SEQ Block 2

What is the role of surgery for her disease and why did the surgeon remove Ms. Monroe's entire colon instead of just the right side?

Surgery to remove the cancer and involved lymph nodes is the only way to cure colon cancer. Her entire colon is at risk for developing colon cancer because every cell in it contains a mutated APC gene (or MYH gene for those who answered MYH-associated polyposis).

Compare and contrast type 1 and type 2 diabetes mellitus with regard to how the diagnosis is made.

The diagnosis of diabetes is made in the same for type 1 or type 2. There are three criteria: (1) A random plasma glucose ≥ 200mg/dL plus typical symptoms of diabetes; (2) A fasting plasma glucose ≥ 126mg/dL; (3) A 75 gm oral glucose tolerance test in which the 2 hour blood sugar is ≥ 200mg/dL.

For each of these three genetic conditions, describe the likely internal urogenital and external genital constitution of the fetus. What pattern of secondary sexual development is expected to occur at puberty for each of these conditions?

(5 ARD) In 5-alpha-reductase deficiency, the Wolffian duct virilizes in a normal male fashion, but the urogenital sinus and genital tubercule persists as female structures. The failure is due to inadequate dihydrotestosterone formation intracellularly in these external genitalia tissues at the time of normal male fetus virilizing. In the 5-alpha-reductase deficiency, the seminal vesicles, ejaculatory ducts, epididymis, and vas deferens, which are all testosterone-dependent, are present, whereas the DHT-dependent structures: external genitalia, urethra and prostate, do not develop along male lines. Affected men have less facial hair and body hair, less temporal hair line recession and no problems with acne. Spermatogenesis, muscle mass, male libido, and deepening of the voice do occur in these men. Antimüllerian hormone activity is present since there are normal testes, and the individual does not have müllerian development (uterus, cervix, and fallopian tubes). (AIS) In androgen insensitivity syndrome, the androgen receptor mediated of Wolffian duct development does not occur. However, antimüllerian hormone activity is present and the individual does not have müllerian development. The vagina is short and it ends blindly. The uterus and fallopian tubes are absent and the testes are normally developed, but abnormally positioned. Testosterone production is normal or is slightly increased. The diagnosis is likely when an individual presents following breast development at puberty with primary amenorrhea, scant or absent pubic and axillary hair and a short vagina, as well as an absent cervix and uterus. Children with inguinal hernias and/or inguinal masses, the possibility of undescended testis should be considered. In congenital insensitivity to androgens, individuals have normal breast development and growths spurt at puberty but do not have menses due to lack of a uterus. On examination, these patients have scant or absent pubic and axillary hair, a short vagina, and an absent cervix and uterus. Children with inguinal hernias and/or inguinal masses should be suspected of having androgen insensitivity syndrome. It is very important to do careful investigation of other family members, since this syndrome follows an X linked recessive pattern. (SRY) With deletion of SRY, the cascade of male sexual differentiation is not initiated, and internal and external sexual development is female. Streak gonads are present instead of ovaries, and internal sex organs are usually female. The external genitals may vary between normal female and normal male, with the majority female. Since there is no gonad, and therefore no antimüllerian hormone, the internal genitalia is female. When SRY is mutated, no secondary sexual (male or female) characteristics develop at puberty, and menstruation does not occur due to lack of estrogen. Since they have no gonads, they do not have a growth spurt and their pubic hair develops scantly. They are usually brought to a doctor due to "infantilism" with lack of secondary sexual characteristics and growth. They need to be placed on "hormones" to develop due to gonadal failure. (5 ARD) 5-alpha-reductase deficiency differs from the complete forms of androgen insensitivity or SRY mutations because at puberty masculinization occurs due to increased secretion of testosterone at puberty. Normal testicular function occurs at puberty and there is some response to endogenous or exogenous androgen and there is growth of the clitoris into a small or sometimes well developed penis, and partial or complete fusion of the labia. At birth, however, the external genitalia are similar to that of androgen insensitivity or SRY mutations. Many of these patients are reared as girls with an enlarged clitoris. At birth steroid levels are normal, ruling out the possibility of adrenal disorders. Patients with 5 ARD retain their fertility and there are reports of live births.

