosomes. Therefore the option B is correct.

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