Height in gryphons is determined by a single locus. You cross a two 9' tall gryphons and get 1/4 10'; 1/2 9' and 1/4 8' tall gryphons. If you across two 7' tall gryphons you get 1/4 8', 1/2 7'; 1/4 6' tall gryphons. pls kindly help to explanation: why In gryphons, alleles of the height lous are semidominant and form an allelic series.

Height in gryphons is determined by a single locus. You cross a two 9' tall gryphons and get 1/4 10'; 1/2 9' and 1/4 8' tall gryphons. If you across two 7' tall gryphons you get 1/4 8', 1/2 7'; 1/4 6' tall gryphons. pls kindly help to explanation: why In gryphons, alleles of the height lous are semidominant and form an allelic series.

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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Please write your complete solution (paper). Albinism in humans is inherited as a simple recessive trait. Determine the genotypes of the parents and their children when a normal male and albino female have 2 normal and 2 albino children.
Give typed full explanation You are studying three linked genes in snapdragons. The flower color locus is in the center. There are 13.8 cM between the flower color locus and the plant height locus. There are 14.5 cM between the flower color locus and the leaf type locus. The coefficient of coincidence is 0.8. Pure-breeding tall, red-flowered plants with fuzzy leaves were crossed to pure-breeding dwarf, blue-flowered plants with smooth leaves. The F1 were testcrossed. Calculate the proportion of the testcross progeny that are expected to be dwarf with red flowers. Round properly to 4 decimal digits.
Please answer fast Gene B codes for purple-colored flowers while the recessive form of the gene (b) results in pink-colored flowers. Another gene, A, when dominant, activates the transcription of gene B, resulting in purple or pink-colored flowers, while the recessive version of aa, results in white-colored flowers. A homozygous purple-colored flowering plant was crossed with a fully homozygous recessive white-colored flowering plant. The F1 flowering plants were self-crossed to generate the F2 generation. Determine the phenotypes and genotypes for each generation. Explain.
TOPIC: GENE INTERACTION AND EPISTASIS In pineapples, leaves may be spiny, spiny-tipped or piping. Pineapples of different phenotypes were used in the following crosses: P1 phenotype piping X spiny P1 genotype ______ ______ F1 phenotype 100% piping F1 genotype ______ Crossing the F1s produced: F2 phenotypes F2 genotypes 95 piping _______ 25 spiny tipped _______ 8 spiny _______
Topic: Trihybrid Cross, Height in merigonias is determined by three unlinked genes that act additively. Each has two alleles, one compltely dominant allele(A,B or C) that makes plants taller and one recessive allele that makes plants shorter. Do the corss AaBbCc X AaBbCc. The fraction of The F1 progenty will be ? Pls explain it with more details. thanks
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?
Using Drosophila notation: A.) Diagram the genotype of a female fly that is recessive for apterus (ap, chromosome 2), heterozygous wild-type for black (b, chromosome 2), recessive for forked (f, x-chromosome), and homozygous wild type for rosy (ry, chromosome 3). B.) Diagram the genotype of a male fly that is heterozygous for clot (ct) eyes (an autosomal gene) and has yellow (y) body color (x-linked gene).
Please explain how I can solve this. Thank you You wanted to draw a linkage map for three genes in the fruit fly (A, B, C) that you suspect are linked. You have done a three-point testcross and found that out of the 1500 progeny in the cross, 612 are a · b · c, 603 are A · B · C, 93 are a · B · C, 85 are A · b · c, 53 are a · b · C, 47 are A · B · c, 5 are A · b · C, and 2 are a · B · c. From this data, you can deduce that the genes are linked and the distance between B and C is:
Help me create a pedigree of this information: Pedigree analysis: Generation 1: Normal parents (AA x AA) Generation 2: Carrier parents (AA x AS) Generation 3: Affected child (AS x AS) Generation 4: Affected grandchild (SS) This pedigree has two normal parents in the first generation. Second generation carriers carry the sickle cell trait from one parent. The disease is 25% more likely to be inherited in the third generation if both parents have the 'S' allele. If both parents have the 'S' allele, their children will have sickle cell anemia in the fourth generation
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