Given the above findings on history and physical examination, describe three tests to confirm the diagnosis of secondary adrenal insufficiency?

1. You could inject synthetic ACTH into the patient and demonstrate a lack of stimulation of cortisol secretion. The adrenal glands would not respond to the injection of synthetic ACTH because they atrophy when they are deprived of the "trophic" effect of ACTH. ("Trophic" effect means that ACTH is needed to maintain the size, growth, and vascularity of the adrenal glands.) 2. You could inject insulin into the patient, causing hypoglycemia. Hypoglycemia is a stress that normally stimulates the hypothalamic-pituitary-adrenal axis. If the pituitary corticotroph cells are not working, then hypoglycemia will not stimulate increased cortisol secretion. 3. Once you have confirmed that the patient has adrenal insufficiency, you must measure blood ACTH levels. In primary adrenal insufficiency the ACTH level will be high due to lack of negative feedback inhibition of cortisol on CRH and ACTH secretion. In secondary (pituitary) adrenal insufficiency, ACTH levels will not be elevated.

In particular, how might a serum thyroglobulin level help to identify why the patient was thyrotoxic?

A differential diagnosis that only included thyrotoxicosis factitia and painless thyroiditis. In painless thyroiditis the thyroid gland is being attacked by the immune system. Thyroid follicular cells are destroyed, releasing their contents, including thyroglobulin, into the blood stream. On the other hand, in thyrotoxicosis factitia the thyroid is not being stimulated by TSH so is inactive. Serum thyroglobulin levels would therefore be low. One would expect both the free T4 and the free T3 to be elevated in painless thyroiditis, since both of these thyroid hormones would be leaking from the thyroid gland.

The lab report on Joan's amniocentesis designates the karyotype shown in Figure 1 as 47,XX,+21. What is a karyotype? What does each component of the report (47 XX and +21) provide to you? Will Joan and Jeremy's child have Down syndrome? What is the most common mechanism causing Down syndrome?

A karyotype is a specific test designed to examine the number and structure of the chromosomes taken from a sample of cells. The standard nomenclature used for a karyotype always includes the total number of chromosomes and the number of sex chromosomes. If the sample shows an abnormal chromosome, the report would also indicate the type of abnormality. Each component of this report shows: 47 = number of chromosomes in the metaphase XX = sex chromosome constitution of the metaphase +21 = indicates that the 47th chromosome is an extra chromosome 21. The karyotype is consistent with a diagnosis of a female fetus with Down syndrome. Down syndrome is a common chromosomal abnormality that occurs in about 1 in 750 live births. About 95% of individuals with Down syndrome have a karyotype with 47 total chromosomes including three complete chromosomes 21. Down syndrome is also caused Trisomy 21, because the symptoms are due to the extra chromosome 21. The most common mechanism causing trisomy 21 is called non-disjunction.

Karen Long is a 32 year old G1, who presented for prenatal care. She had no significant past medical history. She underwent a dating ultrasound because she was unsure about her last menstrual period, on which she was found to be 8 weeks and 1 day by measurement of the crown rump length. The fetal cord insert appeared slightly thickened with bowel extending 8mm into the cord. The patient declined first trimester screening for fetal trisomies, but did desire screening for neural tube defects, so an AFP was performed at 16 weeks gestation. The AFP returned at 4.2 MOM. What is alpha-fetoprotein (AFP) and why was it elevated?

Alpha-fetoprotein is the embryonic and fetal equivalent of serum albumin, an abundant protein in serum. Alpha-fetoprotein is made by the embryonic yolk sac and the fetal liver. Tissue integrity is compromised in many severe congenital anomalies, causing leakage of serum into the amniotic cavity and the maternal serum. Alpha-fetoprotein can be measured in the maternal serum and is elevated in pregnancies with openings of the spinal cord, brain, or abdominal cavity. Elevated alpha-fetoprotein can also be a marker for a hepatoma or teratoma in the non-pregnant adult.

What is alpha-fetoprotein (AFP) and why was it elevated?

