Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 5, Problem 34P
The accompanying pedigree below shows a family in which an autosomal recessive disorder is presents. Family members
Excluding
What is the genotype of each family member, including
What are the synthetic disease gene and VNTR alleles in
What is the chance
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Students have asked these similar questions
Here is a pedigree from a family with a
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The letters A-E are haplotypes (alleles).
dx51
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dx 44
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dx 48 42 dx38
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d.) Are there any individuals in the
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identify the people (using the standard
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pedigree) and suggest an explanation.
The pedigree below shows that inheritance of a disease that is caused by a late onset, dominant, autosomal mutation that is rare, but only 50% penetrant. The gene that is mutated in the disease is linked at a distance of 10 cm to a microsatellite marker that has alleles numbered 1, 2, and 3. The marker alleles detected in each individual are indicated below.
What is the probability that individual A will develop the disease? Explain using an illustration of this occurs.
Brachydactyly type D is human autosomal dominant condition in which the thumbs are abnormally short and broad. In most cases, both thumbs are affected, but occasionally just one thumb is involved. The pedigree above shows a family in which brachydactlyly type D is segregating. Filled circles and squares represent females and males who have involvement of both thumbs. Half-filled in symbols represent family members with just one thumb affected? (Picture attached)
There is evidence of variable expressivity and incomplete penetrance in this family. Which individual is most likely nonpenetrant for the trait?
A) III-11
B) IV-1
C) IV-5
D) III-10
E) II-4
Chapter 5 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 5 - For parts a, b, and c, draw a diagram illustrating...Ch. 5 - 5.2 In a diploid species of plant, the genes for...Ch. 5 - A pure-breeding tall plant producing oval fruit as...Ch. 5 - 5.4 Genes E and H are syntenic in an experimental...Ch. 5 - In tomato plants, purple leaf color is controlled...Ch. 5 - 5.6 In Drosophila, the map positions of genes are...Ch. 5 - 5.7 Genes A, B, and C are linked on a chromosome...Ch. 5 - Gene G recombines with gene T at a frequency of...Ch. 5 - Genes A, B, C, D, and E are linked on a...Ch. 5 - Syntenic genes can assort independently. Explain...
Ch. 5 - 5.11 The recombination frequency between linked...Ch. 5 - On the DrosophilaX chromosome, the dominant allele...Ch. 5 - Researchers cross a corn plant that is pure -...Ch. 5 - 5.14 syndrome is an autosomal disorder affecting...Ch. 5 - 5.15 Three dominant traits of corn seedlings,...Ch. 5 - 5.16 In a diploid plant species, an with the...Ch. 5 - Prob. 17PCh. 5 - The Rh blood group in humans is determined by a...Ch. 5 - 5.19 Genetic linkage mapping for a large number of...Ch. 5 - 5.20 with the genotype form tetrads in the...Ch. 5 -
Gene and gene are genetically linked. Answer...Ch. 5 -
T. H. Morgan’s data on eye color and wing form,...Ch. 5 - Prob. 23PCh. 5 -
The boss in your laboratory has just heard of a...Ch. 5 - In rabbits, chocolate-colored fur (w+) is dominant...Ch. 5 - Prob. 26PCh. 5 - 5.27 In tomatoes, the allele for tall plant height...Ch. 5 - 5.28 Neurofibromatosis is an autosomal dominant...Ch. 5 - A 2006 genetic study of a large American family...Ch. 5 - 5.30 A experiment examining potential genetic...Ch. 5 - A genetic study of an early onset form of heart...Ch. 5 - In experiments published in 1918 that sought to...Ch. 5 - DNA sequence for 10 individuals are Identify the...Ch. 5 - 5.34 The accompanying pedigree below shows a...Ch. 5 - 5.35 Based on previous family studies, an...Ch. 5 - Divide a clean sheet of paper into four quadrant...Ch. 5 - 5.37 For six genes known to be linked on chromosme...
