Seed Type Initial Number of Seeds Number of Seedlings Seedlings Without Purple Pigment Seedlings With Purple Pigment Seedlings With Green Leaves Seedlings With Yellow-Green Leaves Recessive Trait Dominant Trait Dominant Trait Recessive Trait P1 10 10 10 0 10 0 P2 10 9 0 9 5 4 F1 10 10 3 7 3 0 F2 30 30 15 15 15 15 . What are the possible genotypes for the parents based on just the P1 and P2 results? For clarity, use A/a for ANL/anl and Y/y for YRG/yrg

Seed Type Initial Number of Seeds Number of Seedlings Seedlings Without Purple Pigment Seedlings With Purple Pigment Seedlings With Green Leaves Seedlings With Yellow-Green Leaves Recessive Trait Dominant Trait Dominant Trait Recessive Trait P1 10 10 10 0 10 0 P2 10 9 0 9 5 4 F1 10 10 3 7 3 0 F2 30 30 15 15 15 15 . What are the possible genotypes for the parents based on just the P1 and P2 results? For clarity, use A/a for ANL/anl and Y/y for YRG/yrg

Name: SeedTypeInitial Number of SeedsNumber of SeedlingsSeedlings Without Pu
Uploaded: 2024-03-20T06:43:04.000Z
Channel: Bartleby
Description: SeedTypeInitial Number of SeedsNumber of SeedlingsSeedlings Without Purple Pigment Seedlings With Purple Pigment Seedlings With Green Leaves Seedlings WithYellow-Green Leaves RecessiveTraitDominantTraitDominantTraitRecessiveTraitP11010100100P21090

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter3: Transmission Of Genes From Generation To Generation

Section: Chapter Questions

Problem 16QP: More Crosses with Pea Plants: The Principle of Independent Assortment Consider the following cross...

