Phenotype Number Recombinant type sharp 803 parental Answer Bank thin, round 821 DCO SCO parental round 191 parental DCO thin, sharp 185 SCO thin 6. SCO round, sharp DCO thin, round, sharp 28 SCO wild type 26 SCO
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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?RrxRR Genotyperatioofoffspring___________________ Phenotyperatioofoffspring__________________
- ABO Blood TypeThe following pedigree shows the incidence of ABO blood types in a family. dentify the genotypes of the following individuals: Individual Genotype II-1 II-2 II-4 II-5 III-2 III-3Eva’s parents, John and Cher. Are both type AB. John questioned Cher’s fidelity when Eva turned out to be a type O. Assuming that Cher is indeed loyal to her husband, why did Eva turn out to be type O? Help Cher prove her innocence by showing the possible genotypes of John and Cher. Show also the COMPLETE cross that produced Eva.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…
- All possible offspring pheotypes and genotypes in a punnet square for genotypesA 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?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) This woman is phenotypically normal. Does thissurprise you? Why or why not? Under what circumstancesmight you expect a phenotypic effect of such arearrangement?
- A man is brachydactylous (very short fingers; rare autosomal dominant), and his wife is not. Both can taste thechemical phenylthiocarbamide (autosomal dominant;common allele), but their mothers could not.a. Give the genotypes of the couple.If the genes assort independently and the couple hasfour children, what is the probability ofb. all of them being brachydactylous?c. none being brachydactylous?d. all of them being tasters?e. all of them being nontasters?f. all of them being brachydactylous tasters?g. none being brachydactylous tasters?h. at least one being a brachydactylous taster?A. HUMAN PEDIGREE CASE ANALYSIS1. One couple has three children with the following sexes and ages: one son (40 y.o.) and two daughters (35 y.o. and 33 y.o.), all of them have normal pigmentation. Another couple has a son (35 y.o) and a daughter (20 y.o.) and all of them also have normal pigmentation. Both couples have normal pigmentation. The younger daughter from the first couple married the son of the second couple and they had three children. Their eldest daughter (5 y.o.) has normal pigmentation while their only son (3 y.o.) and one daughter (1 y.o.) have albinism. a. Draw the pedigree of this family. Follow protocols in making a pedigree. Provide the genotype of all individuals in the pedigree. Please provide also the gene notation. b. What is the mode of inheritance of this trait? c. Justify your answer in letter (b).d. For their normal daughter, what is the probability that she is a carrier? Show solution. e. If they will have a fourth child, what is the probability that the…How do you figure out the geno and phenotype from the info given? 2 pages out the lab but I figured once explained I’ll be able to do understand and do the rest!
