You are to discover the chromosomal location (autosomal or X linked) and mode of inheritance fortwo new fly mutations coughrotato (cp) and happyhour, (bb)The homozygous couchpotato (cpcp) phenotype is (sedentary) an extreme lack of activityThe homozygous happyhour (hhhh) can drink any other fly under the table so phenotype (drinker)PARENTSCross Male double homozygous mutants cp hh with Female wild wild CP HH are to be crossedBUT FIRST you don't know if either gene is on the X chromosomeParental diploid genotypes could be:male çpcphbbh female CPCPHHHH if Both genes autosomal orMale XcpYhbhh female XCPXCPHHHH if cp on the X chromosome and bh autosomal orMale cpcpXhhY female CPCPXHHXHH if hh on the X chromosome and cp autosomalWhat are the female and male PARENTAL GAMETE GENOTYPES for the above parental crosses if:a) Both genes were autosomal ((REMEMBER BOTH GENDER HAVE ONE GAMETE GENO)b)cp gene on the X chromosome and hh autosomal (REMEMBER MALES WITH HAVE 2 GAMETE GENO)c) hh gene on the X chromosome and cp autosomal (REMEMBER MALES WITH HAVE 2 GAMETE GENO)

Name: You are to discover the chromosomal location (autosomal or X linked)
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Description: You are to discover the chromosomal location (autosomal or X linked) and mode of inheritance fortwo new fly mutations coughrotato (cp) and happyhour, (bb)The homozygous couchpotato (cpcp) phenotype is (sedentary) an extreme lack of activityThe homoz

The above scenario based on Mendelian inheritance. Your answer is attached below.

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You are to discover the chromosomal location (autosomal or X linked) and mode of inheritance for two new fly mutations coughrotato (cp) and happybour, (bb) The homozygous çouchnotato (snan) phenotype is (sedentary) an extreme lack of activity The homozygous happybour (hbhh) can drink any other fly under the table so phenotype (drinker) PARENTS Cross Male double homozygous mutants cp hh with Female wild wild CP HH are to be crossed BUT FIRST you don't know if either gene is on the X chromosome Parental diploid genotypes could be: male çpcphhhh female CPCPHHHH if Both genes autosomal or Male XcpYhhhh female XCPXCPHHHH if cp on the X chromosome and hh autosomal or Male cpcpXhhY female CPGPXHHXHH if hh on the X chromosome and cp autosomal What are the female and male PARENTAL GAMETE GENOTYPES for the above parental crosses if: a) Both genes were autosomal ((REMEMBER BOTH GENDER HAVE ONE GAMETE GENO) b) cp gene on the X chromosome and hh autosomal (REMEMBER MALES WITH HAVE 2 GAMETE GENO)

You are to discover the chromosomal location (autosomal or X linked) and mode of inheritance for two new fly mutations coughrotato (cp) and happybour, (bb) The homozygous çouchnotato (snan) phenotype is (sedentary) an extreme lack of activity The homozygous happybour (hbhh) can drink any other fly under the table so phenotype (drinker) PARENTS Cross Male double homozygous mutants cp hh with Female wild wild CP HH are to be crossed BUT FIRST you don't know if either gene is on the X chromosome Parental diploid genotypes could be: male çpcphhhh female CPCPHHHH if Both genes autosomal or Male XcpYhhhh female XCPXCPHHHH if cp on the X chromosome and hh autosomal or Male cpcpXhhY female CPGPXHHXHH if hh on the X chromosome and cp autosomal What are the female and male PARENTAL GAMETE GENOTYPES for the above parental crosses if: a) Both genes were autosomal ((REMEMBER BOTH GENDER HAVE ONE GAMETE GENO) b) cp gene on the X chromosome and hh autosomal (REMEMBER MALES WITH HAVE 2 GAMETE GENO)

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter7: Development And Sex Determination

Section: Chapter Questions

Problem 12QP: Mutations Can Uncouple chromosomal Sex from Phenotypic Sex Discuss whether the following individuals...