Alpha-fetoprotein is the embryonic and fetal equivalent of serum albumin, an abundant protein in serum. Alpha-fetoprotein is made by the embryonic yolk sac and the fetal liver. Tissue integrity is compromised in many severe congenital anomalies, causing leakage of serum into the amniotic cavity and the maternal serum. Alpha-fetoprotein can be measured in the maternal serum and is elevated in pregnancies with openings of the spinal cord, brain, or abdominal cavity. Elevated alpha-fetoprotein can also be a marker for a hepatoma or teratoma in the non-pregnant adult.

A second ultrasound showed a fetus with bowel outside the abdomen consistent with an omphalocele. Should an amniocentesis be recommended for this patient? Why or why not?

An amniocentesis should be performed on this expectant mother. Fetuses with omphalocele are at an increased risk of having other malformations, and a 30% risk of having a chromosome abnormality such as Trisomy 18. Omphalocele may also be seen in connection with other conditions such as Beckwith-Wiedeman syndrome.

T3 (tri-iodothyronine) is available commercially under the brand name "Cytomel." Occasionally a patient will present to an endocrinologist with a puzzling form of thyrotoxicosis that turns out to be due to the patient's secret ingestion of excessive amounts of thyroid hormone pills. This is a psychiatric illness. The patient takes the excessive thyroid hormone for the secondary gain of having people worry over them and take care of them because they are sick. It is called "thyrotoxicosis factitia." What symptoms and signs of thyrotoxicosis would the patient have? Which ones would be absent compared to the other causes of thyrotoxicosis?

Because of the elevated levels of T3, the patient would be expected to complain of typical symptoms of thyrotoxicosis such as weight loss, restlessness, irritability, heat intolerance with excessive sweating, hyperdefecation, hair loss, tremors, palpitations, and (if a woman) amenorrhea or irregular periods. On physical examination you will find that the patient had a tremor and onycholysis, a condition where the nail cracks or lifts from the nail bed. The skin would be smooth, velvety, warm, and moist. The patient would likely have lid lag. The pulse would probably be fast and the systolic blood pressure would be elevated. No goiter, which is unusual.

Does the genetic mutation causing myotonic dystrophy (DM1) differ from that causing Fragile X? In your answer, explain any similarities and differences between the causative mechanism(s) of two conditions.

Both Fragile X and DM1 involve triplet repeat expansions, but differ in the nucleotide sequence of the repeating unit, and in the location of the expansion within the gene locus. In Fragile X, the (CAG)n trinucleotide repeat is located in the 5' untranslated region of the FMR1 gene, while the (CTG)n trinucleotide repeat in DM1 is in the 3' untranslated region of the DMPK gene. In both diseases, there can be effects that manifest themselves at the RNA level. RNA binding protein sequestration is observed in patients with DMPK full mutations and in older Fragile X premutation males who exhibit tremor/ataxia. In patients with a Fragile X full mutation, the repeat expansion results in increased methylation of the region 5' to the FMR1 gene, leading to cessation of transcription and absence of FMRP in the cells. Methylation patterns of the DMPK gene are not known to change as a result of DM1 (DMPK) repeat expansion. However, it is possible that changes in methylation at the 3' end of the gene (3' UTR) (which would not affect transcription at the 5' end) could alter gene expression, especially if there is an enhancer element in that region. Regulatory regions that control eukaryotic gene expression can exist upstream, downstream or within the regulated gene.

Alan chose to pursue genetic testing, thinking that it would help his children understand their future risk of cancer, as well as their children's risks. Molecular genetic studies were performed on his blood, and on blood and tumor tissue from each eye of his first affected child. Alan's peripheral blood sample revealed a point mutation in a gene residing on chromosome 13q14. One of his daughter's tumors had a small cytogenetically visible deletion in one chromosome 13 (at band 13q14). Describe (in terms of paternal and maternal alleles) the likely findings in his child's peripheral blood, and in the tumor samples. If his second affected child's tumors were tested, how would the results compare to the first child's? Discuss the concept of LOH (loss of heterozygosity) in your answer.

