Question:- 3. Wild type Drosophila has red eyes and gray body*. Autosomal recessive mutations sepia (se) and ebony (e) modify the eye and body color, respectively. What kind of a cross (or crosses) can you set up to determine whether the two genes are linked? You have several lines to choose from: a pure-breeding wild type line (where alleles for both, sepia and ebony are wild type), a double homozygous recessive line, a pure breeding sepia mutant line and a pure breeding ebony mutant line. You don’t have to use all of them. (a) Write down the cross (or crosses, if necessary) which will allow you to determine linkage; include genotypes of both crossed individuals (b) What kind of results (offspring) do you expect if the two genes are linked? (c) What kind of results (offspring) do you expect if the two genes are not linked? For (b) and (c), state the phenotypes, genotypes and numbers.

Question:- 3. Wild type Drosophila has red eyes and gray body*. Autosomal recessive mutations sepia (se) and ebony (e) modify the eye and body color, respectively. What kind of a cross (or crosses) can you set up to determine whether the two genes are linked? You have several lines to choose from: a pure-breeding wild type line (where alleles for both, sepia and ebony are wild type), a double homozygous recessive line, a pure breeding sepia mutant line and a pure breeding ebony mutant line. You don’t have to use all of them. (a) Write down the cross (or crosses, if necessary) which will allow you to determine linkage; include genotypes of both crossed individuals (b) What kind of results (offspring) do you expect if the two genes are linked? (c) What kind of results (offspring) do you expect if the two genes are not linked? For (b) and (c), state the phenotypes, genotypes and numbers.

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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QUESTION:- In Drosophila, assume that the gene for scute bristles (s) is located at map position 0.0 and that the gene for ruby eyes (r) is at position 25.0. Both genes are located on the X chromosome and are recessive to their wild-type alleles. A cross is made between scute-bristled females and ruby-eyed males. Phenotypically wild-type F1 females were then mated to homozygous double mutant males, and 1000 offspring were produced. Give the phenotypes and frequencies expected..
Question:- 1. Why would multi-gene families complicate things in terms of being sure of which member of the family was responsible for a particular phenotype?
Question 3 ) Listen Becker muscular dystrophy is due to a mutation in the same gene that causes the more severe Duchenne muscular dystrophy. This is an X linked recessive condition that causes muscular weakness that begins and a later age and progresses more slowly than Duchenne muscular dystrophy which is often lethal by age 20. One of my former students had a father who had this condition. She did not have the condition. If she has children with a male who does not have the condition, what is the chance that any daughters she would have would have the condition? 100 % 75% O 50% O 25% 0%
Drosophila (fruit-flies) mutants A and B have the same phenotype: the absence of red pigment in the eyes. Homozygous mutants were crossed, and all offspring possessed red eyes. Does the appearance of red eyes in the progeny indicate complementation or a failure to complement? [Select] Do the mutants likely have mutations in the same or different genes? [ Select]
Question 5 [HINT: This problem wants you to calculate gene number from the proportion of individuals in the extreme classes, not from the number of phenotypes.] A farmer has two pure-breeding varieties of tomatoes: Little Pete (average fruit diameter 2cm) and Big Boy (average fruit diameter 14 cm). The farmer crosses Little Pete and Big Boy and obtains F1 plants with fruit averaging 8cm in diameter. When he intercrosses the F1s to obtain F2s, he examines 2000 F2 plants and finds two plants that produce fruit even smaller than Little Pete and three plants that produce fruit even larger than Big Boy. From these data, how many genes can we infer are involved in controlling fruit size in these plants? What is the genotype of Little Pete? What is the genotype of Big Boy? What is the genotype of the F1s? What are the genotypes that are bigger than Big Boy? What are the genotypes that are smaller than Little Pete?
Question:- Based on your selected mode of inheritance, show the genotypes for the following individuals. [Use these symbols for alleles: if it is autosomal, then use the symbols B - dominant, b - recessive (e.g. BB, bb etc.) if it is X-Linked, then X(B) - dominant, X(b) - recessive, and Y for Y-chromosome (e.g. X(B)X(B), X(B)Y etc.) ] I-1 I-2 II-7 II-8 III-10 III-11 III-12 IV-8 IV-9
Topic: Penetrance. Petal number is controlled by a single gene in merigonias. The gene has a completely dominant wild type allele F that makes a plant have five petals and a mutant recessive six petal allele(f). However the six petal trait is only 50% penetrant. You do the cross Ff x Ff. What fraction of the progeny do you have the 6 petals? what is the meaning for 50% penetrant.
Question:- 8. Describe how Morgan’s experiment with the white-eyed mutant fly provided evidence for the chromosomal basis for inheritance. Be sure to include the significance of the x-linked gene.
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?
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?
Question: Black hair of the mouse is produced by dominant gene B and white hair by recessive allele b.Female with black hair crossed with male black hair (both genotypes are heterozygous). Produced individuals 112 and 113, female with white hair and male with black hair, then mating occur between 114 and 113, and another mating occur between 112 and 111. assume that 111 and 114 individuals do not carry the recessive allele, represents the above in a pedigree chart and determine the genotype of each individual.