A cross is made between a heterozygote, +++/abc, and a recessive homozygote, abc/abc. Analysis of the progeny gave the following results: + 450 + + C 10 a + c 70 + b c 210 + b 65 a + + 200 abc 460 ab+ 15 What is the map distance between the a and c genes? O 25.12 map units O 27.71 map units 29.39 map units O 32.53 map units None of the above

A cross is made between a heterozygote, +++/abc, and a recessive homozygote, abc/abc. Analysis of the progeny gave the following results: + 450 + + C 10 a + c 70 + b c 210 + b 65 a + + 200 abc 460 ab+ 15 What is the map distance between the a and c genes? O 25.12 map units O 27.71 map units 29.39 map units O 32.53 map units None of the above

Name: A cross is made between a heterozygote, +++/abc, and a recessive homo
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Description: A cross is made between a heterozygote, +++/abc, and a recessive homozygote, abc/abc. Analysis of the progeny gave the following results:+ + +450+ + C10a + c70+ b c210+ b +65a + +200a bc460ab +15What is the map distance between the a and c genes?O 2

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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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
Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares would you need to complete a Punnett square analysis of this cross?
The text outlines some of the problems Frederick William I encountered in his attempt to breed tall Potsdam Guards. a. Why were the results he obtained so different from those obtained by Mendel with short and tall pea plants? b. Why were most of the children shorter than their tall parents?
In Section 12.3, ''Laws of Inheritance," an example of epistasis was given for the summer squash. Cross white WAvYy heterozygotes to prove the phenotypic ratio of 12 white:3 yellow:l green that was given in the text.
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?
E. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescentwhite seedlings, and 260 yellow seedlings (E. W. Lindstrom. 1921. Genetics 6:91–110). a. Give the genotypes for the green, virescent-white, and yellow progeny. b. Explain how color is determined in these seedlings. c. Is there epistasis among the genes that determine color in the corn seedlings? If so, which gene is epistatic and which is hypostatic?
In a testcross of a corn plant heterozygous for 3 linked, recessive genes (a, b, & d), the following phenotypes & numbers of progeny were obtained (total progeny is 1000). Choose the map that best fits the data. A) A --2.7-- D --4.2-- B B) A --2.4-- D --2-- B C) A --4.2-- B --2.7-- D D) A --4.2-- D --2.7--B E) A --2-- B --2.4-- D F) A --2.7-- B --4.2-- D
EC2. Here is a tetrad produced by mating a H Y strain to an h y strain. a) What is the tetrad type? b) What has recombined with what? EC3. In corn, a dihybrid for the recessive genes a and b is test-crossed. The distribution of the phenotypes is as follows: A B 122A b 118a B 81a b 79 a) Do the genes appear to be sorting independently? Look at map units here. b) Test your hypothesis with a chi-squared test.
An individual heterozygous for four genes, A/a • B/b •C/c • D/d, is testcrossed with a/a • b/b • c/c • d/d, and 1000progeny are classified by the gametic contribution ofthe heterozygous parent as follows:a • B • C • D 42A • b • c • d 43A • B • C • d 140a • b • c • D 145a • B • c • D 6A • b • C • d 9A • B • c • d 305a • b • C • D 310a. Which genes are linked?b. If two pure-breeding lines had been crossed toproduce the heterozygous individual, what would theirgenotypes have been?c. Draw a linkage map of the linked genes, showing theorder and the distances in map units.d. Calculate an interference value, if appropriate
. In common wheat, Triticum aestivum, kernel color is determined by multiply duplicated genes, each with an Rand an r allele. Any number of R alleles will give red, anda complete lack of R alleles will give the white phenotype.In one cross between a red pure line and a white pureline, the F2 was 6463 red and 641 white.a. How many R genes are segregating in this system?b. Show the genotypes of the parents, the F1, and the F2.c. Different F2 plants are backcrossed with the whiteparent. Give examples of genotypes that would give thefollowing progeny ratios in such backcrosses: (1) 1 red :1 white, (2) 3 red :1 white, (3) 7 red :1 white.d. What is the formula that generally relates thenumber of segregating genes to the proportion of redindividuals in the F2 in such systems?