Continuing his experiments, in Miles Morales' genome, the Goblin found similar genes, including one for invisibility which also seems to be dominant.To determine if they were linked to agility, the Goblin set up another cross and got the following results:PhenotypeAgility and InvisibilityAgilityNumber of Progeny12152Invisibility55No Powers125Did the heterozygous parent of the test cross start with the alleles in coupling or repulsion?O Unable to determineO RepulsionO Coupling

Name: Continuing his experiments, in Miles Morales' genome, the Goblin foun
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Description: Continuing his experiments, in Miles Morales' genome, the Goblin found similar genes, including one for invisibility which also seems to be dominant.To determine if they were linked to agility, the Goblin set up another cross and got the following r

In the given case, let the allele associated with invisibility be “I” and allele associated with…

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Continuing his experiments, in Miles Morales' genome, the Goblin found similar genes, including one for invisibility which also seems to be dominant. To determine if they were linked to agility, the Goblin set up another cross and got the following results: Phenotype Number of Progeny Agility and Invisibility Agility Invisibility 121 52 55 No Powers 125 Did the heterozygous parent of the test cross start with the alleles in coupling or repulsion? O Unable to determine Repulsion O Coupling

Continuing his experiments, in Miles Morales' genome, the Goblin found similar genes, including one for invisibility which also seems to be dominant. To determine if they were linked to agility, the Goblin set up another cross and got the following results: Phenotype Number of Progeny Agility and Invisibility Agility Invisibility 121 52 55 No Powers 125 Did the heterozygous parent of the test cross start with the alleles in coupling or repulsion? O Unable to determine Repulsion O Coupling

Concepts of Biology

1st Edition

ISBN:9781938168116

Author:Samantha Fowler, Rebecca Roush, James Wise

Publisher:Samantha Fowler, Rebecca Roush, James Wise

Chapter8: Patterns Of Inheritance

Section: Chapter Questions

Problem 5RQ: Imagine that you are performing a cross involving seed texture in garden pea plants. You cross...

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In sweet pea plant, an allele for purple flowers (P) is dominant when paired with a recessive allele for red flowers (p). An allele for long pollen grains (L) is dominant when paired with a recessive allele for round pollen grain (l). Bateson and Punnett crossed a plant having purple flowers/long pollen grains with one having white/flowers/round pollen grains. All F1 offspring had purple flowers and long pollen grains. Among the F2 generation, the researchers observed the following phenotypes: 296 purple flowers/long pollen grains 19 purple flowers/round pollen grains 27 red flowers/long pollen grains 85 red flowers/round pollen grains What is the best explanation for these results?
More Crosses with Pea Plants: The Principle of Independent Assortment Two traits are examined simultaneously in a cross of two pure-breeding pea-plant varieties. Pod shape can be either swollen or pinched. Pea color can be either green or yellow. A plant with the traits swollen and green is crossed with a plant with the traits pinched and yellow, and a resulting F1 plant is self-crossed. A total of 640 F2 progeny are phenotypically categorized as follows: 360 swollen yellow 120 swollen green 120 pinched yellow 40 pinched green a. What is the phenotypic ratio observed for pod shape? Pea color? b. What is the phenotypic ratio observed for both traits considered together? c. What is the dominance relationship for pod shape? Pea color? d. Deduce the genotypes of the P1 and F1 generations.
More Crosses with Pea Plants: The Principle of Independent Assortment Consider the following cross: P1: AABBCCDDEE aabbccddee F1: AaBbCcDdEe (self-cross to get F2) What is the chance of getting an AaBBccDdee individual in the F2 generation?
Imagine that you are performing a cross involving seed texture in garden pea plants. You cross true-breeding round and wrinkled parents to obtain F1 offspring. Which of the following experimental results in terms of numbers of plants are closest to what you expect in the F2 progeny? a. 8lOroundseeds b. 8lOwrinkledseeds c. 405:395 round seeds:wrinkled seeds d. 610:190 round seeds:wrinkled seeds
Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?
As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progeny
A heterozygous individual has a _______ for a trait being studied. a. pair of identical alleles b. pair of nonidentical alleles c. haploid condition, in genetic terms
Mendel performs a cross using a true-breeding pea plant with round, yellow seeds and a true- breeding pea plant with green, wrinkled seeds. What is the probability that offspring will have green, round seeds? Calculate the probability for the F1 and F2 generations.
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?
Figure 12.16 In pea plants, purple flowers (P) are dominant to white flowers (p) and yellow peas (Y) are dominant to green peas (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares do you need to do a Punnett square analysis of this cross?