1. Genes A and E are independent. Two individuals with the genotypes AaEe x aaEe are crossed (upper case = dominant, lowercase = recessive). a. Draw your Punnett square(s) in the space below. b. On each line, show your calculations and the final answer. ie 1/2 x 1/2 = 1/4 What is the probability of these two individuals having the following: an offspring with the phenotype aE ? an offspring that is heterozygous for both traits? an male offspring with the genotype AaEE ?

1. Genes A and E are independent. Two individuals with the genotypes AaEe x aaEe are crossed (upper case = dominant, lowercase = recessive). a. Draw your Punnett square(s) in the space below. b. On each line, show your calculations and the final answer. ie 1/2 x 1/2 = 1/4 What is the probability of these two individuals having the following: an offspring with the phenotype aE ? an offspring that is heterozygous for both traits? an male offspring with the genotype AaEE ?

Name: 1. Genes A and E are independent. Two individuals with the genotypes
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Description: 1. Genes A and E are independent. Two individuals with the genotypes AaEe x aaEe arecrossed (upper case = dominant, lowercase = recessive).a. Draw your Punnett square(s) in the space below.b. On each line, show your calculations and the final answer

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 3CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...

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A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?
Purple flowers are dominant to white flowers. Identify the phenotypefor the following genotype Ff, FF, ff and determine if the genotype is heterozygous or homozygous. * For each row, you should select two columns. Purple flowers White flowers Heterozygous Homozygous Ff FF ff Brown eyes are dominant to blue eyes. Identify the phenotype for the following genotype BB, Bb, bb and determine if the genotype is heterozygous or homozygous. * 口ロ口
Ob 1. In corn the color purple (P) is dominant to the color yellow (p) and round (R) shape is dominant to wrinkled (r). Your corn is most likely one of two crosses: A. OR B. (shown below) Complete the two crosses using the following Punnett squares and determine the expected phenotypic and genotypic ratios of the offspring. A. A cross between two heterozygous purple/ heterozygous round corn plants:
9. Make a pedigree for each of the following situations. For each individual, write the individual's genotype (when possible) next to the individual's symbol (e.g. O xty, I Gg): a. Two parents do not have cystic fibrosis and they have a daughter with cystic fibrosis and a son who does not have cystic fibrosis. The daughter grows up and she mates with a male who does not have cystic fibrosis. Their only child is a boy and he has cystic fibrosis. b. A man with hemophilia mates with a female without hemophilia. They have one son and one daughter. The daughter has hemophilia and the son does not have hemophilia. The son grows up, and he marries and mates with a female. Their only child is a boy, and he has hemophilia.
6. Using a Punnet square, please calculate the probability of having a Aabb offspring in a cross AaBb x aaBb 7. What is the probability of having a AAbbCcDd offspring in a cross AaBbCCdd x AabbCcDD? (Please show how you derive your answer) 8. In a cross between a blood type AB male and a heterozygous type B female, what are the possible genotype and phenotype for the offspring? What is the phenotypic ratio? :On D. Focus h. 72°F LL F4 PrtScr Insert Home F5 EZ F8 F9 F10 F11 F12 & 3. 6 7 8. 09. E R Y
. a. A mouse cross A/a ⋅ B/b × a/a ⋅ b/b is made, and inthe progeny there are25% A/a ⋅ B/b, 25% a/a ⋅ b/b,25% A/a ⋅ b/b, 25% a/a ⋅ B/bExplain these proportions with the aid of simplifiedmeiosis diagrams.b. A mouse cross C/c ⋅ D/d × c/c ⋅ d/d is made, and inthe progeny there are45% C/c ⋅ d/d, 45% c/c ⋅ D/d,5% c/c ⋅ d/d, 5% C/c ⋅ D/dExplain these proportions with the aid of simplifiedmeiosis diagrams.
A. You cross a true-breeding sunflower, with yellow flowers and black seeds, with another true-breeding sunflower, with white flowers and green seeds. All of the F1 generation have yellow flowers and green seeds. Create allele symbols and genotypes for these genes and create a representation of this cross, which is consistent with the stated premises. Show your work. B. Upon completing an F1xF1 cross of the yellow flower, green seed sunflowers, you observe the following phenotype classes of offspring in the F2: 250 Yellow flowers and black seeds; 250 white flowers and green seeds; 500 yellow flowers and green seeds. A test cross of the yellow flower, green seed F1 generation generates the following offspring: 500 white flowers and green seeds; 500 yellow flowers and black seeds. Do these observations taken together, of the F1, F2, and test cross offspring, respectively, represent violations of Mendel's first law, Mendel's second law, or both? Explain your answer.
Please consider the pedigree below. There are no cases of false paternity. I II III IV в 1 a. Which individual/s definitely has/have Bombay phenotype in the descendants of I-1 and I-2? b. What are the genotypes of individuals II-2 and III-2 at the AB0 and H loci? Please label your answers a and b, Il-2: and Ill-2:.
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?
The following diagram shows a hypothetical diploid cell. The recessive allele for albinism is represented by a, and d represents the recessive allele for deafness. The normal alleles for these conditions are represented by A and D, respectively. Please use the symbols A, a, D and d as appropriate to help me understand your answer. a а D 1. According to Mendel's principal of segregation, what is segregating in this cell? 2. According to Mendel's principle of independent assortment, what is independently assorting in this cell? 3. What is the phenotype of this individual? © Cengage Learning
5 & :56M ******* 24 DIHYBRID CROSSES DRV 0 Stv T alı A @ zladenA 9160p2-id2 bns obidalbaneoviene da II\ MOD YR 21 $59A ... Create a dihybrid cross and determine the expected phenotypic percentages of the offspring of two corn plants both of which are heterozygous for colour and texture (RrTt X RrTt). Don't forget to include clear let statements, and follow the all six steps taught on solving genetics problems. insig moni nellog: bna. zoom