what is the linkage between the ct and s genes? A.15.7 cM B.32.6 cM C.21.3 cM D.43.3 cM In Drosophila melanogaster, cut wings (ct) is recessive to normal wings (ct*), sable body (s) is recessive to gray body (s*), and vermilion eyes (v) isrecessive to red eyes (v*). The following are the progeny resulting from the testcross.v ct s510v* ct s1ct*14of* s*500cts*73ct20ct* s811Total1200what is the linkage between the ct and s genes?O A. 15.7 cMO B. 32.6 cMO C. 21.3 cMD. 43.3 cM

Genes that are sufficiently close together on a chromosome will tend to "stick together," and the…

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what is the linkage between the ct and s genes? A. 15.7 cM B. 32.6 cM C. 21.3 cM D. 43.3 cM

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 2ACQ: Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are...

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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?
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 recessive alleles k (kidney-shaped eyes instead ofwild-type round), c (cardinal-colored eyes instead of wildtype red), and e (ebony body instead of wild-type gray)identify three genes on chromosome 3 of Drosophila.Females with kidney-shaped, cardinal-colored eyes weremated with ebony males. The F1 was wild type. When F1females were testcrossed with kk cc ee males, the following progeny phenotypes were obtained:k c e 3k c + 876k + e 67k + + 49+ c e 44+ c + 58+ + e 899+ + + 4Total 2000a. Determine the order of the genes and the mapdistances between them.b. Draw the chromosomes of the parents and the F1.c. Calculate interference and say what you think of itssignificance
Here are the progeny of this cross: (Note that the categories are not in any particular order.)Fly type # of prog. Phenotype symbols Categorywt eyes black body wt wings 97 grn+ blk crv+Green eyes black body curved wings 709 ParentalGreen eyes wt body wt wings 9Green eyes black body wt wings 162wt eyes wt body wt wings 727wt eyes black body curved wings 12wt eyes wt body curved wings 179Green eyes wt body curved wings 105Total = 2000 9.) Write the phenotype symbols in the right-hand column. The first one has been done for you.10.) Next to that, label all fly categories as parental (NCOs), SCOs, and DCOs. One has been donefor you.11.) After each SCO/DCO label, write which gene got “unlinked” in these offspring.12.) Put these three genes into a genetic map in the proper order.13.) Calculate the genetic distance between the genes and label the map with these distances.14.) Calculate the cross-over interference15.) Return to questions #1-6 above. For question 6, you gave your opinion, but…
The following is a linkage map of chromosome 5 for three genes in tomato: (see image) The cross between the triple heterozygote (Lf J W/ lf j w) and a triple homozygous recessive produced 500 progeny. Assume that there is no interference in the Lf-W region. Give the expected number of individuals for each of the following progeny types and show complete solutions.a. with crossover in the Lf-J and J-W regionsb. with crossover in the Lf-J regionc. with crossover in the J-W regiond. without crossover in the Lf-W region
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?
Three recessive mutations in Drosophila melanogaster, roughoid (ru, small rough eyes), javelin (jv, cylindrical bristles), and sepia eyes (se, dark brown eyes) are linked. A three-point cross was carried out and the following progeny obtained: jv+ ru+ se+ 37 jv+ ru+ se 2 jv+ ru se 14 jv+ ru se+ 146 jv ru se+ 2 jv ru se 35 jv ru+ se 154 jv ru+ se+ 10 a. Determine the order of the genes on the chromosome. b. Determine which progeny contain single crossovers and which contain double crossovers and indicate where among the genes the crossovers occurred. c. Calculate the map distances among the genes. d. Calculate the coefficient of coincidence and interference among the genes
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
Three fruit shapes are recognized in the summer squash (Cucurbita pepo): disk-shaped, elongated, and sphere shaped. A pure disk-shaped variety was crossed to a pure elongated variety. The F1 were all disk-shaped. Among 80 F2 there were 30 sphere-shaped, 5 elongated, and 45 disk-shaped. Keeping in mind this gene interaction, what phenotypic proportions are expected in the progeny of the sphere-shaped in F2 when crossed at random?
In Drosophila, a cross was made between females—all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)—and wild-type males. In the F1, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F2 generation, and 1000 offspring were counted, with the results shown in the following table. Phenotype Offspring sc s v 314 + + + 280 + s v 150 sc + + 156 sc + v 46 + s + 30 sc s + 10 + + v 14 No determination of sex was made in the data. (a) Using proper nomenclature, determine the genotypes of the P1 and F1 parents. (b) Determine the sequence of the three genes and the map distances between them. (c) Are there more or fewer double crossovers than expected? (d) Calculate the coefficient of coincidence. Does it represent positive or negative interference?
Two different female Drosophila were isolated, each heterozygous for the autosomally linked genes black body (b), dachs tarsus (d), and curved wings (c). These genes are in the order d–b–c, with b closer to d than to c. Shown in the following table is the genotypic arrangement for each female, along with the various gametes formed by both. Identify which categories are noncrossovers (NCO), single crossovers (SCO), and double crossovers (DCO) in each case. Then, indicate the relative frequency with which each will be produced.