Answered: An individual whose genotype is AABbCc…

Name: please help me out..... THANK YOU!1. An individual whose genotype is A
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Description: please help me out..... THANK YOU!1. An individual whose genotype is AABbCc is crossed with an individual who is heterozygous for all three of these genes. List all the gametes that the AABbCc parent could produce.

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An individual whose genotype is AABbCc is crossed with an individual who is heterozygous for all three of these genes. List all the gametes that the AABbCc parent could produce.

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

Chapter14: Chromosomes And Human Inheritance

Section: Chapter Questions

Problem 3GP: Human females have two X chromosomes (XX); males have one X and one Y chromosome (XY). a. With...

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Let us practice it again! Analyze the pedigree below to answer the questions that follow. Huntington's disease a disorder in which nerve cells waste away, or disintegrate, is passed down through families. certain parts of the brain Huntington's diseate llustration ereated in htps://pregenygenetion.com/ 1. What members of the family above are affected with the Huntington's disease? 2. Tnere are no carriers ior Huntungton's disease you either have it or you do not. Is Huntington's disease caused-by a dominant or recessive trait? 3. Identify the genotypes of the following individuals using the pedigree above. (homozygous dominant, homozygous recessive, heterozygous). I- 1 II -1: II -3: III - 4 : 4. How many children did individuals I-1 and I-2 have? 5. How many girls did II-1 and II-2 have? How many have Huntington's Disease? 6. How are individuals III-2 and II-4 related? I-2 and III-5?
1. Hemophilia is due to a sex-linked gene. It is recessive and found on the X chromosome. A woman who is a carrier for hemophilia marries a normal man. What will be the possible phenotypes of their children? Use a punnet square. 2. A phenotypically normal man, who has a hemophiliac brother, marries a normal woman, who is not a carrier. What is the probability that any of their children will be hemophiliac? Use a punnet square. 3. Color-blindness is an X-linked, recessive trait. If a normal-sighted woman, whose father was color-blind, marries a colorblind man, what is the probability that they will have a son who is color-blind? Use a punnet square. 4. A man and woman, both of normal vision, have: 1) a color-blind son (#1) who has a daughter of normal vision 2) a daughter (#1) of normal vision who has one color-blind son and one normal vision son 3) another daughter (#2) of normal vision who has five sons, all with normal vision What are the probable genotypes of the…
Consider the following pedigree. 하 3 10 (5 3 2 (a) What pattern of transmission is most consistent with this pedigree? (1) autosomal recessive, (2) autosomal dominant, (3) X-linked recessive, (4) X-linked dominant. (b) If individual V-2 marries a normal individual, and if the condition has a pene-trance of 85 percent, what is the probability that their second child will express the trait? (c) On the third line, what does the diamond with a 10 in the middle mean?
2. .... 3. Mrs. Morales is a carrier of the sex-linked hemophilia allele and Mr. Morales is normal. They actually have 4 daughters and 4 sons. What percentage of each sex will be hemophiliac, carrier, and normal.
1. In humans, hemophilia is a sex linked trait. Females can be normal, carriers, or have the disease. Males will have the disease or not (but they won't be ever carriers) XH XH = female, normal XH Xh = female, carrier Xh Xh = female, hemophilic XH Y = male, normal Xh Y = male, hemophilic a) Show the cross of a man who has hemophilia with a woman who is a carrier. b) What is the probability that their children have the disease
Human females have two X chromosomes XX; males have one X and one Y chromosome XY. a. With respect to X-linked alleles, how many different types of gametes can a male produce? b. A female homozygous for an X-linked allele can produce how many types of gametes with respect to that allele? c. A female heterozygous for an X-linked allele can produce how many types of gametes with respect to that allele?
Human females have two X chromosomes (XX); males have one X and one Y chromosome (XY). a. With respect to X-linked alleles, how many different types of gametes can a male produce? b. If a female is homozygous for an X-linked allele, how many types of gametes can she produce with respect to that allele? c. If a female is heterozygous for an X-linked allele, how many types of gametes can she produce with respect to that allele?
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 if the couple wanted prenatal testing so that a normal fetus could be aborted?
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?
3. The recessive sex-linked gene (h) prolongs the blood-clotting time, resulting in the genetically inherited disease called hemophilia. From the information in the pedigree chart (right), answer the following questions: Hemophilia in humans (a) If 112 marries a normal man, what is the probability of her first child being a hemophiliac? 2 (b) Suppose her first child is actually a hemophiliac. What is the chance that her second child will be a boy with hemophilia? 2 (c) If I14 has children with a hemophiliac man, what is the probability that her first child will be phenotypically normal? (d) If the mother of 12 was phenotypically normal, what phenotype was her father? 4. The phenotypic expression of a dominant gene in Ayrshire cattle is a notch in the tips of the ears. In the pedigree chart on the right, notched animals are represented by the solid symbols. Ear notches in Ayrshire cattle Determine the probability of notched offspring being produced from the following matings: 1 2 (a)…
Hi, I'm having trouble with my study guide for my upcoming genetics exam. If someone could please help with work shown and an explanation it would help so much! Thank you!! 1a. The pedigree below represents inheritance of rare condition (filled symbols used for affected individuals). Test the hypothesis of X-linked dominant inheritance by assigning alleles (A or a) to sex chromosomes of all individuals in generations I and II. Does the X-linked dominant hypothesis agree with the data? It not, indicate all at least 2 individuals by generation and number (e.g. II-8) that are not consistent with the genotype you’ve proposed for the individuals in generation I. 1b. Test the hypothesis of autosomal dominant inheritance by assigning alleles (A or a) to autosomes of all individuals in the pedigree (generations I – IV). Does the autosomal dominant hypothesis agree with the data? It not, indicate all individuals by generation and number (e.g. II-8) that are not consistent with the genotype…