In Drosophila melanogaster, white eyes (w) is an X-linked recessive mutation. Normal eyes are red (W). At the autosomal wing shape locus, normal shaped wings (Vg) are dominant to vestigial wings (vg). The following cross was made: XWXW Vgvg x XWY Vgvg What proportion of the progeny from this cross will be males with white eyes and normal wings? Write your answer as a numerical value rounded properly to 4 decimal digits. Do not write the answer as a fraction or percentage.

In Drosophila melanogaster, white eyes (w) is an X-linked recessive mutation. Normal eyes are red (W). At the autosomal wing shape locus, normal shaped wings (Vg) are dominant to vestigial wings (vg). The following cross was made: XWXW Vgvg x XWY Vgvg What proportion of the progeny from this cross will be males with white eyes and normal wings? Write your answer as a numerical value rounded properly to 4 decimal digits. Do not write the answer as a fraction or percentage.

Name: In Drosophila melanogaster, white eyes (w) is an X-linked recessive m
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Description: In Drosophila melanogaster, white eyes (w) is an X-linked recessive mutation.Normal eyes are red (W). At the autosomal wing shape locus, normal shapedwings (Vg) are dominant to vestigial wings (vg). The following cross was made:XWXW Vgvg x XWY VgvgW

Human Biology (MindTap Course List)

11th Edition

ISBN:9781305112100

Author:Cecie Starr, Beverly McMillan

Publisher:Cecie Starr, Beverly McMillan

Chapter20: Chromosomes And Human Genetics

Section: Chapter Questions

Problem 7CT: The following pedigree shows the pattern of inheritance of red-green color blindness in a family....

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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 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?
In Drosophila melanogaster, white eyes (w) is an X-linked recessive mutation. Normal eyes are red (W). At the autosomal wing shape locus, normal shaped wings (Vg) are dominant to vestigial wings (vg). The following cross was made: X"XW Vgvg x XWY Vgvg What proportion of the progeny from this cross will be males with red eyes and normal wings? Write your answer as a numerical value rounded properly to 4 decimal digits. Do not write the answer as a fraction or percentage. Answer:
In Drosophila melanogaster, red eyes are dominant over white and the variation for this characteristic is on the X chromosome. Vestigial wings (v) are recessive to normal (V) for an autosomal gene. Predict the appearance of offspring of the following crosses: XW/Xw V/v×Xw/Y v/v, Xw/Xw V/v × XW/Y V/v.
A mutant sex-linked trait called “notched” (N) is deadly in Drosophila when homozygous in females. Males who have a single N allele will also die. The heterozygous condition (Nn) causes small notches on the wing. The normal condition in both male and females is represented by the allele n. a) Indicate the phenotypes of the F1 generation from the following cross: XNXn x XnY b) Explain why dead females are never found in the F1 generation no matter which parents are crossed. c) Explain why the mating of female XNXn and a male XNy is unlikely.
You have a Drosophila line that is homozygous for autosomal recessive alleles a, b, and c, linked in that order. Youcross females of this line with males homozygous for thecorresponding wild-type alleles. You then cross the F1 heterozygous males with their heterozygous sisters. You obtain the following F2 phenotypes (where letters denoterecessive phenotypes and pluses denote wild-type phenotypes): 1364 + + +, 365 a b c, 87 a b +, 84 + + c,47 a + +, 44 + b c, 5 a + c, and 4 + b +.a. What is the recombinant frequency between a andb? Between b and c? (Remember, there is no crossingover in Drosophila males.)b. What is the coefficient of coincidence?
Another cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w) and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The F1 females were wild type for all three traits, while the F1 males expressed the yellow-body and white eyes traits. The cross was carried to an F2 progeny and only male offspring were tallied. Based on the data shown here, a genetic map was constructed. a) Diagram the genotypes of the F1 parents. b) Construct a map, assuming the white is at locus 1.5 on the X-chromosome. Phenotype Male offspring y + ct 9 + w + 6 y w ct 90 + + + 95 + + ct 424 y w + 376 y + + 0 + w ct 0
In Drosophila, the white gene located on the X chromosome affects eye color; an autosomal gene, wingless, is on an autosomal chromosome. Use the following allele symbols: Xw+ _ , Xw+Y = wild type red eyes; X-linked dominant allele Xw Xw , XwY = white eyes; X-linked recessive allele Y = Y sex chromosome vg+ = wild type wings; autosomal dominant vg = wingless; autosomal recessive Predict ratios/proportions of genotypes and phenotypes of offspring from the following cross, of a white-eyed male with wild type wings and a wild type red eyed female with wild type wings: indicate sex of offspring along with phenotypes. XwY vg+ vg x Xw+Xw vg+vg
The phenotype of vestigial (short) wings (vg) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive gene for hairy (h) body. Assume that a cross is made between a fly that is homozygous for normal wings and has a hairy body and a fly with vestigial wings that is homozygous for normal body. The wild-type F1 flies were crossed among each other to produce 1024 F2 offspring. Which phenotypes would you expect among the F2 offspring, and how many of each phenotype would you expect? Group of answer choices 192 wild type, 256 vestigial, 64 hairy, and 192 vestigial and hairy All vestigial and hairy. 576 wild type, 192 vestigial, 192 hairy, and 64 vestigial and hairy All wild type 256 wild type; 256 vestigial, 256 hairy, and 256 vestigial and hairy
Another cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w) and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The F1 females were wild type for all three traits, while the F1 males expressed the yellow-body and white eyes traits. The cross was carried to an F2 progeny and only male offspring were tallied. Based on the data shown here, a genetic map was constructed. a) Diagram the genotypes of the F1 parents. b) Construct a map, assuming the white is at locus 1.5 on the X-chromosome *******ANSWER PART B NOT PART A!!!! Phenotype Male offspring y + ct 9 + w + 6 y w ct 90 + + + 95 + + ct 424 y w + 376 y + + 0 + w ct 0
Drosophila females of wild-type appearance but heterozygous for three autosomal genes are mated with malesshowing the three corresponding autosomal recessivetraits: glassy eyes, coal-colored bodies, and striped thoraxes. One thousand (1000) progeny of this cross aredistributed in the following phenotypic classes:Wild type 27Striped thorax 11Coal body 484Glassy eyes, coal body 8Glassy eyes, striped thorax 441Glassy eyes, coal body, striped thorax 29a. Draw a genetic map based on these data.b. Show the arrangement of alleles on the two homologous chromosomes in the parent females.c. Normal-appearing males containing the samechromosomes as the parent females in the precedingcross are mated with females showing glassy eyes,coal-colored bodies, and striped thoraxes. Of 1000progeny produced, indicate the numbers of thevarious phenotypic classes you would expect.