Please kindly assist,thank you. Please consider the pedigree below. There are no cases of false paternity.IAB2IIBAB (A B2 3III1IV12a. Which individual/s definitely has/have Bombay phenotype in this extended family?b. What are the genotypes of individuals II-2 and II-3 at the ABO and H loci?

False paternity is the condition whene a man is wrongly identified as father of any child. Now a…

Answered: Bombay phenotype in this extended…

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter18: Genetics Of Behavior

Section: Chapter Questions

Problem 14QP: A pedigree analysis was performed on the family of a man with schizophrenia. Based on the known...

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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?
Please consider the pedigree below. There are no cases of false paternity. I II III IV в 1 a. Which individual/s definitely has/have Bombay phenotype in the descendants of I-1 and I-2? b. What are the genotypes of individuals II-2 and III-2 at the AB0 and H loci? Please label your answers a and b, Il-2: and Ill-2:.
Please consider the pedigree below. There are no cases of false paternity. I B II A 2 3 III AB (A IV в 1 a. Which individual/s definitely has/have Bombay phenotype in the descendants of I-1 and I-2? b. What are the genotypes of individuals II-2 and III-2 at the ABO and H loci?
Please consider the pedigree below. There are no cases of false paternity. AB 2 I II 2 3 A (AB A)B 1 III 2 3 4 IV B 1 2 a. Which individual/s definitely has/have Bombay phenotype in this extended family? b. What are the genotypes of individuals II-2 and II-3 at the ABO and H loci?
Please consider the pedigree below. There are no cases of false paternity. The ABO blood group phenotypes of individuals who marry into the family are a true reflection of their ABO genotypes. image attached I a. Which individual/s definitely has/have Bombay phenotype in the descendants of I-1 and I-2? b. What are the genotypes of individuals I-1 and II-2 at the ABO and H loci?
The pedigree below shows the phenotypes of the ABO blood groups and Rhesus factors [positive (+) and negative (-)] for several members of a family. I (B+ AB- 1 2 3 4 II O- A+ В- B- AB+ A+ 1 2 4 5 6 a. What are the ABO blood group genotypes of individuals I-1 and I-2? b. Which child/ren of individual I-4 can donate blood to him? c. Which individual in the pedigree can donate blood to all the other individuals in the pedigree?
In this case a family history revealed a genetic basis for the disorder. The pedigree is shown in Fig. 1 Below. Key Ø Female: affected Female: unaffected || IV V 5600 orize 077808 15 10 9 10 CHO વ Male: affected Male: unaffected Deceased Disease status not given Dizygotic twins Monozygotic twins Fig. 1 Disease pedigree. Five generations I, II, III, IV, V are shown. Females are represented by circles, males by squares, dizygotic (non-identical) twins by diagonal lines originating from the same point, Monozygotic (identical) twins by diagonal lines originating from the same point and joined symbols and deceased by a diagonal line through the symbol. Filled symbols indicate that the individual displays the disease phenotype. Unfilled symbols indicate that the individual does not display the disease phenotype. Carriers of the disease are not indicated. Information on disease status is not known for generation I and is omitted for the individuals represented by a symbol with an asterisk.…
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…
While sitting at home during Movement Control Order (MCO) because of pandemic covid19, observe two different traits of a couple in your family (eg. your mom & dad or your sister & her husband or your brother & his wife, etc). Draw a genetic cross that involves cross of the parents with the chosen 2 pairs of their contracting traits. Imagine that the cross obeys the Mendelian Laws, show the cross and gametes production for each generation (P, F1 and F2). By Using a Punnet square as symbolic representation of the results for the cross, determine the phenotypes, genotypes, phenotypic ratio and genotypic ratio of F2 generation in the family.
1. The pedigree chart in Figure 5.29 shows the inheritance of haemopiu family. Study the pattern of inheritance in the pedigree chart, and then answer the questions that follow. о 5 6. 3 8 9 10 11 Key Unaffected male Haemophiliac male О Unaffected female Fig. 5.29 Pedigree chart of a family affected by haemophilia a) What is the genotype and phenotype of individuals 2 and 4? b) (i) How many of the unaffected family members are definitely carriers of the recessive allele? (ii) How are you able to tell which of the family members are carriers? (4) (1) (3) c) (i) If Individual 11 marries a carrier female, what percentage of their sons is likely to be haemophiliacs? (1) (ii) Use a genetic diagram to show how you worked out your answer in i, (6) 2. Why is haemophilia never passed from father to son, even though it is most common in males? (4) 3. Can a mother pass on a sex-linked gene to her daughter? (1) 4. Sipho has red-green colour blindness. One of his grandfathers was also. colour…

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