Fruit flies (Drosophila) were mated. Cross #1 - 2 male red eyes (se+), wild wings(ap+) X 4 female red eyes (se+), wild wings(ap+) Cross 2 - 2 male sepia eyes(se-), apterous wings (ap-) X 4 female red eyes(se+), wild wings (ap+) Write out a dihybrid cross and branch diagram for the F2 generatin

Fruit flies (Drosophila) were mated. Cross #1 - 2 male red eyes (se+), wild wings(ap+) X 4 female red eyes (se+), wild wings(ap+) Cross 2 - 2 male sepia eyes(se-), apterous wings (ap-) X 4 female red eyes(se+), wild wings (ap+) Write out a dihybrid cross and branch diagram for the F2 generatin

Name: Fruit flies (Drosophila) were mated.Cross #1 - 2 male red eyes (se+),
Uploaded: 2024-03-20T06:43:04.000Z
Channel: Bartleby
Description: Fruit flies (Drosophila) were mated.Cross #1 - 2 male red eyes (se+), wild wings(ap+) X 4 female red eyes (se+), wild wings(ap+) Cross 2 - 2 male sepia eyes(se-), apterous wings (ap-) X 4 female red eyes(se+), wild wings (ap+) Write out a dihybrid c

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter2: Cells And Cell Division

Section: Chapter Questions

Problem 20QP: In the following diagram, designate each daughter cell as diploid (2n) or haploid (n).

See similar textbooks

Similar questions

BLACK VESTIGIAL DIHYBRID TESTCROSS In the parental generation, you mate a pure-breeding wild-type female (bl+/blt.vg+/vg+) with a pure-breeding black, vestigial (bl/bl vg/vg) to produce an F1 generation that is all wild-type (bl+/bl vg+/vg). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (bl+/bl vg+/vg) with tester males, which are black, vestigial (bl/bl;vg/vg). The offspring of this dihybrid testcross are: Phenotype Genotype Tester Gamete Dihybrid Gamete Number 440 Wild-type 394 108 135 Black, vestigial Vestigial Black Copy the table into your notes and derive the dihybrid gametes following the example in the previous 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 bl vg gametes. Finally, you can deduce the dihybrid gametes (green) by subtracting the tester gamete…
PURPLE VESTIGIAL DIHYBRID CROSS In the parental generation, you mate a pure-breeding wild-type female (put/pu+;vg+/vg+) with a pure-breeding purple, vestigial (pu/pu;vg/vg) to produce an F1 generation that is all wild-type (pu*/pu;vg+/vg). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (pu*/pu;vg+/vg) with tester males, which are purple, vestigial (pu/pu;vg/vg). The offspring of this dihybrid testcross are: Phenotype Genotype Tester Gamete Dihybrid Gamete Number Wild-type 437 417 77 59 Purple, vestigial Vestigial Purple 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 contribution…
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…
Each sperm of a tomato plant contains seven chromosomes. What are the haploid and diploid number for tomatoes? haploid=7 and diploid=14 haploid=14 and diploid=7 haploid=7 and diploid=49 haploid=3 and diploid=7
Two plants in a cross were each heterozygous for two gene pairs (AB/ab) whose loci are linked and 10 map units (mu) apart. (Recall that 1 mu is equal to 1% recombination between two genes.) Assuming that crossing over occurs during the formation of both male and female gametes and that the A and B alleles are dominant, determine the phenotypic ratio of their offspring. Part D If the two genes are 15 mu apart and the plant is (Ab/aB), what proportion of gametes from a signal plant will be ab? Part E What proportion of the offspring of two plants ( both (Ab/aB)) will be A_B_ if the genes are 15 mu apart? Part F What proportion of the offspring of two plants ( both (Ab/aB)) will be A_bb if the genes are 15 mu apart? Part G What proportion of the offspring of two plants ( both (Ab/aB)) will be aaB_ if the genes are 15 mu apart? Part H What proportion of the offspring of two plants ( both (Ab/aB)) will be aabb if the genes are 15 mu apart? How would I solve these?
Diagram the cross: a male fly expressing white (w) on the x chromosome, vestigial wings (vg) on chromosome 2 and sepia eyes (se) on chromosome 3 with a female that is heterozygous for all three traits.
Draw a Punnett square for the dihybrid cross. There are two known alleles of gene occupying a specific locus in the X chromosome. The gene in question codes for a transcription factor involved in digit development. The mutant allele is dominant and gives rise to an additional but non-functioning little finger (polydactyly) on both hands. A couple have had their DNA sequenced at the region of interest, the male exhibits polydactyly because of the mutation, the female is homozygous wild type at the same locus and therefore has the wild type phenotype. Both have green eyes. In this story; eye colour shows a monogenic autosomal inheritance pattern and the allele for brown eyes shows incomplete dominance with that for blue eyes, the heterozygote phenotype is green eyes. The genes for eye colour and polydactyly show no linkage.
In the following diagram, designate each daughter cell as diploid (2n) or haploid (n).
In a cross between two varieties of corn, gl1gl1Ws3Ws3 (egg parent) * Gl1Gl1ws3ws3 (pollen parent), a triploid offspring was produced with the genetic constitution Gl1Gl1gl1Ws3ws3ws3. From which parent, egg or pollen, did the 2n gamete originate? Is another explanation possible? Explain.
In DrosophilaI, yellow body (y), crossveinless (cv), and forked bristles (f) are all found in the X chromosome, in that order. Yellow is 14 map units from crossveinless, and crossveinlessis 32 map units from forked. A yellow, forked female is crossed to a crossveinless male. An F1 female is crossed to a yelow, crossveinless, forked male, yielding the F2 generation. In the F2, what is the probability of obtaining progeny that express crossveinless and forked, but not yellow? Assume a coefficient of coincidence of 1. In the F2, what is the probability of obtaining progeny that are completely wild type? In the F2, what is the probability of obtaining progeny that express crossveinless, but not yellow or forked?
In fruit flies, normal wings (A) are dominant to dumpy wings (a) and the presence of eyes (B) is dominant to the absence of eyes (b). Create a dihybrid cross. what is the phenotype of each fly
Two plants in a cross were each heterozygous for two gene pairs (AB /ab) whose loci are linked and 30 map units (mu) apart. (Recall that 1 mu is equal to 1% recombination between two genes.) Assuming that crossing over occurs during the formation of both male and female gametes and that the A and B alleles are dominant, determine the phenotypic ratio of their offspring. Part E: What proportion of the offspring of two plants (both (AB/ab ) will be A - B- if the genes are 30 mu apart? Part F: What proportion of the offspring of two plants (both (AB/ab)) will be A - bb if the genes are 30 mu apart? Part G: What proportion of the offspring of two plants (both (AB/ab)) will be aaB- If the genes are 30 mu apart? Part H: What proportion of the offspring of two plants (both (AB/ab)) will be aabb if the genes are 30 mu apart?