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.
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Topic: Gene Locus
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.
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- Miniature wings, X in Drosophila melanogaster result from an X-linked allele that is recessive to the allele for long wings, 9 + X. Match the genotypes for each parent in the crosses. Male parent phenotye long miniature miniature 111 long long Female parent phenotype long long long miniature long m m X X Male offspring phenotypes 231 long, 250 miniature 610 long 410 long, 417 miniature 753 miniature 625 long m X Y Female offspring phenotypes 560 long 632 long 412 long, 415 miniature 761 long 630 long Answer Bank ++ X X Male parent genotype + X Y Female parent genotype + X X m 00Topic: Penetrance. Petal number is controlled by a single gene in merigonias. The gene has a completely dominant wild type allele F that makes a plant have five petals and a mutant recessive six petal allele(f). However the six petal trait is only 50% penetrant. You do the cross Ff x Ff. What fraction of the progeny do you have the 6 petals? what is the meaning for 50% penetrant.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?Tick all the essential steps to demonstrate a genetic linkage between a disease and a molecular marker in humans. identify the alleles of the genetic marker only for diseased individuals in the pedigree enumerate parental type individuals sequence the wild-type and mutant alleles to find the mutation no correct answer calculate a Lod score calculate the recombination frequency between the mutation and the molecular marker identify the alleles of the genetic marker for each individual in the pedigree pedigree analysis cloning the defective gene enumerate recombinant individuals
- 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 _______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.Considering the following chromosome which is represented as a series of genes on each arm separated by the centromere. Describe the type of mutation required to produce each of the mutant chromosomes below. ABCDEFG*HIJKLMN
- Drosohpila Punnet Square of Crosses. I need results of F1 & F2 generation using Punnett Squares for: Make Punnet Squares of the following crosses •Drosophila Female wildtype cross Male White-eye •Drosophila Male wildtype cross Female White-eye •Drosophila Female Wild Type cross Male Scarlet Eye •Drosophila Male Wild Type cross Female Scarlet Eye Also, Which allele is heterozygous and which is homozygous, & which is dominant and which is recessive?uh ec CV + + 10.5 SC 9.1 scute bristles echinus eyes 9.2 ct + crossveinless wings Table 1: phenotype wild-type tapdance feet crossveinless wings tapdance & crossveinless cut wings 15.9 vermilion eyes V + + 66.8 Drosophila X chromosome Use the map provided above for problems 1 & 2. Problem 1: 11.2 10.9 garnet eyes M A new gene is being investigated in fruit flies. The recessive allele of this gene (t) causes the flies' feet to grow tiny tapdance shoes, while the dominant allele (t*) permits wild-type feet to develop. Preliminary studies indicate that this new gene is located on the X-chromosome. You decided to perform a two-point testcross to determine its position relative to the well-established crossveinless wings gene (cv). You cross a female heterozygous for both genes with a testcross male fly and obtain the male offspring results shown in table 1, below. Using this information, answer the following questions: # male offspring 13 405 401 11 forked bristles a) is the original…Calibri 12 - BIU A Section 4: Monohybrid test cross A monohybrid cross is a mating where alleles of just one gene are tracked in the offspring. A test cross is a mating in which two parents (P generation) are crossed to give F1 (first filial generation) offspring, and then two F1 parents are crossed to give F2 (second filial generation) offspring. By tracking offspring through two generations, a test cross can reveal patterns of inheritance for particular genes. 9. In a cross between a black and a white guinea pig, all members of the F1 generation are black. The F2 generation is made of % black and % white guinea pigs. Diagram this cross, and list the genotypes and phenotypes for each generation.