PEDIGREE ANALYSIS— please answer all of them, they are all connected 3. This pedigree is for the autosomal dominant trait achondroplasia (use letter A as allele/s), a rareform of dwarfism.II2IIIVa. What is the most convincing genotype of individual (I-2)?b. What is the most convincing genotype of individual (II-1)?c. What is the probability of having a child with achondroplasia IF the couple l-1 and I-2 will have anotherchild?d. What is the probability of individual (IV) of having achondroplasia?e. If the couple (II-1 and II-2) will be having another offspring, what is the probability of having a son withachondroplasia?2.

(According to Bartleby guidelines, only the first three have been answered. Kindly post the…

3. This pedigree is for the autosomal dominant trait achondroplasia (use letter A as allele/s), a rare form of dwarfism. II 2 II IV a. What is the most convincing genotype of individual (I-2)? b. What is the most convincing genotype of individual (III-1)? c. What is the probability of having a child with achondroplasia IF the couple l-1 and I-2 will have another child?

3. This pedigree is for the autosomal dominant trait achondroplasia (use letter A as allele/s), a rare form of dwarfism. II 2 II IV a. What is the most convincing genotype of individual (I-2)? b. What is the most convincing genotype of individual (III-1)? c. What is the probability of having a child with achondroplasia IF the couple l-1 and I-2 will have another child?

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter12: Mendel, Genes, And Inheritance

Section: Chapter Questions

Problem 3TYK

See similar textbooks

Similar questions

A testcross is a way to determine ________. a. phenotype b. genotype c. dominance
In addition to the two genes in problem 4, assume you now study a third independently assorting gene that has the alleles C and c. For each of the following genotypes, indicate what types of gametes will be produced: a. AA BB CC b. Aa BB Cc c. Aa BB cc d. Aa Bb Cc
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?
Klinefelter syndrome (XXY) can most be easily diagnosed by _______. a. pedigree analysis. b. aneuploidy c. karyotyping d. phenotypic treatment
1) By decomposing the cross, MmNn X Mmnn, into single gene outcomes, the possibility of obtaining a double recessive outcome for the ‘N’ gene is: a) 50%! b) 100! c) 0%! d) 25%! 2) A 16-square Punnett is time-consuming to draw out. Dr. Thompson can easily solve this problem by decomposing her dihybrid crosses into separate monohybrid Punnett squares, which would also be quicker than using a forked line diagram. If so, how many individual monohybrid Punnett squares will she need to draw? a) Four Punnett squares. b) Two Punnet squares. c) One Punnett square. d) Sixteen Punnett squares since there are 16 possible outcomes.
1. A woman who is heterozygous for the brown eye and blue eye alleles marries a man with blue eyes. What are the genotypes and phenotypes of the children? a) What is the genotypic ratio of the above cross (Write number next to genotype)? ___________________________________________________________________ b) What is the phenotypic ratio of the above cross? (Write number next to phenotype)? ___________________________________________________________________
1. Why did the royal family have an issue with the hapsburg jaw/lip? 2. Was it an Autosomal or X-linked Genetic Disorder? 3. Dominant or Recessive?(Explain) Using a Punnett Square predict what the Genotypes are for the persons II-3 & II-(* C/C or C/A mean that they married and had children with a blood cousin).
1. A man with O blood type marries a woman with AB blood type. What are the expected genotypic and phenotypic ratios for the children? a) What is the genotypic ratio of the above cross (Write number next to genotype)? ___________________________________________________________________ b) What is the phenotypic ratio of the above cross? (Write number next to phenotype)? ___________________________________________________________________
1. A human female "carrier" who is heterozygous for the recessive, sex-linked trait causing red-green color blindness (or alternatively, hemophilia), marries a normal male. What proportion of their male progeny will have red-green color blindness (or alternatively, will be hemophiliac)? * a. 100% b. 75% c. 50% d. 25% e. 0% 2. A human female "carrier" who is heterozygous for the recessive, sex-linked trait red color blindness, marries a normal male.What proportion of their female progeny will show the trait? * a. All b. ½ c. ¼ d. 0 e. 3/4 3. Women have sex chromosomes of XX, and men have sex chromosomes of XY. Which of a man's grandparents could not be the source of any of the genes on his Y-chromosome? * a. Father's Mother b. Mother's Father c. Father's Father d. Mother's Mother, Mother's Father, and Father's Mother e. Mother's Mother 4. Male-pattern baldness is an example of a sex-influenced trait. The baldness allele, which causes hair loss, is influenced by the hormones…
1. A mother with blood type A has a child with blood type O, and she claims that Mister X is the father. Mister X denies that he could possibly be the father, because he has blood type B. If you were the judge presiding over this case, which of the following arguments would you find most convincing? a. Mr. X could be the father if his blood genotype is IBIB, and the mother’s blood genotype is IAIA b. Mr. X cannot be the father because a child with blood type O cannot have a parent with blood type B c. Mr. X could be the father if his blood genotype is IBi, and the mother’s blood genotype is IAi d. Mr. X cannot be the father because a parent with blood type B cannot have a child with blood type O e. Mr. X cannot be the father because the human ABO blood types are sex-linked traits 2. A monohybrid cross between two pea plants producing yellow peas, resulting in an F1 phenotypic ratio of 3:1 (three yellow pea plants to one green pea plant) indicates what about the parental pea plants?…