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Part 1= You are to discover the chromosomal location (autosomal or X linked) and mode of inheritance for two new fly mutations couchpotato (cp) and happyhour (hh) The homozygous couchpotato (cpcp) phenotype is (sedentary) an extreme lack of activity The homozygous happyhour (hhhh) can drink any other fly under the table so phenotype (drinker) PARENTS Cross Male double homozygous mutants cp hh with Female wild wild CP HH are to be crossed BUT FIRST you don’t know if either gene is on the X chromosome Parental diploid genotypes could be: male cpcphhhh female CPCPHHHH if Both genes autosomal or Male XcpYhhhh female XCPXCPHHHH if cp on the X chromosome and hh autosomal or Male cpcpXhhY female CPCPXHHXHH if hh on the X chromosome and cp autosomal What are the female and male PARENTAL GAMETE GENOTYPES for the above parental crosses if: a) Both genes were autosomal ((REMEMBER BOTH GENDER HAVE ONE GAMETE GENO) b) cp gene on the X chromosome and hh autosomal…
Unaffected father Camier mother XY Unaffected Afected Carrier Unaffeded Unaffected daugkter U.S. National Lbrany of Mediche Carrier Affected son daughter son In humans, as well as in many other animals and some plants, the sex of the individual is determined by sex chromosomes. The sex chromosomes are one pair of non-homologous chromosomes: XX represents a female, while XY represents a male. When a gene for a specific trait is attached to the X or Y chromosome, we say it is sex-linked, and when it is attached to the X chromosome, we say it is X-linked. Alleles for these linked traits, such as hemophilia or color blindness, crosses, may be recessive or dominant. Hemophilia is an X-linked, recessive trait. The recessive allele for hemophilia is actually a mutated version of the normal alllele but it can still be passed on through generations. Imagine a female is a carrier for hemophilia; her genotype is Xx She is married to a man who does not have hemophilia. What conclusion is NOT valid…
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?
Mutations Can Uncouple chromosomal Sex from Phenotypic Sex Discuss whether the following individuals (1) have male or female gonads, (2) are phenotypically male or female (discuss Wolffian/Mllerian ducts and external genitalia), and (3) are sterile or fertile. a. XY, homozygous for a recessive mutation in the testosterone biosynthetic pathway, producing no testosterone b. XX, heterozygous for a dominant mutation in the testosterone biosynthetic pathway, which causes continuous production of testosterone c. XY, heterozygous for a recessive mutation in the MIH gene d. XY, homozygous fora recessive mutation in the SRY gene that abolishes function
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?
Analysis of X-Linked Dominant and Recessive Traits Suppose a couple, both phenotypically normal, have two children: one unaffected daughter and one son affected with a genetic disorder. The phenotype ratio is 1:1, making it difficult to determine whether the trait is autosomal or X-linked. With your knowledge of genetics, what are the genotypes of the parents and children in the autosomal case? In the X-linked case?
A women is heterozygous, Aa, for classic albinism and this gene maps to chromosome 11, and is a carrier for haemophilia, an X-linked trait (H/h). Provide a diagram of chromosomes 11 and the sexchromosomes (show the alleles associated with albinism and haemophilia) in the cells involved with normal oogenesis for this female. Start with the oogonium and end with the mature ovum; you can ignore all polar bodies. Your diagrams should show the chromatids clearly and all cells must be labelled. Indicatethe total number of chromosomes and chromatids for each cell stage.
G1 phase cat cells will have 19 pairs of chromosomes. In a cat cell in metaphase of mitosis .. A. How many sister chromatids would the cell have? B. How many centromeres would the cell have? C. How many centrosomes would the cell have? Explain what happens in the following stages S phase Prophase Anaphase Cytokinesis What is a checkpoint? Give an example and say what happens at that checkpoint Put the images in the correct order in the cell cycle and name the stages. For the order part, just use the numbers: if the correct order is 1 2 3 4 5, write in that order. 4 3 .. G 0000 5 2 Maphase plait Comono Controles Apter
r blindness is recessive, x-linked trait. Ia marriages between a man who is colorblind and a woman a carrier what percentage of the children can be expected to be sons who are not color blind? What entage of the children can be expected to be daughters who are carriers? What percentage of the children can be expected to be daughters who are not carriers at all? 42 Se
A mutation occurs in the bicoid gene in a female fly. What will result if her eggs are fertilized? 1.Her embryos will be normal if they are fertilized by a male fly that contributed a wild type allele for bicoid 2.Her embryos will be normal if the male that fertilizes her eggs also has that same bicoid mutation 3.All of her embryos will be disrupted irrespective of the genotype of the father 4.Her embryos will develop normally but the adults will be sterile
Explain the sex-linked traits. (S9LT-ld-29.4.4) Solve problems related to sex-linked traits. (S9LT-ld-29.4.4) WHAT I NEED (MATERIALS) • Worksheet • Pens WHAT TO DO (PROCEDURE) PART A Directions: Tell whether the following carry or show the sex-linke recessive spongy-tiny pores trait. Also, tell whether the individuals a male or female. Remember that since most sex-linked traits are recessive, the person w shows the trait can have no X's with big superscripted letters. So, if a m carries the trait, he must also show the trait. XDY XDXD XdY Carrier? (Yes or No) Has the Trait? (Yes or No) Male or Female Phenotype or Appearance of the Individual ided Questions: 1. What chromosomes will result to a normal phenotype (normal spongy po trait12