Purple fur (P) is DOMINANT in monsters. Yellow fur (p) is RECESSIVE. What is the genotype of a PURE PURPLE monster? = What is the GENOTYPE of a HETEROZYGOUS purple monster? What is the GENOTYPE of a YELLOW monster? = Using the punnet square above, make a cross between TWO HETEROZYGOUS PURPLE MONSTERS. POSSIBLE OFF SPRING GENOTYPES PHENOTYPES

Purple fur (P) is DOMINANT in monsters. Yellow fur (p) is RECESSIVE. What is the genotype of a PURE PURPLE monster? = What is the GENOTYPE of a HETEROZYGOUS purple monster? What is the GENOTYPE of a YELLOW monster? = Using the punnet square above, make a cross between TWO HETEROZYGOUS PURPLE MONSTERS. POSSIBLE OFF SPRING GENOTYPES PHENOTYPES

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Description: 188ISAdd-oISal textBIUAE E E ETimes New.1214.Purple fur (P) is DOMINANT in monsters. Yellow fur (p) is RECESSIVE.What is the genotype of a PURE PURPLE monster? =What is the GENOTYPE of a HETEROZYGOUS purple monster?What is the GENOTYPE of a YELLOW m

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter3: Transmission Of Genes From Generation To Generation

Section: Chapter Questions

Problem 18QP: More Crosses with Pea Plants: The Principle of Independent Assortment Determine the possible...

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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?
BbRrppMm X bbRrPpMM Based on the cross provided what is the probability of producing an individual that is dominant for B & M and recessive for R and P?
BbRrppMm X bbRrPpMM: probability of producing an individual that is dominant for B & M and recessive for R and P?
ddTtBbqqAA X ddttBbQqaa:probability of producing an individual that is dominant for B & A and recessive for D, T and Q?
A man, Penoy, whose sister died in early childhood from a recessive lethal disease marries a woman Esmae, who has the same family history. Because Penoy has survived beyond childhood, he does not have the disease, but he may be a carrier (i.e. heterozygous, as may also be the case with Esmae). What is the probability that their first child will suffer from the disease? [Hint: first calculate the probability that Penoy is heterozygous; then determine the probability that both parents are carriers. Remember that he has survived to adulthood when calculating this probability].
Heliodors are either red (R), yellow (Y) or an intermediate phenotype, orange. What is the genotype of the orange heliodor?
Here are the progeny of this cross: (Note that the categories are not in any particular order.)Fly type # of prog. Phenotype symbols Categorywt eyes black body wt wings 97 grn+ blk crv+Green eyes black body curved wings 709 ParentalGreen eyes wt body wt wings 9Green eyes black body wt wings 162wt eyes wt body wt wings 727wt eyes black body curved wings 12wt eyes wt body curved wings 179Green eyes wt body curved wings 105Total = 2000 9.) Write the phenotype symbols in the right-hand column. The first one has been done for you.10.) Next to that, label all fly categories as parental (NCOs), SCOs, and DCOs. One has been donefor you.11.) After each SCO/DCO label, write which gene got “unlinked” in these offspring.12.) Put these three genes into a genetic map in the proper order.13.) Calculate the genetic distance between the genes and label the map with these distances.14.) Calculate the cross-over interference15.) Return to questions #1-6 above. For question 6, you gave your opinion, but…
Punnet square problems A=Codominant; B=Codominant; O=Recessive Mary is homozygous for type A blood. Steve is homozygous for type O blood. If they have children, what are the possible phenotypes and genotypes of their children, and what is the probability of each? Mary and Steve have a son, Brad. Brad’s wife, Samantha is heterozygous for type B blood. If they have children, what are the possible phenotypes and genotypes of their children, and what is the probability of each? Stella loves roses and decides to cross her red rose with her white rose. All of the resulting offspring of this cross are pink roses. What can you say about the red and white alleles as a result of this cross? Stella decides to cross two of the pink roses. What are the possible genotypes and phenotypes of the offspring and the probabilities of each? DNA replication, Transcription and Translation problems It is S phase of the cell cycle, and time to replicate the cell’s DNA. Using the following strand of DNA…
The Delacour family is descended from the Veela race, a semi-human, semi-magical humanoid people reminiscent of the Sirens in Greek mythology with special powers. Imagine that being a Veela and having these powers is a X-Linked Dominant trait that is passed on genetically. (Hint: Use the Punnett Square.) What are the possible genotypes for Fleur and Bill’s daughter, Victoire? What are the genotypic frequencies related to the female children (in percentage format)? What is the likelihood (in percentage format) that Victoire would express Veela powers?
ddTtBbqqAA X ddttBbQqaa Based on the cross provided what is the probability of producing an individual that is dominant for B & A and recessive for D, T and Q?