Match each of the following conditions with the correct choice(s) provided on the right. Some choices may be used more than once. Patau syndrome A. females only Kiinefelter syndrome B. males only C. monosomy D. trisomy karyotype with an extra chromosome Turner syndrome human karyotype with 45 chromosomes E. infant mortality Edwards syndrome F. sterility Down syndrome

Match each of the following conditions with the correct choice(s) provided on the right. Some choices may be used more than once. Patau syndrome A. females only Kiinefelter syndrome B. males only C. monosomy D. trisomy karyotype with an extra chromosome Turner syndrome human karyotype with 45 chromosomes E. infant mortality Edwards syndrome F. sterility Down syndrome

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Description: Match each of the following conditions with the correct choice(s) provided on the right.Some choices may be used more than once.Patau syndromeA. females onlyKlinefelter syndromeB. males onlyC. monosomyD. trisomykaryotype with an extra chromosomeTurn

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 5QP: Achondroplasia is a rare dominant autosomal defect resulting in dwarfism. The unaffected brother of...

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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?
Achondroplasia is a rare dominant autosomal defect resulting in dwarfism. The unaffected brother of an individual with achondroplasia is seeking counsel on the likelihood of his being a carrier of the mutant allele. What is the probability that the unaffected client is carrying the achondroplasia allele?
Match the chromosome disorder to its descriptionin the key. Turner syndrome a. female with undeveloped ovaries and uterus, unable to undergopuberty, normal intelligence, can live normally with hormonereplacementb. XXY male, can inherit more than two X chromosomesc. male or female, mentally impaired, short stature, flat face, stubbyfingers, large tongue, simian palm creased. XXX or XXXX femalee. caused by nondisjunction during spermatogenesis
For each of the terms in the left column, choose thebest matching phrase in the right column.a. phenotype 1. having two identical alleles of a given geneb. alleles 2. the allele expressed in the phenotype ofthe heterozygotec. independent 3. alternate forms of a gene assortmentd. gametes 4. observable characteristice. gene 5. a cross between individuals bothheterozygous for two genesf. segregation 6. alleles of one gene separate into gametesrandomly with respect to alleles of othergenesg. heterozygote 7. reproductive cells containing only onecopy of each geneh. dominant 8. the allele that does not contribute to thephenotype of the heterozygotei. F1 9. the cross of an individual of ambiguousgenotype with a homozygous recessiveindividualj. testcross 10. an individual with two different allelesof a genek. genotype 11. the heritable entity that determines acharacteristicl. recessive 12. the alleles an individual hasm. dihybrid cross 13. the separation of the two alleles of agene into…
Match the chromosome disorder to its descriptionin the key. Jacobs syndrome a. female with undeveloped ovaries and uterus, unable to undergopuberty, normal intelligence, can live normally with hormonereplacementb. XXY male, can inherit more than two X chromosomesc. male or female, mentally impaired, short stature, flat face, stubbyfingers, large tongue, simian palm creased. XXX or XXXX femalee. caused by nondisjunction during spermatogenesis
white b. light black c. medium TA I. X-LINKED OR SEX-LINKED TRAITS A color-blind man marries a woman wit blind. Use aXfor normal vision, and al X for color-blindness, (Remember where sex-linked traits are carried). What is the genotype of the ma al vision, whose father was color- h norm type of the manand woman? Use a Punnett square to show the phenotypes of their progeny 172
Explain the sex-linked traits. (S9LT-ld-29.4.4) Solve problems related to sex-linked traits. (S9LT-ld-29.4.4) WHAT I NEED (MATERIALS) • Worksheet • Pens WHAT TO DO (PROCEDURE) PART A Directions: Tell whether the following carry or show the sex-linke recessive spongy-tiny pores trait. Also, tell whether the individuals a male or female. Remember that since most sex-linked traits are recessive, the person w shows the trait can have no X's with big superscripted letters. So, if a m carries the trait, he must also show the trait. XDY XDXD XdY Carrier? (Yes or No) Has the Trait? (Yes or No) Male or Female Phenotype or Appearance of the Individual ided Questions: 1. What chromosomes will result to a normal phenotype (normal spongy po trait12
The XG locus on the human X chromosome has twoalleles, XG+ and XG. The XG+ allele causes the presence of the Xg surface antigen on red blood cells,while the recessive XG allele does not allow antigento appear. The XG locus is 10 m.u. from the STSlocus. The STS+ allele produces normal activity ofthe enzyme steroid sulfatase, while the recessive STSallele results in the lack of steroid sulfatase activityand the disease ichthyosis (scaly skin). A man withichthyosis and no Xg antigen has a normal daughterwith Xg antigen. This daughter is expecting a child.a. If the child is a son, what is the probability he willlack Xg antigen and have ichthyosis?b. What is the probability that a son would have boththe antigen and ichthyosis?c. If the child is a son with ichthyosis, what is theprobability he will have Xg antigen?
Unaffected father Camier mother XY Unaffected Afected Carrier Unaffeded Unaffected daugkter U.S. National Lbrany of Mediche Carrier Affected son daughter son In humans, as well as in many other animals and some plants, the sex of the individual is determined by sex chromosomes. The sex chromosomes are one pair of non-homologous chromosomes: XX represents a female, while XY represents a male. When a gene for a specific trait is attached to the X or Y chromosome, we say it is sex-linked, and when it is attached to the X chromosome, we say it is X-linked. Alleles for these linked traits, such as hemophilia or color blindness, crosses, may be recessive or dominant. Hemophilia is an X-linked, recessive trait. The recessive allele for hemophilia is actually a mutated version of the normal alllele but it can still be passed on through generations. Imagine a female is a carrier for hemophilia; her genotype is Xx She is married to a man who does not have hemophilia. What conclusion is NOT valid…
a. The ability to taste the chemical phenylthiocarbamideis an autosomal dominant phenotype, and the inabilityto taste it is recessive. If a taster woman with a nontasterfather marries a taster man who in a previous marriagehad a nontaster daughter, what is the probability thattheir first child will be(1) A nontaster girl(2) A taster girl(3) A taster boyb. What is the probability that their first two childrenwill be tasters of either sex?
which genetic disorder matches below descriptions. extra 21st chromosome attaches to chromosome 14. Affected individual exhibit epichanthic folds of eyelids, simian crease in palms, and mental retardation. trisomy 18 Affected individuas have an elongated skull, hair lip, cleft palate, rocker-bottom feet. Life expectancy is about 10 weeks. trisomy 21. Affected indiviudals exhibit epicanthic folds of eyelids, protruding tongue, small, low set ears, and retardation.