Question

In: Biology

In a cross of two flies +/vg Cy/+ +/se +/ab X +/vg +/+ se/se +/ab what...

In a cross of two flies +/vg Cy/+ +/se +/ab X +/vg +/+ se/se +/ab what proportion of the offspring will be mutant in phenotype for all four markers?

ANSWER: 1/64

Please explain the process

Expert Solution

+/vg Cy/+ +/se +/ab

+/vg +/+ se/se +/ab

vg, Cy, se and ab are the mutant alleles for the four genes. Upper case of the first letter in the gene indicates a dominant phenotype. Lower case of the first letter in the gene indicates a recessive phenotype.

Consider the Punnett squares for crosses for each gene separately:

a. +/vg X +/vg

Vestigial gene is inherited in recessive fashion. Hence, two copies of vg should be present to see mutant phenotype.

		+/vg
		+	vg
+/vg	+	+/+ Homozygous dominant Wild type phenotype	+/vg Heterozygous Wild type phenotype
	vg	+/vg Heterozygous Wild type phenotype	vg/vg Homozygous recessive Mutant phenotype

Vg/Vg is the mutant phenotype.

Probability of Vg/Vg is 1/4.

b. Cy/+ X +/+

Cy gene (curly wings) is dominant inherited gene and a mutant phenotype will be seen with only one mutant allele (Cy/+ or Cy/Cy).

		Cy/+
		Cy	+
+/+	+	Cy/+ Heterozygous Curly wing mutant phenotype	+/+ recessive Homozygous Wild type phenotype (no curly wings)
	+	Cy/+ Heterozygous Curly wing mutant phenotype	Homozygous recessive Wild type phenotype (no curly wings)

Probability of mutant phenotype (Cy/+) = 2/4= 1/2

c. +/se X se/se

Se is sepia brown phenotype gene and is inherited in recessive fashion.

		+/se
		+	se
Se/se	se	+/se Wild type phenotype Heterozygous	se/se sepia brown Homozygous recessive
	se	+/se Wild type phenotype Heterozygous	se/se sepia brown Homozygous recessive

Probability of mutant phenotype= se/se= 2/4=1/2.

d. +/ab X +/ab

The ab (abrupt) gene is recessive. Hence, two copies of mutant ab alleles are required to inherit the phenotype.

		+/ab
		+	ab
+/ab	+	+/+ Homozygous dominant Wild type phenotype	+/ab Heterozygous Wild type phenotype
	ab	+/ab Heterozygous Wild type phenotype	ab/ab Homozygous recessive Mutant phenotype

Probability of mutant phenotype= ab/ab= 1/4

Hence,

Probability of obtaining progeny flies that are mutants for all genes in the +/vg Cy/+ +/se +/ab X +/vg +/+ se/se +/ab cross= Multiplication of individual probabilities for each gene

Probability of obtaining progeny flies that are mutants for all genes in the +/vg Cy/+ +/se +/ab X +/vg +/+ se/se +/ab cross= 1/4 X 1/2 X 1/2 X 1/4= 1/64

Thus, probability of offspring’s that are mutant for all genes= 1/64.

gladiator answered 1 year ago

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