Show me how to do this punnet square problem 5. (8 points) In a species of grain, the allele for round grains (G) is completely dominant over the allelefor wrinkled grains (g). Height in the same plant is controlled by a gene that exhibits incompletedominance, such that homozygous dominant (HH) plants are tall, heterozygous plants (Hh) are mediumheight, and homozygous recessive (hh) plants are short. You cross a true-breeding plant that is tall withround grains with a medium height plant with wrinkled grains.HHGGa. What is the genotype of the true-breeding tall parent with round grains?6List the genotype(s) of the gametes this parent can produce. ay, gHb. What is the gentoype of the medium height parent with wrinkled grains?// List the genotype(s) of the gametes this parent can produce.c. Complete a Punnett square for this cross.(NOTE: a small Punnett square is sufficient to solve this problem)HJay, 66 = roun!y3 = urinisindjg Hhý HgHighshGGHyGHGg H H = .1 |Gy.H.H = 1буни.G g HhGgd. What phenotypes are possible in the resulting offspring?기Gy.thHhHHGH, HHHhHH = tollHh= nodionhh = shortGG HH x gg HhMHGyни។ودا وHg, hgHhigh thGH gHgH Gg H | gg H Hgh Gylh g ghhggGHg (GyshH GHHH

Name: Show me how to do this punnet square problem 5. (8 points) In a specie
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Description: Show me how to do this punnet square problem 5. (8 points) In a species of grain, the allele for round grains (G) is completely dominant over the allelefor wrinkled grains (g). Height in the same plant is controlled by a gene that exhibits incomplete

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5. (8 points) In a species of grain, the allele for round grains (G) is completely dominant over the allele for wrinkled grains (g). Height in the same plant is controlled by a gene that exhibits incomplete dominance, such that homozygous dominant (HH) plants are tall, heterozygous plants (Hh) are medium height, and homozygous recessive (hh) plants are short. You cross a true-breeding plant that is tall with round grains with a medium height plant with wrinkled grains. HHGG a. What is the genotype of the true-breeding tall parent with round grains? GHH /List the genotype(s) of the gametes this parent can produce. Cog, gH b. What is the gentoype of the medium height parent with wrinkled grains? ½/2 List the genotype(s) of the gametes this parent can produce. GH HH #h th c. Complete a Punnett square for this cross. (NOTE: a small Punnett square is sufficient to solve this problem) H "y, GG = roun! = wrinkled g High G y GH GH gH G₂ HH₂.A Gy.H.H| 21 sh ✔ >1 водни Gy Hh What phenotypes are possible in the resulting offspring? GyHh Gg Hh HH = Hh= hh= GG HH Gy HH y H و ودا و high th GH OH | G₂HV | ₁HH | Hg, hg toll nasion short дун х ду gy ни G H 9 Gylsh HGH HH

5. (8 points) In a species of grain, the allele for round grains (G) is completely dominant over the allele for wrinkled grains (g). Height in the same plant is controlled by a gene that exhibits incomplete dominance, such that homozygous dominant (HH) plants are tall, heterozygous plants (Hh) are medium height, and homozygous recessive (hh) plants are short. You cross a true-breeding plant that is tall with round grains with a medium height plant with wrinkled grains. HHGG a. What is the genotype of the true-breeding tall parent with round grains? GHH /List the genotype(s) of the gametes this parent can produce. Cog, gH b. What is the gentoype of the medium height parent with wrinkled grains? ½/2 List the genotype(s) of the gametes this parent can produce. GH HH #h th c. Complete a Punnett square for this cross. (NOTE: a small Punnett square is sufficient to solve this problem) H "y, GG = roun! = wrinkled g High G y GH GH gH G₂ HH₂.A Gy.H.H| 21 sh ✔ >1 водни Gy Hh What phenotypes are possible in the resulting offspring? GyHh Gg Hh HH = Hh= hh= GG HH Gy HH y H و ودا و high th GH OH | G₂HV | ₁HH | Hg, hg toll nasion short дун х ду gy ни G H 9 Gylsh HGH HH

Biology: The Unity and Diversity of Life (MindTap Course List)

14th Edition

ISBN:9781305073951

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter13: Observing Patterns In Inherited Traits

Section: Chapter Questions

Problem 7GP: Several alleles affect traits of roses, such as plant form and bud shape. Alleles of one gene govern...