Both of Alan's affected children will carry the same point mutation as Alan on their paternal allele of the RB gene. Both of his unaffected children are almost certain to contain non-mutated maternal and paternal alleles for RB. When peripheral blood from any of them is tested, these will be the only findings. In the tumor tissue tested from the first affected child the results will be as follows: The paternal allele in each sample will be identical to Alan's mutation. The remaining allele will carry a second mutation different from Alan's, and will almost certainly be different between the two tumors. The second allele mutation most commonly is a deletion, as was seen in one tumor from one of his affected children. Because the somatic mutations occur independently in each malignant clone, the mutation is likely to be different in the two separate tumor specimens. However, since recombination with the paternal allele, and chromosome duplication and non-disjunction are other possible mechanisms for loss of the functional (maternal) allele, there is some chance that one or both of the second mutations will be identical to the first (i.e. Alan's mutation). Statistically, however, they are likely to be different. Loss of heterozygosity is the term used to describe the appearance of the somatic mutation in the second allele. The germline is "functionally" heterozygous in that there is a non-functional allele which was inherited, and a wild-type allele. A somatic mutation occurs at the wild-type allele, and now the genotype is NO LONGER heterozygous (both alleles mutated, not necessarily with the same mutation).

Jennifer Jones is a 28 year old woman who is 16 weeks pregnant. Her obstetrician requested an ultrasound examination to evaluate her baby because on examination her uterus was larger than would have been expected based on her last menstrual period. The ultrasound examination revealed a normal female fetus with no congenital anomalies. Since Ms. Jones has two cousins with Down syndrome, she felt very strongly about having an amniocentesis to exclude the possibility of Down syndrome in this pregnancy. An amniocentesis was done at the time of the ultrasound. Approximately three weeks later she received a call from her obstetrician, congratulating her that the baby had a normal male karyotype. This immediately triggered some confusion since she was told previously that she was going to have a baby girl. In fact, she had already picked a name for her. A repeat ultrasound was immediately arranged, now at 20 weeks gestation. After a very detailed examination, the ultrasonographer continued to report a normal female intrauterine gestation based on a detailed ultrasound visualization of the external genitalia. Ms. Jones' family history was explored in more detail. She has a brother who was born with hypospadias which was surgically corrected. Three genetic conditions which result in female genital development in a fetus with a normal male karyotype are 5-alpha-reductase deficiency, androgen insensitivity syndrome (AIS), and SRY mutations. How common are these conditions? Briefly describe the underlying genetic alterations in each disorder and describe a potential mechanism that could contribute to the prevalence of these genetic conditions?

Congenital insensitivity to androgens, or androgen insensitivity syndrome, is caused by mutations in the androgen receptor gene, located on the X chromosome. The trait is inherited as X-linked recessive condition. Given that these deficiencies in females do not affect fertility, this contributes to the prevalence of 2-5/100,000 for complete androgen insensitivity syndrome and are based on estimates derived from otherwise healthy phenotypic females found to have histologically normal inguinal or abdominal testes. 5 alpha-reductase deficiency (5-ARD) is caused by an autosomal recessive gene mutation in the 5 -ARD type 2 gene. This is an intersex condition in which genetic males have female-appearing external genitalia due to an absence of dihydrotestosterone effect on the external genitalia. The 5-ARD enzyme is responsible for conversion of testosterone to dihydrotestosterone. These individuals have normal internal genitalia and normal wolffian structures but are undervirilized and display ambiguous external genitalia. Affected individuals may have hypospadias or a micropenis. Typically, SRY mutations are sporadic, although familial transmission has been documented in one-third of all identified SRY mutations. This syndrome occurs in approximately 1 in 20,000 newborn males. Most cases are caused by a de novo translocation of genetic material, including the SRY gene, between the X and Y chromosomes. SRY can be detected by fluorescent analysis in approximately 80% of individuals with this syndrome.

You meet with the patient and his wife in your longitudinal medical school clinic. His wife told you that she looked up his symptoms on the Internet and thinks that he has "low blood glucose". How would you explain the definition of "low blood glucose" or hypoglycemia to this patient? In your answer discuss whether a solitary glucose measurement is enough for a diagnosis of hypoglycemia, and if not, what other elements are required.

Hypoglycemia is blood glucose is less than 45 mg/dl or 40 mg/dl. Hypoglycemia is considered to be present, if an individual has low blood glucose level in presence of symptoms of hypoglycemia, and these symptoms are relieved by correcting blood glucose values. This triad of 1. Symptoms 2. Low blood glucose 3. Resolution of symptoms on correction of blood glucose is what is used to make a correct diagnosis of hypoglycemia.

