a) Aristapedia is an autosomal allele for an antennae shape that is also lethal dominant. Dominant lethal alleles are deadly to the organism when homozygous. When a dominant lethal allele is selected in the Design view of Fly Lab, the designated fly becomes heterozygous for the lethal trait, (a homozygous fly would not complete development.) Lethal traits often deviate from Mendelian ratios. Wild type flies do not carry the allele for aristapedia. Predict what the outcome of a cross between a wild-type fly and one with aristopedia. b) Then Perform the cross and determine if your prediction is correct using statistical analysis. Summarize your results and indicate whether your prediction is confirmed. c) Suggest a reason why a typical punnett square prediction for a heterozygous cross might not be accurate in this case.

a) Aristapedia is an autosomal allele for an antennae shape that is also lethal dominant. Dominant lethal alleles are deadly to the organism when homozygous. When a dominant lethal allele is selected in the Design view of Fly Lab, the designated fly becomes heterozygous for the lethal trait, (a homozygous fly would not complete development.) Lethal traits often deviate from Mendelian ratios. Wild type flies do not carry the allele for aristapedia. Predict what the outcome of a cross between a wild-type fly and one with aristopedia. b) Then Perform the cross and determine if your prediction is correct using statistical analysis. Summarize your results and indicate whether your prediction is confirmed. c) Suggest a reason why a typical punnett square prediction for a heterozygous cross might not be accurate in this case.

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 1CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...

See similar textbooks

Similar questions

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 if the couple wanted prenatal testing so that a normal fetus could be aborted?
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?
Pedigree attached shows an autosomal recessive genetic disease. G is the normal allele and g is the disease-causing allele. Individual 1’s father is heterozygous (*) and his mother is homozygous dominant. Other individuals in the pedigree may be carriers, but are not marked. The question mark (?) indicates that you do not yet know anything about this individual’s phenotype with regard to the disease. part a) What is the probability that individuals 1 and 2 will have a child (5) who is a boy with the disease (the child is unborn and the sex is not yet known)? a)1/8 b)1/4 c)0 d)1/16 part b) What is the probability that the daughter (6) that individual 3 and 4 just had will have the disease? a)1/8 b)1/6 c)1/4 d)1/12
X‑linked, recessive diseases, such as hemophilia, are extremely rare in the population. However, many women are carriers and show no sign of the disease. The pedigree illustrates the inheritance of an X‑linked, recessive disease. Determine whether the unknown individuals are affected by the disease, unaffected by the disease, or carriers of the X‑linked recessive allele. Unaffected individuals are not carriers of the X‑linked recessive allele.
A recessive maternal effect mutant in zebrafish, called ichabod, results in embryos lacking heads that are non-viable. You have been instructed to identify females that are homozygous for the ichabod mutant allele. At your disposal are a tank of wild-type fish (males and females), a tank of male and female parental fish that are all heterozygous for the ichabod mutant allele (ichabod/+), and a tank of F1 fish derived from a cross between a heterozygous male and heterozygous female (ichabod/+). Which of the following would be a way to identify females that are homozygous mutant, i.e. ichabod/ichabod? Select all answers that would work.
A woman who is heterozygous for gene B has brown eyes. B is a dominant allele for brown pigmentation, while b is recessive allele for blue pigmentation. The woman has a patch of blue color in her left eye. Give THREE (3) reasons how this might occur.
Albinism, lack of pigmentation in humans, results from an autosomal recessive gene designated a. Two parents with normal pigmentation have an albino child. What is the probability that their next child will be albino? What is the probability that the next child will be an albino girl? If the child is normal, what is the probability that it will be a carrier (heterozygous) for the albino gene?
None
A recessive maternal effect mutant in zebrafish, called ichabod, results in embryos lacking heads that are non-viable. You have been instructed to identify females that are homozygous for the ichabod mutant allele. At your disposal are a tank of wild-type fish (males and females), a tank of male and female parental fish that are all heterozygous for the ichabod mutant allele (ichabodl+), and a tank of F1 fish derived from a cross between a heterozygous male and heterozygous female (ichabodl+). Which of the following would be a way to identify females that are homozygous mutant, i.e. ichabodlichabod? Select all answers that would work. Cross F1 females to F1 males and observe their offspring. Crosses that produce headless offspring came from a homozygous female. a. Cross F1 males to females from the parental tank and observe their offspring. 25% of these crosses should produce headless offspring. b. Cross F1 females to F1 males to make the F2 generation. Cross F2 females to F2 males and…
In ogres, nose type results from a genetic maternal effect where the autosomal allele for a warty nose (w*) is dominant over the allele for a smooth nose (w). Shrek's third cousin once removed, Shrekina, has a warty nose with a w*w* genotype. She marries Golothm, a male, smooth nosed ogre with a ww genotype and they have a warty nosed daughter, Mary. Mary marries Blutto, a smooth nosed male with a ww genotype to produce a son and a daughter. What would be the expected genotypes and ratio of the F2 ogres (Mary and Blutto's children)? O 1 ww+ : 1 ww O 1 XwXw+ :1 XwXw O 1 ww: 2 w+w: 1 w+w+ O 1 ww+ : 1 w+w+ O 1 XwXw+ :1 YXw+
A standard three-point mapping is conducted for recessive mutations in autosomal genes purple eye (pr), curved wing ( c) and black body (b). Their wild type alleles are also used for genetic mapping. An F1 Drosophila female heterozygous for purple eye (pr), curved wing (c) and black (b) is crossed to a triply homozygous mutant male. The observed numbers and phenotypes of the offspring are as follows: 360 pr c b 380 pr+ c+ b+ 104 pr c+ b 96 pr+ c b+ 30 pr c b+ 20 pr+ c+ b 6 pr c+ b+ 4 pr+ c b PROVIDE THE FOLLOWING: A) State the order of genes on this chromosome. B) Calculate map distances between the gene pairs: pr-c, pr-b, c-b. Show calculations, state the number of map units and which gene pairs they refer to.