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1. In the milotypes of wheat, 4 recessive genes for height have been identified. They have been designated as h1, h2, h3, and h4. A plant homozygous recessive for these 4 pairs of genes is 21inches tall, while a plant homozygous dominant for these 4 pairs of genes is 45 inches tall. Assume that these alleles contribute quantitatively, cumulative and equally for the height difference. A plant breeder crossed plants with genotypes of H1H1H2H2h3h3h4h4 and h1h1h2h2h3h3H4H4. Both F1 and F2 generations were observed. a. What are the heights of the parents? b. What would be the heights of the F1 plants? c. Enumerate the range of heights that you can expect in the F2 generation. Also, indicate the expected number of plants of each height. d. What would be the expected mean height of the F2 progeny?
8. Kernel color in wheat is controlled by 2 pairs of genes (AABB). Determine the color of the offspring with the following genotypes: (Note: 4 contributing alleles – red; 3 contributing alleles – medium red; 2 contributing alleles – intermediate red; 1 contributing allele – red; and 0 – white AAbb x AaBb AABb x Aabb aaBb x Aabb AABb x aabb AABb x AaBb A wheat plant producing medium red seed is crossed with another plant producing intermediate red seed (refer to problem no. 8). How many individuals will be? Red Medium red Intermediate red Light red White
1. Alleles of the gene that determines seed coat patternsin lentils can be organized in a dominance series:marbled > spotted = dotted (codominant alleles) >clear. A lentil plant homozygous for the marbled seedcoat pattern allele was crossed to one homozygous forthe spotted pattern allele. In another cross, a homozygous dotted lentil plant was crossed to one homozygous for clear. An F1 plant from the first cross wasthen mated to an F1 plant from the second cross.a. What phenotypes in what proportions are expectedfrom this mating between the two F1 types?b. What are the expected phenotypes of the F1 plantsfrom the two original parental crosses?
1. In pea plants, the traits below exhibit the following dominance pattems: Trait: 1. Seed shape (R) 2. Flower color (P) 3. Color of seed coat (Y) 4. Form of ripe pods (I) 5. Color of unripe pods (G) 6. Position of flowers (A) 7. Length of stem (T) Dominant Expression: Round Purple Yellow Inflated Green Axial Tall Recessive Expression: Wrinkled White Green Constricted Yellow Terminal Short Record the genotypes for pea plants with the following descriptions: A. A plant with yellow pods : B. A plant heterozygous for green pods C. A plant homozygous for round seeds D. A plant heterozygous for purple flowers: E. A plant homozygous for yellow seeds: F. A plant with white flowers: G. A plant with green seeds H. A short pea plant.
10. In pea plants, seed shape and seed color are controlled by genes located on different chromosomes. Seeds may be round (R) or wrinkled (r), with the allele for round seeds being dominant. Alleles for seed color are yellow and green, with the green allele (y) recessive to the yellow (Y) allele. If you cross an individual that is homozygous round and yellow with an individual that is homozygous for wrinkled and green, what is the genotype of the F1 individuals? Set up a Punnett square for the dihybrid cross.
5A. In shorthorn cattle, the polled condition is dominant over horned. Also, the heterozygous condition (RW) of red coat (R) and white coat (W) is roan, an example of codominance. (The roan coat contains both red and white fur because neither allele is dominant over the other). If a homozygous polled, white male is bred to a horned, red female, what will be the appearance of the F1 generation? Horned: r: Roan Coat: a. Possible genotypes for Polled condition Coat:_ ossible geno c. Male genotype: d. Female genotype: Male possible gametes Female possible gametes e. f. Genotypic Ratio: g. Phenotypic Ratio 5B, Breed two of the F1 generation to determine the possibilities in the F2 generation. Horned: a. Possible genotypes for Polled condition b. Possible genotypes for Red coat: White Coat:_ Roan Coat: c. Male genotype: d. Female genotype Male possible gametes Female possible gametes: f. Genotypic Ratio: g. Phenotypic Ratio
