1IIIIVLactose tolerant femaleLactose intolerant female3Recessive3Recessive6Lactose tolerant maleLactose intolerant maleDominant89 107 8DominantFigure 2. Pedigree of Family A. (Adapted from a figure in Enattah, N. S., et al.2002 Nature Genetics 30: 233-237.)1. Based on the pedigree above, which of the following terms are appropriate fordescribing the inheritance of the lactose-intolerance trait (filled-in symbols)?Check all that apply.X-linkedFamily A911? Unknown phenotypeDeceasedX-linked2. Which of the following terms are appropriate for describing the inheritanceof the lactose-tolerance trait (empty symbols)? Mark all that apply.AutosomalAutosomalYou decide you need more data to find out if your model for inheritance oflactose intolerance holds up. You examine three additional pedigrees from theFinnish study.

Name: 1IIIIVLactose tolerant femaleLactose intolerant female3Recessive3Rece
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Description: 1IIIIVLactose tolerant femaleLactose intolerant female3Recessive3Recessive6Lactose tolerant maleLactose intolerant maleDominant89 107 8DominantFigure 2. Pedigree of Family A. (Adapted from a figure in Enattah, N. S., et al.2002 Nature Genetics 30: 2

Lactose intolerant people are those who do not contain lactase enzyme in them. As a result lactose…

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Biology

IV ctose tolerant female ctose intolerant female I Recessive Lactose tolerant male Lactose intolerant male Recessive 8 Dominant 7 jure 2. Pedigree of Family A. (Adapted from a figure in Enattah, N. S., et al 02 Nature Genetics 30: 233-237.) Dominant 9 10 ased on the pedigree above, which of the following terms are appropriate scribing the inheritance of the lactose-intolerance trait (filled-in symbols)? eck all that apply. 8 X-linked 9 ? Unknown phenotype Deceased Which of the following terms are appropriate for describing the inheritance he lactose-tolerance trait (empty symbols)? Mark all that apply. X-linked Autosomal Autosomal decide you need more data to find out if your model for inheritance of se intolerance lds up. You examine three additional pedigrees from the sh study.

IV ctose tolerant female ctose intolerant female I Recessive Lactose tolerant male Lactose intolerant male Recessive 8 Dominant 7 jure 2. Pedigree of Family A. (Adapted from a figure in Enattah, N. S., et al 02 Nature Genetics 30: 233-237.) Dominant 9 10 ased on the pedigree above, which of the following terms are appropriate scribing the inheritance of the lactose-intolerance trait (filled-in symbols)? eck all that apply. 8 X-linked 9 ? Unknown phenotype Deceased Which of the following terms are appropriate for describing the inheritance he lactose-tolerance trait (empty symbols)? Mark all that apply. X-linked Autosomal Autosomal decide you need more data to find out if your model for inheritance of se intolerance lds up. You examine three additional pedigrees from the sh study.

Biology: The Unity and Diversity of Life (MindTap Course List)

14th Edition

ISBN:9781305073951

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter13: Observing Patterns In Inherited Traits

Section: Chapter Questions

Problem 11GP: A single allele gives rise to the Hbs form of hemoglobin. Individuals who are homozygous for the...

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14:27 46 lll 33% GENETICS...IGNMENT ASSIGNMENT 1 1. Using the central dogma of molecular biology, explain the terms replication, transcription and translation. 2. A heterozygous fair male had a conjugal interaction with a homozygous dark female. Using an appropriate genetic diagram, explain the genotypic and the phenotypic appearance of the second felial generation of these parents. Note: Darkness is dominant over fairness. III
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?
A single allele gives rise to the Hbs form of hemoglobin. Individuals who are homozygous for the allele (HbS/HbS) develop sickle-cell anemia (Section 9.6). Heterozygous individuals (HbA/HbS) have few symptoms. A couple who are both heterozygous for the HbS allele plan to have children. For each of the pregnancies, state the probability that they will have a child who is: a. homozygous for the HbS allele b. homozygous for the normal allele (HbA) c. heterozygous: HbA/HbS
4. Phenylketunoria (PKU) is an inborn error of metabolism of the amino acid phenylalanine. The following pedigree is of an affected family. I II III 1 2 3 5 a. What is the mode of inheritance of PKU? b. Which persons are known to be heterozygous for PKU? Identify from I, II, and III generations. c. What is the probability that III-2 9 is heterozygous? d. If III-3 and III-4 marry, what is the probability that their first child will have PKU?
5 & :56M ******* 24 DIHYBRID CROSSES DRV 0 Stv T alı A @ zladenA 9160p2-id2 bns obidalbaneoviene da II\ MOD YR 21 $59A ... Create a dihybrid cross and determine the expected phenotypic percentages of the offspring of two corn plants both of which are heterozygous for colour and texture (RrTt X RrTt). Don't forget to include clear let statements, and follow the all six steps taught on solving genetics problems. insig moni nellog: bna. zoom
BB AB AB A. None B.3 с.1 D.2 ВВ OE. The father AB In the pedigree presented above, an autosomal dominant disease which causes significant visual loss and eventual blindness, is segregating in the family. The disease gene causing this sight loss looks as if it may be linked to a marker locus. The alleles of this marker locus that are present in this family are allele A and allele B. Are there any recombinant individuals in this pedigree? AB ВВ AB ВВ AB ВВ AB AB ВВ
1. The pedigree below shows the incidence of rare, autosomal dominant disorder called Ehlers-Danlos disease. The pedigree covers three generations of a particular family and also shows individual genotypes at a potential marker locus (M). a) Indicate the phase of all gen II and III individuals. DdM1M3 ddM2M6 II DDM3M6 ddM4M5 III DdMзM4 DdMЗМ5 DDM3M4 ddM3M5 DDM3M4 ddM5M6 DDM3M4 ddM4M6 ddM5M6 ddM5M6 b) Which, if any, of the gen III individuals are recombinants? c) Calculate the LOD score as a test of physical linkage between the marker (M) and the disease locus. d) What do you conclude about linkage between D and M?
31 1. In guinea pig, fur color is determined by one gene, and fur length by another independently segregating gene. The black fur allele (B) is dominant to brown fur allele (b) and the short allele (S) is dominant to long fur allele (s). From progeny listed in the following table, provide the probable genotype for the parents of each cross. Phenotype of Progeny Black short black long brown short Bio3110 Genetics Additional problems (Chapter 2): Parents Black short X Black short Black short X Brown long Black long X Black long 90 17 0 32 16 35 29 14 0 brown long 8 15 2. A mouse with the genotype BbccEeFf is mated with another mouse with BBCcEEFF genotype. What is the probability of obtaining the following genotype in the progeny? a. BBCcEeFF b. BbccEEFf C. BbccEeFF 12 3. Determine the type of gametes that are formed from the following individuals with the given genotype? A. AaBbCc B. AABBCc C. AABbCc
FAlpQLSfiOhfAvlhxzCSiUll_6rt-nU5b0WI73UmWOxkOw8OCwk01ng/formResponse B 1 2 Bb x Bb b 4 The fur in both parents in this cross is * 1 B B Bb x Bb b 3 4 brown black O homozygous dominant homozygous recessive 3. 近