An individual has the following genotype. Gene loci (A) and (B) are 15 m.u. apart. What are the correct frequencies of some of the gametes that car be made by this individual? 9. a BI O A. Ab = 7.5%; AB = 42.5% O B. ab = 25%; aB = 50% O C. AB = 7.5%; aB = 42.5% O D. aB = 15%; Ab = 0% E. aB 70%; Ab = 15%

An individual has the following genotype. Gene loci (A) and (B) are 15 m.u. apart. What are the correct frequencies of some of the gametes that car be made by this individual? 9. a BI O A. Ab = 7.5%; AB = 42.5% O B. ab = 25%; aB = 50% O C. AB = 7.5%; aB = 42.5% O D. aB = 15%; Ab = 0% E. aB 70%; Ab = 15%

Name: An individual has the following genotype. Gene loci (A) and (B) are 1
Uploaded: 2024-03-20T06:43:04.000Z
Channel: Bartleby
Description: An individual has the following genotype. Gene loci (A) and (B) are 15 m.u. apart. What are the correct frequencies of some of the gametes that carbe made by this individual?BlaO A. Ab = 7.5%; AB = 42.5%B. ab = 25%; aB = 50%O C. AB = 7.5%; aB = 42.5

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...

See similar textbooks

Similar questions

_30*_SP23 - General Biology I (for maj A 1:1 phenotypic ratio will occur when which of the following crosses is done? us page Select one: F1 O a. Ttx Tt O b. TT x tt O c. TT X TT TTx Tt tt x tt tt x Tt GgTt x GgTt GGTT x ggtt GGtt x ggTT O d. e. O f. O g. Oh. Oi. F2 # 80 F3 LA $ 000 000 F4 % F5 MacBook Air < F6 & F7
Answer the following questions. 1. Construct a map for the genes d,e,f. Assume that: d and e = 3%; e and f = 5%. Give 2 arrangements of the genes/maps. 2. If d and f = 2%, what is the correct arrangement of the genes d,e,f? 3. Consider the fourth gene "g". if g and e = 1.5%, give two possible arrangements. 4. If d and g = 1.5 % give the correct order of the four genes %3D
Name: 2. Some ladybugs have 10 black spots on their shells and some have 4. When true breeding 10 spot individuals are crossed with true breeding 4 spot individuals, the offspring have 7 spots. a. Propose two distinct explanations for this finding. Explain the nature of spot inheritance in each case. D. rew bacteria erred them figure above would of14N7 then ely 2 moldon omrod abitqaq ratlsmmem sge of delw toiisoibom s s neu not oomod aid to soubnup ogusl onomod odi to slevel ismon oouborg o o consu ud od souboini bluow uoy dairlw yd za00oq sdi mialex b. Propose an experiment that would distinguish between these possibilities. sor proieins
30*_SP23 - General Biology I (for majors page A 9:3:3:1 phenotypic ratio will occur when which of the following crosses is done? Select one: O a. Tt x Tt O b. TTx TT O c. TT x tt d. GGTT x ggtt O e. tt x Tt O f. GgTt x GgTt O g. GGtt x ggTT Oh. tt x tt Oi. TTx Tt 20 000 000 MacBook Air
30*_SP23 - General Biology I (for majo us page Ö A 3:1 phenotypic ratio will occur when which of the following crosses is done? Select one: O a. Ttx Tt O b. O c. O d. O e. O f. GgTt x GgTt GGtt x ggTT GGTT x ggtt TT x tt tt x Tt tt x tt Oh. TT x Tt Oi. TT x TT g. O g. 000 MacBook Air
Dihybrid Cross gray hair dominant In rabbits, is dominant to white hair. Also in rabbits, black red eyes. These letters the eyes are to represent genotypes of the rabbits: GG = gray hair Gg = gray hair gg = white hair BB = black eyes Bb = black eyes bb = red eyes 1. What the phenotypes (descriptions) of rabbits that have the are following genotypes? Ggbb ggBB ggbb GgBb
30*_SP23 - General Biology I (for maj s page A 1:1 phenotypic ratio will occur when which of the following crosses is done? Select one: O a. Tt x Tt b. TT x tt O c. TTx TT O d. TTx Tt O e. tt x tt O f. tt x Tt O g. GgTt x GgTt h. GGTT x ggtt GGtt x ggTT O i. MacBook Air
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. 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. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?
Select all true statements about the Xist gene. pls help asap with the following and explain how you get to your answer.
1_30*_SP23 - General Biology I (for majors 4 I of evious page A 1:2:1 genotypic ratio will occur when which of the following crosses is done? F1 Select one: O a. GgTt x GgTt O b. TTx TT O c. tt x tt O d. O e. O f. Og. Oh. Oi. TTx Tt tt x Tt GGtt x ggTT GGTT x ggtt TT x tt Tt x Tt F2 80 F3 F4 F5 MacBook Air F6 A F7