In peas, green pods(G) are dominant to yellpw pods(g) and inflated pods(I) are dominant to constricted pods(i). Cross an individual heterozygous for both trairs with an individual that has yellow, constricted pods. Using the five steps, show the possible phenotypes and their ratios(as fractions) for this cross. Punnet square.

In sweet pea plants, the trait for purple flowers (P) is dominant to the trait for red flowers (p).Similarly, the trait for long pollen (L) is dominant to the trait for round pollen (l).Two heterozygotes are crossed, yielding the following frequencies for the F1 generation:293 purple long plants;19 purple, round plants ;28 red, long plants ;87 red, round plantsUse the chi-square test to determine if these results are due to independent assortment. detailed explanation ?

Fruit flies having one allele for curly wings (Cy) and one allele for normal wings (Cy+) develp curly wings.  Fruit flies homozygous for normal wings (Cy+) develop normal wings. When two curly-winged flies were crossed, 204 curly-winged and 98 normal-winged flies.  In fact, all crosses between curly-winged flies produce nearly the same curly-winged to normal-winged ratio.  How would you explain these results?   The Cy allele is epistatic to the Cy+ allele. The Cy+ allelle shows partial dominance over the Cy allele. The Cy allele is dominant for the curly phenotype and is lethal when homozygous.  The Cy+ allele is dominant over the Cy allele and is lethal when homozygous.    The Cy+ and Cy alles are co-dominant.

In watermelons, bitter fruit (B) is dominant over sweet fruit (b), andyellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. Q. If an F1 plant is backcrossed with the sweet, unspotted parent, what phenotypes and proportions are expected in the offspring?

In radishes, color and shape are each controlled by asingle locus with two incompletely dominant alleles.Color may be red (RR), purple (Rr), or white (rr) andshape can be long (LL), oval (Ll), or round (ll). Whatphenotypic classes and proportions would you expectamong the offspring of a cross between two plantsheterozygous at both loci?

In garden pea plants, axial flower location is dominant (A) to terminal flower location (a). If a plant that is heterozygous for axial flower location is crossed with a plant that has terminal flower location, what is the genotypic ratio of the offspring?

Three corn seed traits are C for red, c for white; S for plump, s for shrunken; W for normal, w for waxy. Use the data from the testcross to map the distances and order of these three loci. Show calculations and illustrate the gene map. (Hint: drawing out the chromosomes and gametes will help.)              Parents:         CCssWW (CsW/CsW)          X       ccSSww (cSw/cSw)              Trihybrid:      CcSsWw (CsW/cSw) X       ccssww (csw/csw)   Test cross offspring   Seed trait Gamete from trihybrid Number Red, shrunken, normal CsW 2777 White, plump, waxy cSw 2708 R ed, plump, waxy CSw 116 White, shrunken, normal csW 123 Red, shrunken, waxy Csw 643 White, plump, normal cSW 626 Red, plump, normal CSW 4 White, shrunken, waxy csw 3 Total number of progeny:   7000

D D 1 2 d In a particular plant, leaf color is controlled by gene locus D. Plants with at least one allele D have dark green leaves, and plants with the homozygous recessive dd genotype have light green leaves. A true-breeding, dark-leaved plant is crossed with a light-leaved one, and the F1 offspring is allowed to self-pollinate. The predicted outcome of the F2 is diagrammed in the Punnett square shown in the figure, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square. Which of the boxes marked 1-4 correspond to plants with a heterozygous genotype? O 1 only O 1, 2, and 3 2 and 3 2, 3, and 4 4) 3.

In the fruit berry nispero, Eriobotrya japonica. Has the recessive mutant genes p (purpura = fluorescent fruit), a (amarillis = yellow fruit), and o (oval = acidic fruit) on chromosome 6. The following data represents the phenotypes and number obtained in the testcross mating of an F1 that is heterozygous for all three genes (loci) x homozygous recessive for all three genes (loci). Number Phenotypes PAO 91 PAo 427 PaO 6 Pao 106 PAO 120 pAo рао pao 7 323 62 A) What is the sequence of these three genes on chromosome 6? B) What were the genotypes of the homozygous parents that were used to create the F₁ heterozygote? C) Construct a map giving the locations and distances in map units. D) What is the coefficient of coincidence?

In a certain plant, the dominant form of gene B codes for blue fruit, while the recessive form results in pink fruit. The dominant form of another gene, E, inhibits the activity of the enzyme coded for by gene B, resulting in white fruit, while the recessive form is unable to inhibit this enzyme and results in colored (i.e., blue or pink) fruit. A doubly-homozygous dominant white-fruited plant is crossed with a pink-fruited plant. The F1 progeny were then self-crossed to generate the F2 generation. Determine the ratios of genotypes and phenotypes for each generation.