P +/+· +/+· +/+ x _p/p· v/v · b/b Gametes +.+·+ p · v · b a. Determine which genes are linked. b. Draw a map that shows distances in map units. c. Calculate interference, if appropriate. Recombinant for Progeny phenotypes Class F, gametes Numbers p-b p-v v-b gre sen pla pur res bro gre res pla pur sen bro pur res pla gre sen bro gre res bro pur sen pla 1 • +. + 3,210 3,222 1,024 1,044 p • v · +• v. + b R p · + · b p• v +: +· b 4 R 690 R 678 R 7 +• v• b 72 R R 8 p:+· + 60 R R Total 10,000 1,500 2,200 3,436 RRRR

P +/+· +/+· +/+ x _p/p· v/v · b/b Gametes +.+·+ p · v · b a. Determine which genes are linked. b. Draw a map that shows distances in map units. c. Calculate interference, if appropriate. Recombinant for Progeny phenotypes Class F, gametes Numbers p-b p-v v-b gre sen pla pur res bro gre res pla pur sen bro pur res pla gre sen bro gre res bro pur sen pla 1 • +. + 3,210 3,222 1,024 1,044 p • v · +• v. + b R p · + · b p• v +: +· b 4 R 690 R 678 R 7 +• v• b 72 R R 8 p:+· + 60 R R Total 10,000 1,500 2,200 3,436 RRRR

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 16QP: Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior...

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The Figueroa family has a genetically inherited trait called ocular albinism—lack of pigment in the eye—which is caused by a 516-bp deletion in an exon of a single gene. DNA for this gene was amplified from each member of the family and run on the gel shown below. The thickness of the band indicates the relative amount of DNA. Use the pedigree and the gel to determine the most likely mode of inheritance for this disease. a) x-linked dominant b)x-linked recessive c)autosomal dominant d) autosomal recessive e) cant be determined
A RFLP is discovered that is linked to the gene for Duchenne’s muscular dystrophy (DMD). DMD is an X-linked, recessive trait. The RFLP is 2 map units from the gene for DMD. Consider the following pedigree and Southern blot using a probe that hybridizes to the RFLP. Which band/s is/are associated with DMD? What is the genotype for individuals 3 and 4? (Remember, this is an X linked disease, so use X’s and Y’s to denote). Individual 9 married a man who does NOT have muscular dystrophy, and she is pregnant. DMD is an X-linked trait. What is the probability for their child to have DMD? An amniocentesis is performed and it is determined that 9’s child in utero has only a 10 kb band that hybridizes to the same probe used above. What can you say about the child now?
A. Look at the pedigree, and DISREGARD individual II-8 for the moment. Is the pattern of inheritance of Unetan syndrome dominant or recessive? You may assume that the gene is FULLY-PENETRANT in this family. Please give two specific reasons that support your conclusion. B. Now, looking at BOTH the pedigree AND at the Southern blot, is this trait autosomal, X-linked, or Y-linked? Please give two specific reasons that support your conclusion. Once again, disregard II-8 for the moment. One of your two reasons must refer specifically to evidence present in the Southern blot. C. Define the gene alleles associated with Unetan syndrome. Your alleles MUST be consistent with the pattern of inheritance, AND your genetic notation must be consistent with that used throughout the course. Unetan syndrome allele: ________ Normal allele: ________
A genetic disorder is caused by a LOF mutation (Bm) and epigenetic imprinting of gene B. Through pedigree analysis of many families, researchers have observed the following results: Female carrier B+Bm x B+B+ --> B+Bm and Mother's genotype Father's genotype Children's genotype Children's phenotype Cross 1: B+Bm B+B+ B+Bm , B+B+ some affected, some unaffected Cross 2: B+B+ B+Bm B+Bm , B+B+ always unaffected Based on these results, gene B is imprinted on the: a. maternal b. paternal
When expressing X-linked recessive gene notation, how should the notation be written. Should I use XaXa for the recesive and Xa+Xa for heterozygotes? Is there a better/more correct way to write this?
A normal chromosome A B C D * E F G H I J K has mutated toA B C F E * D G H I J K J K.This is an example of a _________________ chromosomal mutation. ( * denotes the centromere) Group of answer choices Pericentric inversion of DEF and tandem duplication of JK. Pericentric inversion of DEF and displaced duplication of JK Paracentic inversion of DEF and displaced duplication of JK. Paracentric inversion of DEF and tandem duplication of JK.
Using the data below please identify the following: Identify which of the siblings in generation II has a recombinant chromosome. Determine which of his/her children is at risk of developing adenomatous polyposis based on the new data. Table 2 RFLP data for a family suffering from APC syndrome. III1 III2 III3 II1 II2 III4 III5 III6 I1 I2 II3 II4 III7 II5 II6 III8 III9 III10 II7 II8 III11 III12 III13 II9 II10 A A A A A A A A A A A A B B B B B B B B B B B B B B B B B B B C C C C C C C C D D D D D D D
In Neurospora, the mutant stp exhibits erratic stop-andstart growth. The mutant site is known to be in the mtDNA. If an stp strain is used as the female parent in a crosswith a normal strain acting as the male, what type ofprogeny can be expected? What about the progeny fromthe reciprocal cross?
In a diploid plant species, an F1 with the genotype Gg Ll Tt is test-crossed to a pure-breeding recessive plant with the genotype gg ll tt. The offspring genotypes are listed in the table. Genotype Number Gg Ll Tt 621 Gg Ll tt 3 Gg ll Tt 64 Gg ll tt 109 gg Ll Tt 103 gg Ll tt 67 gg ll Tt 7 gg ll tt 626 1600 Calculate the recombination frequency between G and T pair of genes. A. 0.227 B. 0.139 C. 0.454 D. 0.233
PLEASE ANSWER THE FOLLOWING LETTERS: a,b,c, and d Examine the pedigree of the McGraw family shown below. Certain individuals in this family are affected by a brain condition that makes them more susceptible to vertigo. As a genetic counselor, you interview the family and draw DNA samples. You discover that the condition is caused by a mutation that changes the sequence 5’GCATTC3’ to 5’GAATTC3’ introducing an EcoRI cut site. You decide to amplify a 1200bp fragment from the DNA that spans this mutation and then digest it with EcoRI. You run the results on a gel next to a marker that shows bands at 2000bp, 1200bp, 900bp, 800bp, and 400bp. Some individuals from the pedigree are identified on the gel.
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?