Her2 and EGFR1 are both classified as receptor tyrosine kinases. In the case of Her2, the ligand that binds and activates it (if any) is not well understood. On the other hand, EGFR1 binds and is activated by epidermal growth factor (EGF). Activation of the EGFR1 receptor can activate several intracellular signaling cascades that ultimately lead to changes in the transcription of numerous genes involved in proliferation. One of these cascades is commonly referred to as the Ras/Raf/Mek pathway. List the changes that take place in the components of this pathway upon ligand binding, starting with the receptor itself, and including Ras, Raf, MEK, Map kinase, and nuclear transcription factors (Don't worry about the specific transcription factors - just address them generally).

EGF binds to the extracellular portion of the receptor leading to dimerization of two receptor subunits. Upon dimerization, a conformational change in the protein leads to activation of its kinase activity. This leads first to autophosphorylation of the receptors intracellular domain, and subsequently to binding of guanine nucleotide exchange factors (for example GRB2 and SOS) that promote activation of the G-protein, Ras. Ras, in its inactive state is bound to GDP. Upon interacting with the activated receptor and bound adaptors, GDP is exchanged for GTP, activating Ras. Ras-GTP activates the kinase Raf-1 which in turn phosphorylates the kinase MEK. MEK phosphorylates MAP kinase, which in turn regulates several transcription factors, also by phosphorylation. (The details of this pathway are difficult to sort out for the novice. An acceptable answer should include receptor dimerization, autophosphorylation, GDP/GTP exchange, and some series of phosphorylation events, ultimately impacting the activity of transcriptional regulators).

What are the risks of future cancers for the affected children and for the unaffected children?

Even without testing, the unaffected children are very unlikely to carry a mutation at the RB locus, as the development of retinoblastoma in patients who harbor a gene mutation usually occurs in younger children. If the penetrance of retinoblastoma is 95-98%, then each unaffected offspring has a 1 to 2.5% (1/2 of 2 to 5%) chance of also carrying the paternal point mutation in the RB gene. Their risks to be affected and to develop later onset retinoblastoma is then about 2.5% (0.025 * .95 = 0.02375). Therefore, the risk for his unaffected children to developing cancer either retinoblastoma or a related cancer may be higher than the general population, but is much lower than individuals who are known to harbor the gene mutation. The affected children, on the other hand, have a substantial risk of developing additional cancers during their lifetime. The literature states that RB mutation carriers have a high risk of developing osteosarcomas, brain tumors and melanomas. In more recent series, these cancers seem to occur predominantly in people who underwent radiation as treatment for their retinoblastoma. Compared with the general population, hereditary retinoblastoma survivors have higher mortality from lung cancer, bladder cancer and all other epithelial cancers combined. In one study, cancer incidence over a lifetime approached 70% and mortality was approximately 60%. Both of these figures are at least two-fold higher than for the general population.

Joan and Jeremy's DNA was amplified using the PCR technique for a marker located near the centromere of chromosome 21 (21q11.1). Figure 2 shows the number of repeats for each allele for Joan, Jeremy and their fetus. Which parent contributed the extra chromosome 21 in the fetus? In your answer, explain the segregation of the alleles.

Fetus has two copies of Joan's STR (allele 2 with 14 repeats) and one copy of Jeremy's (allele 2 of 12 repeats). Additional chromosome 21 is maternal in origin. Joan's age places her at increased for a child with Down syndrome. A 38-year-old woman has a 1/175 chance and this risk increases each year so that, at age 45, the risk is 1/30.

