Codominant Genetics # 1 1.In Andalusian chickens, B is the gene for black plumage. B' is the gene for white plumage. The genes show codominance. The heterozygous condition results in blue plumage. Show the crosses, Punnet squares, genotypic ratios and phenotypic ratios expected from the following crosses: a. black x blue genotypic ratio- phenotypic ratio- b. blue x blue genotypic ratio- phenotypic ratio- c. blue x white genotypic ratio- phenotypic ratio-

Codominant Genetics # 1 1.In Andalusian chickens, B is the gene for black plumage. B' is the gene for white plumage. The genes show codominance. The heterozygous condition results in blue plumage. Show the crosses, Punnet squares, genotypic ratios and phenotypic ratios expected from the following crosses: a. black x blue genotypic ratio- phenotypic ratio- b. blue x blue genotypic ratio- phenotypic ratio- c. blue x white genotypic ratio- phenotypic ratio-

Name: NamePeriodCodominant Genetics # 11.In Andalusian chickens, B is the g
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Description: NamePeriodCodominant Genetics # 11.In Andalusian chickens, B is the gene for black plumage. B' is the genefor white plumage. The genes show codominance. The heterozygouscondition results in blue plumage. Show the crosses, Punnet squares,genotypic ra

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 10GP: Red-flowering snapdragons are homozygous for allele RI. White-flowing snapdragons are homozygous for...

