Forensics Q3 3. You are trying to determine whether any of the children of Nancy are fathered by TomYou analyze the DNA from 2 genes and get the following dataGene 1NancyTomPaulaCindeeRoss RalphGene 2NancyTomPaulaCindeeRoss RalphBased on this information(1) Which offspring can not be the offspring of Tom?(2) Which offspring could be the offspring of Tom?(3) Which offspring must be the offspring of Tom?

1. **Offspring who cannot be the offspring of Tom (assuming a typo in the provided data corrected…

Answered: 3. You are trying to determine whether…

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 3CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...

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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?
9.) In humans, when a dark-skinned person reproduces with a very light-skinned person, their children may range in color from dark to brown. This occurs because skin color is controlled by more than one gene (the exact # of genes is not known). For your convenience, a table is presented below that shows several possible genotypes & phenotypes: Genotypes Phenotypes AABB Very dark skin AABb or AaBB Dark skin AaBb, AAbb, or aaBB Medium brown skin Aabb or aaBb Light skin aabb Very light skin If a man with dark skin whose genotype is AaBB reproduces with a woman who has light skin (aaBb), what are the possible skin colors that their children will have? Develop a Punnett square and list the possible genotypes & phenotypes.
4. (1) How many females and males are affected in this pedigree? (2) Who is the proband? II (3) How many generations are included? 1 2 3 4 5 6 7 (4) What kind of inheritance pattern is the disease- causing allele in this pedigree? Why? 1 2 3 4 5 7 8 9. (5) Who have the same genotype as the proband? Why? 5. The figure below shows a pedigree of a genetic disorder caused by the X-linked recessive allele X'. Determine the genotype or possible genotypes of everyone in generations I, II and III. II 3 5 8. IV 1 2 3 4 56 7 8 9 10 1 12 13 14 15 16 17 18 19 20
d/e/1FAIPQLSCDH_No2]MEaHUdrt5NWa2uY6AaKa7Bhy4xfhwplF-CsVAbyA/formResponse M Gmail YouTube O Maps Which phrase best describes a human with the chromosomes represented in the karyotype below? * 1 3 4 5 X r8 K* XX从 6. 8 9. 10 X* ** ** 11 12 13 14 15 16 17 18 19 20 XX 44 21 22 XX O A female who exhibits Down syndrome A male who exhibits Down syndrome O A female who does not exhibit Down syndrome O A male who does not exhibit Down syndrome 8.
You are an allele on a single uncondensed chromosome in a Yellowfever-carrying mosquito (Aedes aegypti) diploid cell (2n = 6). Demonstrate in as much detail as possible (1) how you will get to each genetically identical diploid cell of the multicellular offspring; (2) how you will get to the next generation of sexually reproduced progeny. Help, I no understand. Perhaps you will?
3. The recessive sex-linked gene (h) prolongs the blood-clotting time, resulting in the genetically inherited disease called hemophilia. From the information in the pedigree chart (right), answer the following questions: Hemophilia in humans (a) If 112 marries a normal man, what is the probability of her first child being a hemophiliac? 2 (b) Suppose her first child is actually a hemophiliac. What is the chance that her second child will be a boy with hemophilia? 2 (c) If I14 has children with a hemophiliac man, what is the probability that her first child will be phenotypically normal? (d) If the mother of 12 was phenotypically normal, what phenotype was her father? 4. The phenotypic expression of a dominant gene in Ayrshire cattle is a notch in the tips of the ears. In the pedigree chart on the right, notched animals are represented by the solid symbols. Ear notches in Ayrshire cattle Determine the probability of notched offspring being produced from the following matings: 1 2 (a)…
9 Genet x K Kami 9 Point x K point- x 9 Band, X S Week x 9 Geog x My Qu X STI In X > (2 x R Read X Discu X A web.kamihq.com/web/viewer.html?file=https%3A%2F%2Ftuscaloosacity.schoology.com%2Fattachment%2F1716642933%2Fsource%2F775235e0ba17ab191f792e5f.. K * : = 3+Branches+of+Go. O STI InformationNo. E Schubert = Pod 2 7th Master S. Other bookmarks Каmi Student Upgrade O O A My Drive Kami Export - Scan Feb 26, 2021.pdf A Turn In 100% JT Use the following information for questions 10-12: In dogs, the gene for fur color has two alleles. The dominant allele (F) codes for grey fur and the recessive allele (f) codes for black fur. 10) The female dog is heterozygous. The male dog is homozygous recessive. What is the chance their offspring have grey fur? T 14 - 11) The female dog has black fur. The male dog has black fur. What is the chance their offspring has a heterozygous genotype? 12) The female dog is heterozygous. The male dog is heterozygous. What is the chance their offspring is…
Chromosomes and Genetics Name 1. Edward and Jenna had 4 children together. They have a few "interesting" traits, shall we say! Look at the chromosome combinations that each child received and list their genotype and phenotype. Key: D= long eyelashes B = brown eyes F= freckles G = vampire teeth T= tall W = werewolf fur d = short eyelashes b = red eyes f= no freckles g = no vampire teeth t = short w = no werewolf fur
9.) In humans, when a dark-skinned person reproduces with a very light- skinned person, their children may range in color from dark to brown. This occurs because skin color is controlled by more than one gene (the exact # of genes is not known). For your convenience, a table is presented below that shows several possible genotypes & phenotypes: Genotypes Phenotypes Very dark skin Dark skin AABB AABB or AaBB AаBb, AAЬЬ, or aaBB Aabb or aaBb Medium brown skin Light skin Very light skin aabb If a man with dark skin whose genotype is AABB reproduces with a woman who has light skin (aaBb), what are the possible skin colors that their children will have? Develop a Punnett square and list the possible genotypes & phenotypes.
A woman who sought genetic counseling is found to be heterozygousfor a chromosomal rearrangement between the second andthird chromosomes. Her chromosomes, compared to those in anormal karyotype, are diagrammed on the next page:(a) What kind of chromosomal aberration is shown?
7.4 5577 20/ CV* 5. (a) Complete the following diagram to show the process by which gametes are formed. amand NOR NO devane 8 8. 072 404 PACLIGNO YOU Forsch chenyle VI SAMNING Zeytmos Lectopic of 12 mesecer in 200AU g Chor WERY IN 12120/ ZAINA Stage Act PE Stage B 2X P Stage C AUT2 Xyadden che d S www.w opov 02/01 88 8 us ml 151/3/0/00 COADCA CLAS Stage D Janis Stage E

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