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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. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?arrow_forwardA 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?arrow_forwardA proband female with an unidentified disease seeks the advice of a genetic counselor before starting a family. Based on the following data, the counselor constructs a pedigree encompassing three generations: (1) The maternal grandfather of the proband has the disease. (2) The mother of the proband is unaffected and is the youngest of five children, the three oldest being male. (3) The proband has an affected older sister, but the youngest siblings are unaffected twins (boy and girl). (4) All the individuals who have the disease have been revealed. Duplicate the counselors featarrow_forward
- The pedigree shows inheritance of the autosomal recessive trait cystic fibrosis in a family. Family members who are known to have the disease are represented by black shading. I 1 2 3 4 II 1 3 4 6 8 7 III 1 2 3 4 5 6 7 Part A: Select the best answer from the boxes below to indicate the likely genotypes of the following individuals, where F indicates the wild-type (usual) allele and f indicates the cystic fibrosis allele. The first part of their number (I, II or III) indicates their generation or row, whilst the second part of their number (1-8) indicates the individual within that generation or row. |-4 Il-3 Il-8 III-3arrow_forwardThere are six types of agglutinogen named C,D, E and c,d,e.the first three are dominant and last three are recessive.discussarrow_forwardPedigree attached shows an autosomal recessive genetic disease. G is the normal allele and g is the disease-causing allele. Individual 1’s father is heterozygous (*) and his mother is homozygous dominant. Other individuals in the pedigree may be carriers, but are not marked. The question mark (?) indicates that you do not yet know anything about this individual’s phenotype with regard to the disease. part a) What is the probability that individuals 1 and 2 will have a child (5) who is a boy with the disease (the child is unborn and the sex is not yet known)? a)1/8 b)1/4 c)0 d)1/16 part b) What is the probability that the daughter (6) that individual 3 and 4 just had will have the disease? a)1/8 b)1/6 c)1/4 d)1/12arrow_forward
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- In this case a family history revealed a genetic basis for the disorder. The pedigree is shown in Fig. 1 Below. Key Ø Female: affected Female: unaffected || IV V 5600 orize 077808 15 10 9 10 CHO વ Male: affected Male: unaffected Deceased Disease status not given Dizygotic twins Monozygotic twins Fig. 1 Disease pedigree. Five generations I, II, III, IV, V are shown. Females are represented by circles, males by squares, dizygotic (non-identical) twins by diagonal lines originating from the same point, Monozygotic (identical) twins by diagonal lines originating from the same point and joined symbols and deceased by a diagonal line through the symbol. Filled symbols indicate that the individual displays the disease phenotype. Unfilled symbols indicate that the individual does not display the disease phenotype. Carriers of the disease are not indicated. Information on disease status is not known for generation I and is omitted for the individuals represented by a symbol with an asterisk.…arrow_forwardPlease consider the following pedigree. Assume that people who marry in to the family do not carry the allele. Assume complete penetrance. I II III 3 IV 1 2 a. Is it possible for the inheritance pattern for the trait illustrated in this pedigree to be as a result of each of the following? Answer yes or no. (i) an autosomal recessive allele (AR) (ii) an autosomal dominant allele (AD) (iii) a X-linked recessive allele (XR) (iv) a X-linked dominant allele (XD) b. Based strictly on the characteristic patterns of inheritance that define the four different options in (a), give a definitive motivation for the most likely mode of inhertance.arrow_forwardBoth hemophilia (h) and favism (gd) are inherited as X-linked recessive traits. Hemophilia is an inherited disorder of blood clotting, and favism is an inherited hemolytic anemia caused by absence of the enzyme glucose-6-phosphate dehydrogenase. A phenotypically normal woman is known to have the X chromosome genotype h + / + gd. The frequency of recombination between h and gd is 16%. What proportion of sons born to this woman are expected to be phenotypically normal with respect to both hemophilia and favism?arrow_forward
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