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Mendelian GeneticsF1 Cross: Yellow, Round x Green, RoundGgWw x ggWWCharacter: Pea color & shapeUse Punnett square and fork-line method to check the F2.Show and interpret all the possible genotypes and phenotypes of the offsprings.Mendelian GeneticsF1 Cross: Tall, White, Axial x Dwarf, Violet, TerminalDdwwAA x ddWWaaCharacter: Stem height, Flower color & positionUse fork-line method to check the F2.Show and interpret all the possible genotypes and phenotypes of the offsprings.B.A.I. For A and B,1. Identify the type ofinheritance. Justifyyour answer.2. Decode the genotypesof the individuals inthe pedigree. (Useletter A forrepresentation ofalleles.)3. List down all affected
A pure-breeding strain of squash that produced diskshaped fruits (see the accompanying illustration) wascrossed with a pure-breeding strain having long fruits.The F1 had disk fruits, but the F2 showed a new phenotype, sphere, and was composed of the followingproportions:Longlong 32 sphere 178 disk 270Sphere DiskPropose an explanation for these results, and showthe genotypes of the P, F1, and F2 generations.
. An allotetraploid species has a genome composed oftwo ancestral genomes, A and B, each of which havea basic chromosome number (x) of seven. In thisspecies, the two copies of each chromosome of eachancestral genome pair only with each other duringmeiosis. Resistance to a pathogen that attacks the foliage of the plant is controlled by a dominant allele atthe F locus. The recessive alleles Faand Fbconfersensitivity to the pathogen, but the dominant resistancealleles present in the two genomes have slightly different effects. Plants with at least one FAallele areresistant to races 1 and 2 of the pathogen regardlessof the genotype in the B genome, and plants with atleast one FBallele are resistant to races 1 and 3 of thepathogen regardless of the genotype in the A genome.What proportion of the self-progeny of an FA Fa FB Fbplant will be resistant to all three races of the pathogen?
A researcher crosses two white-flowered lines ofAntirrhinum plants as follows and obtains the followingresults:pure line 1 × pure line 2↓F1 all whiteF1 × F1↓F2 131 white29 reda. Deduce the inheritance of these phenotypes; useclearly defined gene symbols. Give the genotypes ofthe parents, F1, and F2.b. Predict the outcome of crosses of the F1 with eachparental line.
TOPIC: GENE INTERACTION AND EPISTASIS Two pure-breeding (homozygous) lines of shrubs were crossed as follows: P1 phenotype red leaves X green leaves P1 genotype ___________ _____________ F1 phenotype 100% green leaves with white flecks F1 genotype ___________ Crossing the F1s produced: F2 phenotypes F2 genotypes 123 green leaves w/ white flecks ____ 44 green leaves ____ 63 red leaves ____
A snapdragon with pink petals, black anthers, and longstems was allowed to self-fertilize. From the resultingseeds, 650 adult plants were obtained. The phenotypesof these offspring are listed here.78 red long tan26 red short tan44 red long black15 red short black39 pink long tan13 pink short tan204 pink long black68 pink short black5 white long tan2 white short tan117 white long black39 white short blacka. Using P for one allele and p for the other, indicatehow flower color is inherited.b. What numbers of red : pink : white would havebeen expected among these 650 plants?c. How are anther color and stem length inherited?d. What was the genotype of the original plant?e. Do any of the three genes show independentassortment?f. For any genes that are linked, indicate the arrangements of the alleles on the homologous chromosomes in the original snapdragon, and estimate thedistance between the genes
. A true-breeding strain of Virginia tobacco has dominantalleles determining leaf morphology (M), leaf color(C), and leaf size (S). A Carolina strain is homozygousfor the recessive alleles of these three genes. Thesegenes are found on the same chromosome as follows:M C S6 m.u. 17 m.u.An F1 hybrid between the two strains is now backcrossedto the Carolina strain. Assuming no interference:a. What proportion of the backcross progeny willresemble the Virginia strain for all three traits?b. What proportion of the backcross progeny willresemble the Carolina strain for all three traits?c. What proportion of the backcross progeny will havethe leaf morphology and leaf size of the Virginiastrain but the leaf color of the Carolina strain?d. What proportion of the backcross progeny will havethe leaf morphology and leaf color of the Virginiastrain but the leaf size of the Carolina strain?
Round seed is dominant over wrinkled. Yellow cotyledon is dominant overgreen. Full pod is dominant over constricted. The genes for these traits arefound on separate chromosomes. If a trihybrid plant is self-fertilized, whatproportion of the progeny will have round seeds in constricted pods?Express answer in fraction form.
The purple colour results from the formation of a purple pigment in the skin of the kernel. The gene locus for the colour trait is designated “R”. The wrinkled phenotype arises from a decrease in the amount starch produced within the kernel. The gene locus for the texture trait is designated “S”. Present the data as monohybrids (Xx, XX, xx) for all five corn cobs andfor the plant. Present the colour and texture phenotypes with their associated genotypes in separate tables. Your first table should include the corn ID, the number of purple kernels, the number of yellow kernels, and the phenotypic ratio between the two kernel colours. Present the texture results in the same format to a separate table. What is the dominant phenotype for kernel colour in your assigned cornplant? What is the dominant phenotype for kernel texture in your assigned cornplant?
Waxy endosperm (wx), virescent seedling (v), and shrunken endosperm (sh) are all recessive mutants in corn. Waxy and shrunken are both on chromosome IX, located 18 map units apart. Virescent is on chromosome V and is not linked to the other loci. A strain of corn that is homozygous for waxy and virescent is crossed to a strain that is homozygous for shrunken endosperm. The F1 were entirely wild type. An F1 individual was then test crossed to a strain that is homozygous for waxy, virescent, and shrunken. Of 1,000 offspring, how many of the offspring will be virescent and shrunken, but not waxy? Using the information from Problem 1, how many of the offspring will be waxy, virescent, and shrunken?
In maize trisomics, n + 1 pollen is not viable. If adominant allele at the B locus produces purple colorinstead of the recessive phenotype bronze and a B b btrisomic plant is pollinated by a B B b plant, whatproportion of the progeny produced will be trisomicand have a bronze phenotype?
On rare occasions, an organism may have three copies of achromosome and therefore has three copies of the genes on thatchromosome (instead of the usual number of two copies). Forsuch a rare organism, the alleles for each gene usually segregateso that a gamete will contain one or two copies of the gene. Let’ssuppose that a rare pea plant has three copies of the chromosomethat carries the height gene. Its genotype is TTt. The plant is alsoheterozygous for the seed color gene, Yy, which is found on adifferent chromosome. With regard to both genes, how manytypes of gametes can this plant make, and in what proportions?(Assume that it is equally likely that a gamete will contain oneor two copies of the height gene.)

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Seed Type	Initial Number of Seeds	Number of Seedlings	Seedlings Without Purple Pigment	Seedlings With Purple Pigment	Seedlings With Green Leaves	Seedlings With Yellow-Green Leaves
			Recessive Trait	Dominant Trait	Dominant Trait	Recessive Trait
P₁	10	10	10	0	10	0
P₂	10	9	0	9	5	4
F₁	10	10	3	7	3	0
F₂	30	30	15	15	15	15