Arabidopsis thaliana is a diploid plant model organism with 2n 10. Please select the number options to match the followingnumber of copies of each gene an Arabidopsis thaliana1 leaf cell hasnumber of chromosomes an Arabidopsis thaliana leaf cellcontains1. two2. fivenumber of chromosomes an Arabidopsis thalianagamete cell contains3. tenpairs of homologous chromosomes an Arabidopsis thaliana2leaf cell contains

Diploid number of chromosomes means that chromosomes are present in pairs. Haploid number of…

number of chromosomes an Arabidopsis thaliana leaf cell contains 1. two 2. five number of chromosomes an Arabidopsis thaliana - gamete cell contains 3. ten pairs of homologous chromosomes an Arabidopsis thaliana -- leaf cell contains

number of chromosomes an Arabidopsis thaliana leaf cell contains 1. two 2. five number of chromosomes an Arabidopsis thaliana - gamete cell contains 3. ten pairs of homologous chromosomes an Arabidopsis thaliana -- leaf cell contains

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

Chapter14: Chromosomes And Human Inheritance

Section: Chapter Questions

Problem 12SQ

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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
ala Marked out of 1 Select one: OA. A Tissue Sample of Unknown Organism OB. B OC C D. D d. b. Mitosis must be carefully regulated to ensure the normal distribution of chromosomes to the daughter cells. A ring-shaped protein molecule known as cohesin attaches to the centromere of a chromosome and holds sister chromatids together to prevent their premature separation. Enzymes detach cohesin molecules from the centromere immediately before the sister chromatids segregate. -based on Nature, 2006 In which of the cells labelled in the diagram above are enzymes most likely acting to detach cohesin molecules from the centromere? Megee, Paul, 2006. Chromosome guardians on duty. Nature 441, no. 7089 (May 4): 35-36.
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?
ISlate edu/ d2l/le/content/5003190/viewContent/44248878/View Google Tranx 4 My Drive-G X 4. Suppose that a parent Drosophila is e ca* ca The gamete frequency is as follows: e'ca e ca 16% e'ca е са 31% 14% 29% a. Circle the recombinant gametes. b. What is the map distance between the ebony and claret genes?
The plant Haplopappus gracilis has a 2n of 4. A diploid cellculture was established and, at premitotic S phase, aradioactive nucleotide was added and was incorporatedinto newly synthesized DNA. The cells were then removed from the radioactivity, washed, and allowed to proceed through mitosis. Radioactive chromosomes or chromatids can be detected by placing photographic emulsionon the cells; radioactive chromosomes or chromatids appeared covered with spots of silver from the emulsion.(The chromosomes “take their own photograph.”) Drawthe chromosomes at prophase and telophase of the firstand second mitotic divisions after the radioactive treatment. If they are radioactive, show it in your diagram. Ifthere are several possibilities, show them, too.
In the plant, Haplopappus gracile there is one long pair and one short pair of chromosomes. In the diagrams below, anaphase of individual cells in meiosis or mitosis in a plant that is heterozygous for the genes, A and B (which are on separate chromosomes) are shown. The lines represent chromosomes or chromatids and the points of the "V" are the centromeres. For each case, indicate if the cell represents meiosis I, meiosis II, mitosis, or impossible situation. Provide a brief (one sentence) reason for your decision. 1. A A a В a B 9. A a B A a 2. а В В a A a B 9. B. a A 3.
Corn has a chromosome number of 2=20. Supposing there are different aneuploidy/polyploidy in corn, provide the correct answer in the table below. 1. Type of aneuploidy/polyploidy Formula Chromosome Chromosome Types of gametes (n, n+1, n+2, n-1, n-2) number Configuration 1. monosomic 2. trisomic 3. double trisomic 4.nullisomic 5.tetrasomic 6. Autotetraploid 7. allotetraploid
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?
Let us practice it again! Analyze the pedigree below to answer the questions that follow. Huntington's disease a disorder in which nerve cells waste away, or disintegrate, is passed down through families. certain parts of the brain Huntington's diseate llustration ereated in htps://pregenygenetion.com/ 1. What members of the family above are affected with the Huntington's disease? 2. Tnere are no carriers ior Huntungton's disease you either have it or you do not. Is Huntington's disease caused-by a dominant or recessive trait? 3. Identify the genotypes of the following individuals using the pedigree above. (homozygous dominant, homozygous recessive, heterozygous). I- 1 II -1: II -3: III - 4 : 4. How many children did individuals I-1 and I-2 have? 5. How many girls did II-1 and II-2 have? How many have Huntington's Disease? 6. How are individuals III-2 and II-4 related? I-2 and III-5?
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?