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Genetic Crosses that Involve 2 Traits In rabbits, black hair is dominant to brown hair. Also in rabbits, long straight ears are dominant to floppy ears. These letters represent the genotypes and phenotypes of the rabbits: BB = black hair EE = long ears Ee = long ears ee = floppy ears %3D %3D Bb = black hair bb = brown 1. A male rabbit with the genotype BBee is crossed with a female rabbit with the genotype bbEe The square Is set up below. Fill it out and determine the phenotypes and proportions in the offspring. How many out of 16 have black hair Ве Ве Ве Ве and long ears? bE How many out of 16 have black hair and floppy ears? be How many out of 16 have brown hair and long ears? bE How many out of 16 have brown hair be and floppy ears?
dd-ons Help в I U A Calibri 12 三三1 |:三 6. Consider a guinea pig with a homozygous genotype and a white fur color phenotype. a. What is the probability this parent will produce a gamete with the dominant allele? b. What is the probability this parent will produce a gamete with the recessive allele? C. If 31 sperm cells are collected from this guinea pig, how many would you expect to have the recessive allele (as determined by sequencing the gene)? !!!
31 1. In guinea pig, fur color is determined by one gene, and fur length by another independently segregating gene. The black fur allele (B) is dominant to brown fur allele (b) and the short allele (S) is dominant to long fur allele (s). From progeny listed in the following table, provide the probable genotype for the parents of each cross. Phenotype of Progeny Black short black long brown short Bio3110 Genetics Additional problems (Chapter 2): Parents Black short X Black short Black short X Brown long Black long X Black long 90 17 0 32 16 35 29 14 0 brown long 8 15 2. A mouse with the genotype BbccEeFf is mated with another mouse with BBCcEEFF genotype. What is the probability of obtaining the following genotype in the progeny? a. BBCcEeFF b. BbccEEFf C. BbccEeFF 12 3. Determine the type of gametes that are formed from the following individuals with the given genotype? A. AaBbCc B. AABBCc C. AABbCc
ces Edit e 9. In the F2 Generation, both the male and female pea plants are heterozygous for round/wrinkles seeds. 1)Draw a Punnett square that shows this cross. 2)What are the enetics... possible genotypes? 3)What are the possible phenotypes? 10. In pea plants, the allele for smooth pods (P) is dominant over the allele for pinched pods (p). Construct a Punnett Square that shows a cross between an PP plant and an Pp plant. Predict what percent of the offspring are likely to have smooth pods.
Name Legrning Target: I will incorporate ratios Period into the genetics problem. Genetics Worksheet # 2 1. In pea plants, smooth peas (S) are dominant to wrinkled peas (s). Work the following crosses. a. homozygous dominant x recessive Genotypes: Phenotypes: Genotypic ratio: Phenotypic ratio: b. heterozygous x heterozygous Genotypes: Phenotypes: Genotypic ratio: Phenotypic ratio: 2, A man and his wife are both heterozygous for brown eyes (B). They have six children all of whom have blue eyes> How can this happen? List the genotypes of the parents and the children, Show the Punnett square, Parents: Kids: What is the chance that the next child will have blue eyes?
Direction: Answer the following problems: 1. Predict the phenotypic ratios of offspring when a homozygous white cow is crossed with a roan bull. Illustrate using a Punnett square. 2. Complete the Punnett square for a cross between a black chicken (BB) and a white chicken (WW). Give the phenotype of the offspring in the F1 generation. Кey BB – black ww- white F1 BW – checkered Phenotype:
Drosohpila Punnet Square of Crosses. I need results of F1 & F2 generation using Punnett Squares for: Make Punnet Squares of the following crosses •Drosophila Female wildtype cross Male White-eye •Drosophila Male wildtype cross Female White-eye •Drosophila Female Wild Type cross Male Scarlet Eye •Drosophila Male Wild Type cross Female Scarlet Eye Also, Which allele is heterozygous and which is homozygous, & which is dominant and which is recessive?
O e. Parent 2: Parent 1: Parent 2: ¡Ai QUESTION 9 Bi QUESTION 8 Let's assume that, in dragons, red scales (B) are dominant to green scales (b), and long tongues (S) are dominant to short tongues (s). The genes that determine these characteristics assort independently. A homozygous red, long-tongued dragon is crossed with a homozygous green, short-tongued dragon. If an F1 dragon is crossed to a homozygous green and homozygous short-tongued dragon, what phenotypes and proportions are expected in the offspring? O a. 100% green and long-tongued O b. ½ green and short-tongued & ½ red and long-tongued O c. ½ red and short-tongued & ½ green and long-tongued O d.% red and long-tongued, % red and short-tongued, ½ green and long-tongued, % green and short-tongued O e. 9/16 red, long-tongued, 3/16 green, long-tongued, 3/16 red, short-tongued, 1/16 green, short-tongued Save and Submit to save and submit. Click Save All Answers to save all answers. 000 MacBook Air
Paragraph Styles DIHYBRID GENETICS PROBLEMS Name In garden peas, tall (T) vine is dominant over short (t) vine, and round pea pod (R) is dominant over wrinkled pea pod (r). Cross a homozygous tall, round pea plant with a homozygous short, wrinkled pea plant. Show the genotypes and phenotypes of the F1 offspring, and the probability of each. 1. Genotype of the homozygous tall, round pea plant Genotype of the homozygous short, wrinkled pea plant. Genotypes Phenotypes acer
Highlight Draw ソeu र In guinea pigs, short hair is dominant over long hair. If a short haired SS guinea pig is erossed with a long haired ss guinea pig, what are the possible genotypes and phenotypes of their offspring and the percent chance of each? 3. S Fill in the table: Genotype Phenotype Chance
This exercise primarily addresses o MLO 1: Correctly assign and interpret allele symbols according to the conventions of the fruit fly research community. o MLO 2: Correctly assign genotypes and phenotypes to parents and offspring in genetic crosses given a mode of inheritance. Autosomal Recessive Inheritance We learned in the Linkage Mapping lab module the conventions for representing alleles in fruit flies. Recall that we use one to three letters, and the case (upper or lower) is determined by whether the mutant allele is recessive or dominant. Additionally, we add a "+" as a superscript to indicate the wild-type ("normal") allele. For example, if a new mutant phenotype is recessive, we can represent the recessive allele as m. Then the wild-type allele is mt, and m+ > m. Based on the rule of dominance, we can predict phenotypes based on genotypes: mm Wild type Wild type Mutant m+m mm
Miniature wings, X in Drosophila melanogaster result from an X-linked allele that is recessive to the allele for long wings, 9 + X. Match the genotypes for each parent in the crosses. Male parent phenotye long miniature miniature 111 long long Female parent phenotype long long long miniature long m m X X Male offspring phenotypes 231 long, 250 miniature 610 long 410 long, 417 miniature 753 miniature 625 long m X Y Female offspring phenotypes 560 long 632 long 412 long, 415 miniature 761 long 630 long Answer Bank ++ X X Male parent genotype + X Y Female parent genotype + X X m 00