Chapter 10: Human Genetics

Terms in this set (290)

The answer is E.

Klinefelter syndrome is most commonly caused by a meiotic nondisjunction event during parental gametogenesis that results in a 47,XXY karyotype. Variants include 46,XY/47,XXY mosaicism and 48,XXXY. In general, patients with higher numbers of X chromosomes are more likely to have more severe manifestations. The disorder is usually not diagnosed until puberty when the characteristic physical signs begin to develop. The major features are as follows:
1. Klinefelter syndrome causes primary testicular failure due to hyalinization and fibrosis of the seminiferous tubules. This resume in small, firm testes and azoospermia (infertility). Leydig cell dysfunction also occurs and leads to testosterone deficiency. Gonadotropin (FSH, LH) levels are increased secondary to gonadal failure .
2. Testosterone deficiency results in development of a eunuchoid body habitus. Patients have tall stature and gynecomastia. Facial and body hair is sparse or absent and muscle mass is decreased.
3. Mild intellectual disability is seen in some patients. although the majority have normal intelligence. Psychosocial abnormalities (eg, lack of insight, poor judgment) are also common.

(Choice A) Arachnodactyly, scoliosis and aortic root dilation are signs of Marfan syndrome, which occurs due to an inherited defect of the extracellular matrix protein fibriilin.

(Choice B) Macroorchidism, large jaw, and intellectual disability are soon in patients with fragile X syndrome, an X-linked disorder caused by mutations in the fragile X mental retardation 1 gene.

(Choice C) In females, loss of an X chromosome (45,XO karyotype) results in Tumor syndrome, which presents with short stature, broad chest, and primary amenorrhea.

(Choice D) Preder-Willi syndrome is characterized by short stature, hypotonia, intellectual disability, and obesity. The most common cause is a microdeletion affecting the paternal chromosome 15q11 ~13 critical region.

Educational Objective:
47,XXY is the most common genotype causing Klinefelter syndrome. Patients present with tall stature, small, firm testes azoospermia; and gynecomastia. Mild intellectual disability is seen in some patients, and the severity generally increases with each additional X chromosome.
The answer is E.

The neuromuscular lesions, ragged red skeletal muscle fibers, and lactic acidosis in these family members together suggest mitochondria! encephalomyopathy. Mitochondrial disorders follow a maternal inheritance pattern, as an embryo's mitochondria are inherited from the ovum only.

Mitochondria are responsible for ATP production via oxidative phosphorylation, which is why mitochondrial defects tend to cause lactic acidosis and primarily affect tissues with the highest metabolic rates (e.g., neural tissue, muscular tissue). Though many mitochondrial proteins are coded for in the nuclear genome. mitochondria also contain their own genome, which is also vulnerable to mutations. Defects in the mitochondrial genome may occur in any number of the mitochondria
within a cell, and the severity of mitochondrial diseases is often related to the proportion of abnormal to normal mitochondria within a patients cells Heteroplasmy describes the condition of having different organellar genomes (eg mutated and wild-type) within a single cell. For mitochondrial diseases, patients with more severe disease are those with a higher proportion of defective mitochondrial genomes within their cells.

Educational Objective:
The presence of lactic acidosis and ragged red skeletal muscle fibers histologically suggest a mitochondrial myopathy. There may be variable clinical expression of mitochondrial DNA defects in different affected family members due to heteroplasmy, which is the coexistence of both mutated and wild-type versions of mitochondrial genomes in an individual cell.
The answer is C.

The probability of having a child affected by an autosomal recessive disorder is given by the product of the following individual probabilities:

P(affected child given carrier parents) x P(carrier mother) x P(carrier father)

In this example:
P(affected child given carrier parents) = 1/4. as the child of 2 carriers of a recessive condition has a 1/4 chance of being affected
P(carrier mother) = 1, as the patient must be a carrier given that her first son has Pompe disease
P(carrier father) can be calculated using Hardy-Weinberg analysis

Hardy-Weinberg analysis can be used to relate gene/allele, disease. and carrier frequencies if 1 of these values is known:

Gene/allele frequency: By convention, p = frequency of normal (dominant) allele. and q = frequency of mutant (recessive) allele in the population of interest
Disease frequency: As homozygous recessive individuals (with the disease) must have 2 copies of the recessive (mutant) allele, the frequency of homozygous recessive individuals (disease frequency) = q x q = q^2
Carrier frequency: Heterozygous individuals (disease carriers) have only 1 mutant allele (genotype is either pq or qp), so in general, cancer frequency = 2pq. For rare autosomal recessive disorders, p = 1, therefore, the probability of being a carrier approximates to 2x frequency of the mutant allele or 2q.

