Two fruit flies (Drosophila melanogaster) were crossed. The cross was between a homozygous red-eyed, wingless female and a white-eyed male heterozygous for wings. Recall that red eyes (R) is dominant over white eyes (r) and is inherited on the X chromosome (sex-linked) and as such, eye colour alleles should be written as superscript. In addition, the autosomal trait wings (W) is dominant over wingless (w). 1. List the female genotype 2. List the male genotype 3. Construct a Punnett Square for this dihybrid cross on a piece of paper, to determine what the offspring of such a cross would be with respect to sex, eye colour & wings. 4. Using the information from your Punnett Square answer the following: a. What genotypes did you get (list all genotype combinations; if you have more than one of the same kind, you do NOT have to re-list it.) b. What are the phenotypic ratios of the offspring with respect to sex, eye colour and wings?

Two fruit flies (Drosophila melanogaster) were crossed. The cross was between a homozygous red-eyed, wingless female and a white-eyed male heterozygous for wings. Recall that red eyes (R) is dominant over white eyes (r) and is inherited on the X chromosome (sex-linked) and as such, eye colour alleles should be written as superscript. In addition, the autosomal trait wings (W) is dominant over wingless (w). 1. List the female genotype 2. List the male genotype 3. Construct a Punnett Square for this dihybrid cross on a piece of paper, to determine what the offspring of such a cross would be with respect to sex, eye colour & wings. 4. Using the information from your Punnett Square answer the following: a. What genotypes did you get (list all genotype combinations; if you have more than one of the same kind, you do NOT have to re-list it.) b. What are the phenotypic ratios of the offspring with respect to sex, eye colour and wings?

Name: Two fruit flies (Drosophila melanogaster) were crossed. The cross was
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Description: Two fruit flies (Drosophila melanogaster) were crossed. The cross was between a homozygous red-eyed, wingless female and a white-eyed male heterozygous for wings.Recall that red eyes (R) is dominant over white eyes (r) and is inherited on the X chro

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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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?
Bar (B) is a dominant sex-linked mutant of D. melanogaster. Held-out wings and ebony body are recessive, autosomal mutants on chromosome III, mapping 12 map units apart. A Bar-eyed, ebony-bodied male was crossed to a held-out female, and the resulting F1 progeny WERE INBRED to produce the F2. At what frequency do you expect wild-type flies from this cross? (Reminder: Crossing over does not occur in the male). Based on the information in Problem 3, how often would you get a bar ebony but not held-out progeny?
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 gene for wing type is known to be autosomal with vestigial wings (vg) known to the recessive to normal, long wings (+). You are already familiar with the white eye gene (w) from lecture with the normal color being red (+). Answer the following questions assuming an initial cross (P1) between a pure breeding white eyed, long winged female and a pure breeding red eyed vestigial winged male. a) What are the parental genotypes? b) What phenotypes would occur in the F1 generation and in what proportions? c) What would the phenotypes and phenotypic ratios would occur in F2 generation?
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
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?
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
Explain why it is possible for the proband in the following pedigree to have children of blood types A, B, and AB. Considering epistatic genes, what are the possible genotypes of II-2?
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, an X-linked recessive mutation, (s) causes irregular wing margins. Give the genotypes, phenotypes and phenotypic ratio of the F1 and F2 offspring in the following crosses a. scalloped female crossed with a normal male b. scalloped male crossed with a homozygous normal female. Compare these results to those that would be obtained if the scalloped gene is autosomal.
The pedigree below shows the phenotypes of the ABO blood groups and Rhesus factors [positive (+) and negative (-)] for several members of a family. I (B+ AB- 1 2 3 4 II O- A+ В- B- AB+ A+ 1 2 4 5 6 a. What are the ABO blood group genotypes of individuals I-1 and I-2? b. Which child/ren of individual I-4 can donate blood to him? c. Which individual in the pedigree can donate blood to all the other individuals in the pedigree?