(A) Poly T polymorphism in intron 8 consists of three alleles, in which there can be 5T's, 7T's, or 9T's. The 9T allele is the most common and leads to normal splicing of the CFTR transcript. Normal splicing also occurs with the 7T allele, but exon 9 is skipped if the 5T allele is present. (B) Both the poly T polymorphism and an R117H mutation are associated with CBAVD or mild CF. The R117H mutation combined on the same chromosome (in cis) with a 7T allele acts as a mild CF mutation. The R117H mutation has a greater (more severe) phenotypic effect if it is in cis with a 5T allele. If there is also a CF mutation on the opposite chromosome (trans), R117H when in cis with 7T results in CBAVD but when in cis with 5T results in mild CF. CBAVD may also occur when R117H (in cis with either a 7T or 9T allele) is in trans with a 5T allele. CBAVD may also result with 5T homozygosity alone, though it is unknown whether either 5T chromosome also carries another, undetected CF mutation. This patient (above) and his wife come to you for an explanation of the genetic test. You inquire about any lung or sinus symptoms, and you find that he has always had "a sinus infection". You find this illustration (Fig 1) from Human Genetics and Genomics by Bruce R. Korf (Third Edition). What do you expect the results of the CF test to show? Does the molecular event in the illustration explain his clinical picture? In your answer, discuss the concepts illustrated in the figure as well as any genotype-phenotype correlations.

Genetic testing for CF can either be done by direct sequencing of the exons and intron-exon boundaries of the gene or by testing the patient for a series of known, more commonly observed CFTR mutations grouped into a test panel. The panel is usually the first step in testing. Currently available panels test for as few as 23 mutations to nearly 100 mutations. Because the test was "positive", this patient should have one or two identifiable mutations in the CFTR gene. Not all mutations will result in the same phenotype, as some are associated with severe disease, like the delta F 508 allele, and some are considered mild, like the R117H allele. Atypical CF cases can occur when a severe mutation is coupled with a mild mutation. Mutations in the CF gene have been found in about 10% or more of males with obstructive azospermia, although the percentage of identifiable mutations differs with the population tested. For those with congenital bilateral absence of the vas deferens (CBAVD), up to 70% have two identifiable mutations, while a lesser proportion of mutations is found in patients with a unilateral CAVD. The test would have also made special mention of the "T tract" alleles in the polypyrimidine tract in intron 8 of the CFTR gene that alters the efficiency of pre-mRNA splicing. Of the three well-studied T track alleles (5T, 7T, and 9T), the 9T track is most efficient, and the 5T and the 7T are less efficient, leading to a decrease in the full length CFTR mRNA and a decrease in functional CFTR protein. The severity of disease depends on whether a T tract occurs in cis with another CFTR variant on the same allele. For example, patients with a R117H mild mutation and the 7T tract on the same allele may have milder disease compared to patients with the R117H mutation and the 5T tract. Note to students: Both classic and non-classic forms of CF are considered to be autosomal recessive traits. However, not all patients with atypical CF have two identifiable mutations in the CFTR gene, as illustrated in Part 1 of this SEQ. This may be due to an unknown/unidentifiable mutation in the other "normal" allele, a mutation in another gene that modifies the risk for disease, or that the single mutation is causative. Whether this means that some heterozygote carriers of CF gene mutations are at risk for mild forms of disease is not known at the present time but might be a consideration when relating CF testing results to patient phenotypes.

Jimmy's mother wants to have another child, but is worried that she shouldn't. She wants to know what her options are and if you can help her do the right thing. What information do you need to provide? What information does the patient need to provide? What are the limits of your recommendation?

Good ethics begin with good facts. The physician needs to share the reproductive options (and associated risks) with Jimmy's mom. Jimmy's mother needs to share her values and opinions regarding the available options. This sharing of information is necessary to determine what reproductive options are most appropriate and in alignment with the patient's values. Options include: natural pregnancy, patient needs to understand that there is a 50% chance her child would also have DM1 mutations (autosomal dominance), that it is caused by triplet repeat expansion in the DMPK gene that vary but most likely get more severe with each generation, and that there is some uncertainty as to what will happen. There is also the possibility of a natural pregnancy followed by CVS or amniocentesis, these tests would allow you to determine if the fetus carries the DM1 mutations. These tests would allow the patient to terminate the pregnancy or have knowledge of the child's condition before birth. The mother would need to share her views on termination and the value of knowing the diagnosis in utero. The mother could also seek assisted reproductive services. IVF: The patient could use her own eggs and select the embryos not carrying the mutation through PGD. She would need to determine what to do with the unused embryos. The patient could also use donated eggs or embryos. There is also adoption. The patient should know about associated costs as this will determine the feasibility of any of these options. The counseling the physician provides should be both supportive and non-directive. The goal is to match the best available medical option to the patient's personal values and beliefs. You could also refer to a genetic counselor if available. A helpful resource is the National Society of Genetic Counselors (NSGC)

Again referring to her reading on the Internet, his wife asks whether he is diabetic and needs insulin. You remind her that diabetic patients have high glucose levels that could require the use of insulin to regulate and lower the glucose levels. You later investigate the role between glucose and insulin. Do the levels of serum insulin and glucagon change in a normal person having hypoglycemia? How does measuring the level of serum insulin help in determine the cause of hypoglycemia?

