Analysis of linkage involving many other families in addition to the one illustrated for Question 8 led to the conclusion that this SNP marker is 12 cM from the gene responsible for Leber congenital amaurosis. Alleles of this SNP marker are also shown in the pedigree for the unborn child in generation IV, whose future phenotype is unknown. Based on this child's genotype, and assuming that the mutation in this family causing Leber congenital amaurosis is fully penetrant, what is the probability that this child will have the trait? 44% 88% 12%

Theodor Boveri predicted that malignancies would often be associated with chromosomal mutation. What lines of evidence substantiate this prediction?

Analyzing Karyotypes 1. Originally, karyotypic analysis relied only on size and centromere placement to identify chromosomes. Because many chromosomes are similar in size and centromere placement, the identification of individual chromosomes was difficult, and chromosomes were placed into eight groups, identified by the letters A to G. Today, each human chromosome can be readily identified. a. What technical advances led to this improvement in chromosome identification? b. List two ways this improvement can be implemented. c. What clinical information does a karyotype provide?

Analyzing Cloned Sequences A base change (A to T) is the mutational event that created the mutant sickle cell anemia allele of beta globin. This mutation destroys an MstII restriction site normally present in the beta globin gene. This difference between the normal allele and the mutant allele can be detected with Southern blotting. Using a labeled beta globin gene as a probe, what differences would you expect to see for a Southern blot of the normal beta globin gene and the mutant sickle cell gene?

Achondroplasia is an autosomal dominant form of dwarfism caused by a single gene mutation. Calculate the mutation rate of this gene given the following data: 10 achondroplastic births to unaffected parents in 245,000 births.

Equalizing the Expression of X Chromosome Genes in Males and Females Individuals with an XXY genotype are sterile males. If one X is inactivated early in embryogenesis, the genotype of the individual effectively becomes XY. Why will this individual not develop as a normal male?

You learned in the chapter that an STR locus is a locus where alleles differ in the number of copies of a short, tandemly repeated DNA sequence. PCR is used to determine the number of alleles present, as shown by the size of the DNA fragment amplified. In the Figure below are the results of PCR analysis for STR alleles at a locus where the repeat unit length is 9 bp, and alleles are known that have 5 to 11 copies of the repeat. Given the STR alleles present in the adults, state whether each of the four juveniles could or could not be an off-spring of those two adults. Explain your answers.

The results of a paternity test are shown in the table below. Numbers indicate the number of short tandem repeats for loci tested. Whos the daddy? How sure are you?

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?