5A. In shorthorn cattle, the polled condition is dominant over horned. Also, the heterozygous condition (RW) of red coat (R) and white coat (W) is roan, an example of codominance. (The roan coat contains both red and white fur because neither allele is dominant over the other). If a homozygous polled, white male is bred to a horned, red female, what will be the appearance of the F1 generation? a. Possible genotypes for Polled condition: Horned: b. Possible genotypes for Red coat:White Coat: Roan Coat: c. Male genotype Male possible gametes: Female genotypeFemale possible gametes: e. f. Genotypic Ratio: g. Phenotypic Ratio:
2 (a) The trait of hen- versus cock-feathering is a sex-limited trait controlled by a single gene. HH and Hh males always exhibit the hen-featuring trait. Only hh males show the cock-feathering trait. Starting with two heterozygous fowls that are hen- feathered, describe how you would obtain a true-breeding line that always produces cock-feathered males. [Ciri berhulu avam hatin
1. In pea plants, the traits below exhibit the following dominance patterns: Recessive Expression: Wrinkled White Green Constricted Yellow Terminal Short Trait: 1. Seed shape (R) 2. Flower color (P) 3. Color of seed coat (Y) 4. Form of ripe pods (1) 5. Color of unripe pods (G) 6. Position of flowers (A) 7. Length of stem (T) Dominant Expression: Round Purple Yellow Inflated Green Axial Tall Record the genotypes for pea plants with the following descriptions: A. A plant with yellow pods : B. A plant heterozygous for green pods : C. A plant homozygous for round seeds : A plant heterozygous for purple flowers : E. A plant homozygous for yellow seeds : F. A plant with white flowers : G. A plant with green seeds : H. A short pea plant: :
6. In certain plant species such as tomatoes and petunias, a highly polymorphic incompatibility gene Swith more than 100 known alleles prevents self-fertilization and promotes outbreeding. In this form ofincompatibility, a plant cannot accept sperm carrying an allele identical to either of its own incompatibility alleles. If, for example, pollen carryingsperm with allele S1of the incompatibility genelands onto the stigma (a female organ) of a plantthat also carries the S1allele, the sperm cannot fertilize any eggs in that plant. (This phenomenonoccurs because the pollen grain on the stigma cannotgrow a pollen tube to allow the sperm to unite withthe egg.)For the following crosses, indicate whether anyprogeny would be produced, and if so, list all possiblegenotypes of these progeny. a. ♂ S1S2× ♀ S1S2b. ♂ S1S2× ♀ S2S3c. ♂ S1S2× ♀ S3S4d. Explain how this mechanism of incompatibilitywould prevent plant self-fertilization.
4. Albinism is lethal in plants, but many plants produce albino offspring. If albino plants C always die before reproducing, how does albinism persist? Migh ne 5. Mendel crossed pea plants producing round seeds with those producing wrinkled seeds. Of 7,324 F2 seeds, 5,464 were round, and 1,860 were wrinkled. Using the symbols W and w for the alleles, (a) symbolize the original P₁ cross; Slover don pet (b) determine the parental gametes; 69 (c) show the F₁ offspring; (d) show a cross between two F₁ plants; dong berbella" 10. Jogolų ar preu 39 woy Your Mirb & ever ascohes berloste di ritod 8 201 artW.20 meylo evit bins tight pdchse sent sitt 10 euogysomod Ceadolise borbeis blirib o svad Jism adoles Parl (e) symbolize the F₁ gametes; and, odb seit (f) summarize the expected F2 results for phenotypes, genotypes and phenotype ratios. mun 100 3161 16 gr nation ritiw nam A (6) His neblinks Sylimstedt phow yd showenis uby sloy 6 or be 160 (0) endonce 100 to eno elouosis na (b) wari hot…
. The production of pigment in the outer layer of seedsof corn requires each of the three independently assorting genes A, C, and R to be represented by at leastone dominant allele, as specified in Problem 64. Thedominant allele Pr of a fourth independently assortinggene is required to convert the biochemical precursorinto a purple pigment, and its recessive allele pr makesthe pigment red. Plants that do not produce pigmenthave yellow seeds. Consider a cross of a strain of genotype A/A ; C/C ; R/R ; pr/pr with a strain of genotypea/a ; c/c ; r/r ; Pr/Pr.a. What are the phenotypes of the parents?b. What will be the phenotype of the F1?c. What phenotypes, and in what proportions, willappear in the progeny of a selfed F1?d. What progeny proportions do you predict from thetestcross of an F1?