If a person with normal physiology becomes hypoglycemic after fasting, the insulin levels are suppressed and the glucagon levels are elevated. In other words, the metabolism starts producing available energy and stops storing energy. If a person with hypoglycemia has a normal or elevated serum insulin level, then an abnormality in insulin regulation has occurred. One cause of hypoglycemia is an excess secretion of insulin. Example, islet cell carcinomas or excess exogenous insulin.

What course of action, if any, should be taken for each condition? If action should be taken describe why. 5ARD AIS SRY

In 5-alpha-reductase deficiency, gender assignment will dictate course of treatment. If female gender is assigned, then gonadectomy is necessary to avoid neoplasia and virilization that is certain to appear at puberty. If a male gender is assigned with a possibility of a functional penis as an adult, then testosterone or dihydrotestosterone therapy should be considered. In androgen insensitivity syndrome, gonadectomy is performed after puberty, allowing growth to be completed since no virilization occurs and this allows for a slow transition into puberty. Since gonadal cancers happen much later in life (20-30's) gonadectomy can be delayed until growth is completed. In SRY deletion/mutation, gonadectomy is necessary to avoid neoplasia, and hormone replacement is necessary.

Compare and contrast the developmental defects in gastroschisis and omphalocele. In your answer, relate both anomalies to normal mid-gut development.

In gastroschisis, midgut herniation, rotation and retraction occurs normally, then the midgut herniates through a rupture of the body wall. In omphalocele, the retraction of the midgut fails, and the midgut persists outside the abdomen after 11 weeks. In omphalocele, the herniated midgut normally remains closely covered by the amnion.

Daniel, a 54 year old ex-smoker, has been taking oral glucocorticoids (i.e. steroids) for 6 months to treat his chronic obstructive airway disease (COPD). Now that his pulmonary symptoms have improved, his physician wants to stop the glucocorticoids. However, every time he goes off these medications, Daniel feels extremely weak, dizzy and has a drop in his blood pressure. How would the above clinical scenario be different from a patient with isolated ACTH deficiency. What would you expect blood levels of cortisol (glucocorticoids) and ACTH (adrenocorticotrophic hormone) to be? What would you expect the levels of corticotropin releasing hormone (CRH) to be in the hypophyseal portal system.

In isolated ACTH deficiency, there is destruction of the pituitary corticotroph cells, so blood ACTH levels will be low. Because of the lack of ACTH stimulation on the cells of the adrenal cortical zona fasciculata there will be little cortisol secretion, adrenal cortex would atrophy after sometime and blood cortisol levels will be low. Because of the lack of negative feedback inhibition from cortisol on the hypothalamus, CRH levels in the hypophyseal portal system will be elevated. ACTH also exerts a "short loop" feedback inhibition on CRH secretion, so the absence of ACTH is another reason for the increased CRH levels in the hypophyseal portal system. So in isolate ACTH deficiency, CRH level would be high; it would be suppressed when hypothalamus is suppressed by exogenous steroid intake.

Severe lung disease and failure to thrive are classic features associated with cystic fibrosis (CF). However, it is now clear that there is a wider range of phenotypes from mutations in the cystic fibrosis transmembrane regulator (CFTR) gene. Milder phenotypes are called "non-classic" or "atypical" CF and may include chronic sinusitis, asthma, recurring pancreatitis, or infertility. The following questions are about the genetics and molecular biology of "non-classic" forms of CF. 74 patients with non-classic CF were tested for mutations in the CFTR gene. These patients had little to no lung disease, sweat chloride levels that range from less than 40 mmol/L (normal) to 90 mmol/L (abnormal), and normal pancreatic function. The testing was done by direct sequencing of the CFTR gene. Of these 74 patients, 29 were found to have a previously reported mutation in each CFTR allele (compound heterozygotes), 15 were found to have only one previously reported mutation or only one novel (previously unreported) mutation, and 30 had no identifiable mutations. For these 30 patients, in whom no mutations in the CFTR gene were detected, what other kinds of cellular factors, when altered, could give rise to a phenotype equivalent to that of CF? In your answer, assume that other forms of pulmonary disease have been excluded and that the CFTR gene sequencing examined the entire coding and non-coding sequences.

