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

Solutions

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

Next > < Previous

Related Solutions

Suppose that we cross two flies: One is a female that is b+b vg+vg; the other...

Suppose that we cross two flies: One is a female that is b+b vg+vg; the other is a male that is homozygous recessive. The female has grey (wild-type) body and normal (wild-type) wing length. The male is black-bodied with vestigial wings. The female's parents (I'm going to refer to them as the grandparents) were a true-breeding grey, normal fly and a true-breeding black, vestigial fly. A) What is the frequency of phenotypes you would expect to see in their offspring?...

In lab you are going to be using drosophila flies with sepia mutations (SE, se) causing...

In lab you are going to be using drosophila flies with sepia mutations (SE, se) causing brown eyes found on chromosome 3 and apterous mutation (AP, ap) on chromosome 2 causing loss of wings. These are recessive mutations. Red eye is the wild-type phenotype. Assuming you have a female heterozygous at both loci. Draw the chromosomes as they go through meiosis (Make sure you show somatic cell, prophase I, metaphase I, metaphase II, gametes.) Draw all possible gametes. What percentage...

View the example chi square calculated for a monohybrid cross between two heterozygous flies Phenotype Observed...

View the example chi square calculated for a monohybrid cross between two heterozygous flies Phenotype Observed Expected Deviation (observed - expected)^2/Expected Brown Body Color 580 600 Black body Color 220 200 Total Flies = 800 Chi Square =? What is the chi square value and the Deviation? Answer this based on the values: The Chi square value is ____ (more or less) than the critical value _______. The variation between the observed and expected is ______(due or not due) to...

Suppose that you are working with flies that are segregating three traits. You cross a homozygous...

Suppose that you are working with flies that are segregating three traits. You cross a homozygous truncated, vestigial, shaven fly with a homozygous wild-type, apterous, spineless fly and the F1 flies are all wild-type, apterous, spineless. You next backcross the F1 flies to a truncated, shaven, vestigial parent. The progeny from this type of cross are shown below: truncated apterous shaven 94 truncated vestigial shaven 425 truncated apterous spineless 46 truncated vestigial spineless 19 wild-type apterous shaven 16 wild-type vestigial...

A parental cross between homozygous cinnabar flies (cnbw+/cnbw+) and homozygous brown flies (cn+bw/cn+bw) is done. Use...

A parental cross between homozygous cinnabar flies (cnbw+/cnbw+) and homozygous brown flies (cn+bw/cn+bw) is done. Use a Punnett square to show how the F1 cross will not produce homozygous double mutant flies in the F2 generation.

1)Two good (x and y) are complements. The cross-price elasticity of x with respect to the...

1)Two good (x and y) are complements. The cross-price elasticity of x with respect to the price of y is a. positive. b. negative. c. zero. d. None of the above. 2) For an inferior good, the income and substitution effects a. Work together. b. Work against each other. c. Can work together or in opposition to each other depending upon their relative magnitudes. d. Always exactly cancel each other. 3) Elasticity measures a. The slope of a demand curve....

x^2=ln(cy^2) Find the perpendicular orbits of the family of curves

Imagine that you as a fruit fly researcher cross true breeding, wild-type flies to true breeding,...

Imagine that you as a fruit fly researcher cross true breeding, wild-type flies to true breeding, eyeless flies in the P generation. The resulting F1 generation is made up of 100% wild-type flies with normal eyes. Based on Mendelian genetics, what phenotype ratio would you expect to see for a F1 monohybrid cross? How would you go about determining the genotype ratio in the F2 generation? In cactus, the relationship between Gene S and Gene N is known to be...

1. What is Self-Employment(SE) Tax? 2. What schedule is SE tax reported on? 3. What portion...

1. What is Self-Employment(SE) Tax? 2. What schedule is SE tax reported on? 3. What portion of SE tax is deductible on page 1 of Form 1040 (for AGI deduction)? What line number? 4. IRS is enforcing penalties for failure to issue 1099 to payee. Why is 1099 important to IRS? and the payee?

2. Flies inoving two populations are isolated from other populations of flies. Population S lives on...

2. Flies inoving two populations are isolated from other populations of flies. Population S lives on a very small island and has 103 individual population size. Population L lives on a continent and has 108 individual population size. Assuming that the island population was found by a small group of organisms that drifted on a log from the continent to the island. No interchange of alleles were found between populations since that time. a. Would heterozygosity be expected for most...

Subjects