In this example, we are given q= disease frequency = 1/40.000. Therefore, q = square root (1/ 40,000) = 1/200 and P(carrier father) as 2q = 2 x (1/200) = 1/100.

In sum, the probability of this patient having a second affected child is: (1/4 x 1 x 1/100) = 1/400.
The answer is B.

This child has many of the characteristic features of Down syndrome (DS). The abnormalities found in DS result from extra genetic material from chromosome 21. Three cytogenetic abnormalities can lead to DS:

1. Meiotic nondisjunction accounts for nearly 95% of DS cases. Meiotic nondisjunction (failure of homologous chromosomes to separate during meiosis) of chromosome 21 occurs in the ovum, resulting in the inheritance of 3 copies in one daughter cell (trisomy) and 1 copy in the other daughter coil (monosomy). Nondisjunction is almost always of maternal origin, and increased maternal age is a risk factor.
2. Unbalanced Robertsonian translocations account for 2%-3% of DS cases. These individuals have 46 chromosomes, but an extra arm of chromosome 21 is attached to another chromosome (translocation). Approximately one-third of these cases are due to a balanced translocation in one parent. These balanced translocations are associated with high recurrence risk. Genetic counseling for the parents is indicated if a translocation is identified in the infant.
3. Mosaicism accounts for <2% of cases. Affected individuals have 2 cell: lines 1 with a normal genotype, and 1 with trisomy 21. The proportion of affected cells determines the severity of DS features.

Educational Objective:
Common findings in Down syndrome include cognitive impairment, facial dysmorphism. and cardiac defects; 95% of cases are caused by the presence of an extra chromosome 21 (trisomy) resulting from nondisjunction. Unbalanced Robertsonian translocations or mosaicism are less common causes.
The answer is D.

The male patient described above has a history of excessive bleeding and hemarthroses, suggesting a diagnosis of hemophilia A or B. Both of these diseases are X-linked recessive coagulation factor deficiencies. The probability that his sister will give birth to an affected child can be calculated by multiplying the following probabilities:

- The probability (p1) that the sister is a carrier = 0.5. The patients father does not carry the mutation on his X chromosome, because he would be affected by the disease if he did. That means the mother carries the mutation on one of her two X chromosomes. This gives the daughter e 50% chance of having inherited the mutated x chromosome end thus being a carrier.
- The probability (p2) that the offspring of a female carrier will inherit the X chromosome with the hemophilia gene = 0.5. Assuming the daughter is a carrier, the probability of passing on the hemophilia-carrying version of her two x chromosomes is 50%, since only one of her two x chromosomes is passed to her offspring.
- The probability (p3) that his sister will have a male child = 0.5. If the sister's child is female then the child could be a carrier of the disease, but would not be affected by it if a male child inherits the mutated X chromosome. he will have the disease

The probability that the sister will have an affected son is thus the probability that all three of the above events take place (i.e., the product of their individual probabilities):
p1 x p2 x p3 = 1/2 x 1/2 x 1/2 = 1/8
The answer is E. The presence of consanguinity (double line in the figure) is a red flag for autosomal recessive inheritance because, although disease-causing alleles are rare, the probability of a homozygous individual escalates dramatically when the same rare allele descends through two branches of a family. Using a lowercase r to denote the retinitis pigmentosa allele, the affected male (individual II-2 in the pedigree) has a genotype of rr. His prospective mate has a very low risk to be a carrier for this rare disease, making her genotype RR. Their children will all have genotypes Rr, making them carriers but not affected. Retinitis pigmentosa is another disease manifesting genetic heterogeneity, with autosomal dominant, autosomal recessive, and X-linked recessive forms. Carriers of autosomal recessive diseases are heterozygotes with one normal and one abnormal allele. Many autosomal recessive diseases involve enzyme deficiencies, indicating that 50% levels of enzymes found in heterozygotes are sufficient for normal function. The probability that an affected individual will encounter a mate who is a carrier is approximately twice the square root of the disease incidence. This figure derives from the Hardy-Weinberg law. Since most recessive diseases have incidences lower than 1/10,000, the risk for unrelated mates to be carriers is less than 1/50, and the chance of having an affected child is less than 1/50 × 1/4 = less than 1/200. Disorders that are fairly common in certain ethnic groups, such as cystic fibrosis, are exceptions to this very low risk.