Human Karyotype B (questions #7 to #12) 7. The diploid number is 46. 8. The haploid number is 23 only. 9. There are 23 homologous pairs of chromosomes. 10. There are no trisomies in this karyotype. 10 11 12 11. The sex is male for this person. 아 13 14 15 16 17 18 12. This person may experience birth defects. te 19 20 21 22

Human Karyotype B (questions #7 to #12) 7. The diploid number is 46. 8. The haploid number is 23 only. 9. There are 23 homologous pairs of chromosomes. 10. There are no trisomies in this karyotype. 10 11 12 11. The sex is male for this person. 아 13 14 15 16 17 18 12. This person may experience birth defects. te 19 20 21 22

Biology 2e

2nd Edition

ISBN:9781947172517

Author:Matthew Douglas, Jung Choi, Mary Ann Clark

Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark

Chapter13: Modern Understandings Of Inheritance

Section: Chapter Questions

Problem 11RQ: The genotype XXY corresponds to Klinefelter syndrome Turner syndrome Triplo-X Jacob syndrome

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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 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?
Use keyboard only to enter your answer below. ALL WORKING MUST BE SHOWN Problem 1) mutation in a gene on chromosome 15 that codes for an enzyme. The disease is an inherited autosomal recessive condition which is Tay-Sachs disease is caused by loss of function found amongst Ashkenazi Jews of Central European origin. In this population, 3 in 5,200 children are born with the disease. What proportion of the population are carriers (heterozygotes) for this disease?
3 (Lecture 13 Study Guide, Question #10) Genetic traits on the X-chromosome are called X-linked. Some X-linked conditions are referred to as X-linked dominant, such as Fragile X Syndrome. Consider the following genetic cross: A father with Fragile X Syndrome and a mother with normal phenotype. Perform a Punnett Square. Select the possible genotypes of their offspring below. xFxf xFY
C B 2089202222 a b d B А. a e The figure above shows images of five different cells. Each box contains an image that represents a cell of an AaBb dihybrid that is in some stage of meiosis or mitosis. The lines represent chromatids. The letters on the lines represent alleles. The shaded region shows the shape of the cell. Refer to this figure as you consider the questions below. Each box contains an image that represents a cell of an AaBb dihybrid that is in some stage of meiosis or mitosis. The lines represent chromatids. The letters on the lines represent alleles. The shaded region shows the shape of the cell. Refer to this figure as you consider the questions below. Each box represents a cell in meiosis or mitosis. The lines represent chromatids. The letters next to the lines represent alleles. The shaded region shows the shape of the cell. The A gene controls an X-linked trait. The B gene controls an autosomal trait. Refer to this figure as you consider questions X-Y below + 8)…
Short Answer 3. The pedigree shown is of a family with an X-linked recessive disease. The sex chromosomes are indicated. Individual II-2 is the result of a non-disjunction event in one of the parents. I || 1 2 XHXH XhY XhY 12 XhXhY A) If the nondisjunction event that gave rise to Il-2 occurred in the father, in which phase of meiosis did it happen? Explain your reasoning. B) Upon further testing, it is discovered that Individual 1-1 has a mild case of the disease. Please give a short explanation as to how that is possible.
Match the term with the correct definition or example Question 27 options: Product of a cross between true breeding plants that make yellow peas and true breeding plants that make green peas. A description of the alleles present in an individual at one or more loci. Example: MM or Mm or mm. Example: 11q1. 4-q2. 1", meaning it is on the long arm of chromosome 11, somewhere in the range from sub-band 4 of region 1 to sub-band 1 of region 2. Under a microscope, these look exactly the same. The centromere, banding, long arm, short arm, overall length, gene locations, and order of genes are exactly the same. However, nucleotide sequences in genes on the two chromosomes can differ. the unit of heredity, the sequence of nucleotides in the exon regions specifies the sequence of amino acids in a particular polypeptide Brown allele in eye color Red eye in fruit flies Ability to synthesize alcohol dehydrogenase…
(X^X^) and one male (X"). 12. The following are the gene order on the chromosomes of an individual who is heterozygous for this translocation. translocations (a.unbalanced breciprocal C. Robertsonian MN OP QR N067 89 34 5P QR 3 4 5 6 7 8 9 a. 0 LMNOP QR LMN067 89 345P QR 3456789 hmm Draw the synapsis (Prophase/Metaphase I) configuration dark spots b. What type of translocation is depicted by these chromosomes and what are the consequences of this chromosome rearrangement to the individual? this type of translocation is unbalanced and the major consequence here is that the chromosome number is not reciprocated leading to all sorts of problems with non-disjunction.
KARYOTYPE #8 ZWK99032 KEY 14 15 17 19 21 22 Y Is this karyotype male or female? What kind of error (if any). Name of syndrome 9. 3. 20 OXEO gerte
Arial 11 BIUA 田回▼ 三=三|三|: 12 II 4. Below is a diploid cell in meiosis. a. Label one set of homologous chromosomes, one set of sister chromatids and one set of heterologous chromosomes. b. How many alleles of the 'A' alleles are present in the cell at Prophase 1? c. How many copies of the 'A' genes are present in each cell at Prophase II? d. How many 'a' alleles are present in Anaphase 1? e. How many 'F' alleles are present in each cell in Metaphase I? f. How many 'f alleles are present in each cell in Metaphase ll? g. How many 'F' genes are present in each gamete? h. How many chromosomes will be present in each gamete? MacBook Air
GENETIC DISORDERS 2 3 1. 2 3 XK XK XK X* XX xX XX KX Xe XX KK X* 3 K* K XX 8 9 10 11 12 13 14 15 8 9 10 11 12 13 14 15 ?8 ת A XX XX XX X 称家a M 16 17 18 19 20 21 22 XX 16 17 18 19 20 21 22 XXY Figure 2 Figure 1 1. Suppose a child was found to have the chromosome pattern shown in Figure 1 above. a. Is the child a male or female? b. Explain your answer. c. Down syndrome is caused by one extra autosome in each cell. What pair of chromosomes has an extra chromosome? d. How did this child get an extra chromosome? 2. Suppose a child was found to have the chromosome pattern shown in Figure 2 above. a. Is the child a male or female? b. Explain your answer. e. Which chromosome is the extra chromosome, an X or Y7 d. How did the child get an extra chromnseme?