Use this information for the first 5 questions In the fruit fly Drosophila melanogaster, tan body color (B) is dominant to black (b), and dark red eyes (E) is dominant to bright orange (e). A black fly, heterozygous for eye color (Fly #1) is crossed to an orange eyed fly, heterozygous for body color (Fly #2). Complete a Punnett Square for the dihybrid cross of Fly #1 and Fly #2. What proportion of offspring are... Tan & dark red: Tan & orange: Black & dark red: Black & orange:

Use this information for the first 5 questions In the fruit fly Drosophila melanogaster, tan body color (B) is dominant to black (b), and dark red eyes (E) is dominant to bright orange (e). A black fly, heterozygous for eye color (Fly #1) is crossed to an orange eyed fly, heterozygous for body color (Fly #2). Complete a Punnett Square for the dihybrid cross of Fly #1 and Fly #2. What proportion of offspring are... Tan & dark red: Tan & orange: Black & dark red: Black & orange:

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter11: The Basic Principles Of Heredity

Section: Chapter Questions

Problem 2TYU: The F1 flies described in question 1 were mated with brown-eyed flies from a true-breeding line....

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In Drosophila melanogaster body color is controlled by one gene while wing shape is controlled by a second gene. Gray body color is dominant to black body color, and normal wings are dominant to vestigial wings. Flies homozygous for gray body color and vestiial wing are crossed with flies homozygous for black body color and normal wings Compare the possible F2generation genotypes and phenotypes and proportions if these two traits are autosomally linked in comparison to the non-linked How does your answer change if one of the original parents is homozygous for gray body color and normal wings while the other has black body color and vestigial wings
Table (see attached image) Diagram the cross showing the segregation of genes responsible for eye-color in Drosophila. Indicate all applicable ene relationships. Indicate the genotypes, phenotypes and gametes of individuals involved in all crosses. Indicate the genotypic and phenotypic ratios of the F2 generation.
In Drosophila melanogaster, cinnabar eye (cn) and vestigial wing (vg) are simple recessive traits. A pure breeding cinnabar fly is crossed with a pure breeding vestigial wing fly and have offspring that are all wild type. If the genes are closely linked and no crossing over is found to occur, then what would the F2 offspring phenotypes be when testcrossed? O 1/2 wild type and 1/2 double mutant 1/2 cinnabar and 1/2 vesitigial wing O all wild type O 1/4 wild type, 1/4 cinnabar, 1/4 vestigial and 1/4 double mutant
In Drosophila melanogaster, ebony body color is determined by the e allele. The e+ allele produces the wild-typehoney-colored body. In heterozygotes, the body is honey-colored except for a dark marking called the trident seen on the thorax. The e+ allele is thus considered to be incompletely dominant to the e allele.a. When female e+ e+ flies are crossed to male e+ e flies, what is the probability that progeny will have the tridentmarking? Animals with the marking mate among themselves. Of 100 progeny, how many would be expected to have a trident, how many would have ebony bodies, and how many would have honey-colored bodies?
Phenotypically wild-type F1 female Drosophila, whose mothers had light eyes (lt) and fathers had straw (stw) bristles, produced the following offspring when crossed to homozygous light eyed, straw bristled males: Phenotype Number light-straw 140 wild-type 160 light 360 straw 340 Total 1000 Compute the map distance between the light and straw loci. Group of answer choices 70 map units 3 map units 7 map units 0.03 map units 30 map units
Female Drosophila with cinnabar eye (cn) and vestigial wings (vg) were mated to males with roof wings (rf). The F1 were all wild-type. When the F1 females were test crossed with males homozygous for all three traits the following result were obtained. For this problem, rf cn vg is for roof wing / cinnabar eye / vestigial wing rf + cn + vg + all wild (no definite phenotype described) Give the genotype of the offspring (this time simply follow the given sequence above and separate the genes with single space) F2: Phenotype Frequency Genotype cinnabar, vestigial 382 roof 401 cinnabar 3 roof, vestigial 4 roof, cinnabar, vestigial 59 Wild 67 roof, cinnabar 44 vestigial 40 2. Give the phenotype and genotype of the female parent: phenotype: ____ genotype: _______ 3. Give the phenotype and genotype of the male parent: phenotype: _______ genotype: _______ 4. What…
In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn+ and ct+ is crossed to a sn ct male. The F1 flies are interbred. The F2 males are distributed as follows sn ct 36 sn ct+ 13 sn+ ct 12 sn+ ct+ 39 What is the map distance between sn and ct?
In the fruit fly Drosophila melanogaster, the following genes and mutations are known:Wing size: recessive allele for tiny wings t; dominantallele for normal wings T.Eye shape: recessive allele for narrow eyes n;dominant allele for normal (oval) eyes N.For each of the four following crosses, give thegenotypes of each of the parents.Male FemaleWings Eyes Wings Eyes Offspring1 tiny oval × tiny oval 78 tiny wings, oval eyes24 tiny wings, narrow eyes2 normal narrow × tiny oval 45 normal wings, oval eyes40 normal wings, narrow eyes38 tiny wings, oval eyes44 tiny wings, narrow eyes3 normal narrow × normal oval 35 normal wings, oval eyes29 normal wings, narrow eyes10 tiny wings, oval eyes11 tiny wings, narrow eyes4 normal narrow × normal oval 62 normal wings, oval eyes19 tiny wings, oval eyes
PURPLE VESTIGIAL DIHYBRID TESTCROSS In the parental generation, you mate a pure-breeding wild-type female (put/put.vg+/vg+) with a pure-breeding purple, vestigial (pu/pu;vg/vg) to produce an F1 generation that is all wild-type (put/pu;vg+/vg). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (put/pu;vg+/vg) with tester males, which are purple, vestigial (pu/pu;vg/vg). The offspring of this dihybrid testcross are: Phenotype Wild-type Purple, vestigial Vestigial Purple Genotype Tester Gamete Dihybrid Gamete Number 437 417 77 59 Copy the table into your notes and derive the dihybrid gametes following the example in the first section. The columns in blue (phenotypes and numbers of offspring) are what you can see and count. The genotypes of the testcross offspring (orange) must be deduced from the phenotypes and knowing that the tester contributed pu vg gametes. Finally, you can deduce the dihybrid gametes (green) by subtracting the tester gamete…
Female Drosophila with cinnabar eye (cn) and vestigal wings (vg) were mated to males with roof wings (fr). The F1 were all wild type. When the F1 females were tested and crossed with male homozygous for all thrre traits, the following result were obtained. For this problem, rf, cn, vg is for roof wing / cinnabar eye / vestugal wing rf + cn + vg all wild type (no definite phenotype described). What is the value of coincidence and interference?
In Drosophila, the sepia mutation (se, chromosome 3, position 26) results in dark brown eyes, while cinnabar (cn, chromosome 2, position 57.5) results in bright orange-red eyes. True breeding, wild type females are mated with true breeding males homozygous recessive for both traits. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1
Female Drosophila with cinnabar eye (cn) and vestigial wings (vg) were mates to males with roof wings(fr). The F1 were all wild type ñ. When the F1 females were tested and crossed with males homozygous for all three traits the following result were obtained Give the phenotype of the offspring ( this time simply follow the given sequence above and separate the genes with single space):