It is possible that these patients have a defect in other genes that have a role in CFTR biosynthesis, stability or function. Biosynthesis of the CFTR gene could include additional genes that encode specific factors necessary for proper transcription of the gene. Also, there are likely to be specific protein factors that maintain the stability of transcribed CFTR mRNA. If the genes coding for these factors are altered by a mutation, aberrantly low levels of both CFTR mRNA and protein could result. Factors that modify the CFTR protein may be important. Since CFTR is glycosylated as well as phosphorylated, enzymes and cofactors necessary for the post-translational modification of CFTR protein could also cause a CF-like phenotype and mutations in genes coding for these enzymes and cofactors could also be a cause of non-classic CF. Another possibility for developing a CF like phenotype could be mutations other genes with similar function to CF. As of yet, CF is thought to be a monogenic disease and no other genes have been linked to this phenotype.

Gayle (age 17) and Kiesha (age 16) are two female patients with Turner syndrome that you regularly follow in your adolescent clinic. After recently seeing them on the same day, you notice that these patients have several similar features, such as their short stature, facial structure, and absence of menarche. Both patients have the standard 45,X karyotype. However, Gayle has severe developmental delays when compared to Kiesha and other patients you read about in the Turner syndrome literature. In fact, Gayle has very poor speech development and is in special classes compared to Kiesha who is doing well in regular school. You ask the mother of each of your patients about their family medical histories. Kiesha is one of 4 siblings. She has two full younger brothers and one older half-sister through her mother. Her siblings are all in regular school and have no birth defects or physical disabilities. Her sister menstruated at the normal age. Her parents live at home and stated that there were no diseases running through the family. Gayle is an only child. She has two male first cousins on her mother's side that have been diagnosed with Fragile X syndrome. The sister of these boy cousins is in special school. Their mother, the patient's only aunt, has no health problems. Gayle's mother is healthy, but reported that she struggled in school. Using your knowledge of the X chromosome and inheritance patterns, propose an explanation of the phenotypic differences between Gayle and Kiesha.

Kiesha appears to have the expected phenotype of Turner syndrome associated with her 45,X karyotype. This includes short stature and primary amenorrhea. Her family history is negative for developmental delay and both of her full brothers are normal. Gayle has short stature and primary amenorrhea consistent with 45,X karyotype. Developmental delay can be seen in a small minority of Turner syndrome patients. However, the family history of Fragile X in her maternal male cousins and learning difficulties in her mother, suggests that she may have both Fragile X syndrome and Turner syndrome. Her aunt is an obligate carrier of a mutation in FMR1, which means that her mother has up to a ½ chance of also carrying the same mutation. Based on her position in the family tree, Gayle then has a ¼ chance of carrying the same mutation. If she inherited her single X chromosome with the mutation from her mother, she would not have a normal FMRP expression, and would thus have Fragile X.

In addition to skewed X inactivation, geneticists have considered at least six other mechanisms that can cause females to develop X-linked disease. List and explain at least three of these additional processes.

Manifesting X-linked disorders in females can be caused by several situations, broadly categorized into situations in which there are two X chromosomes and situations in which there is an X and Y chromosome. If the phenotypic female has two X chromosomes, then Skewed X inactivation. See the previous question. Inheritance of a recessive trait. This implies that a) an abnormal paternal and maternal allele are inherited and b) that the father is affected as he only has one X chromosome. Color-blindness is a common example of this mechanism. De novo (new) mutation on either the paternal or maternal X chromosome, and an inherited mutation on the other. If the new mutation is on the paternal chromosome, you would not expect that the father would also manifest disease. Maternal isodisomy X. Disomy is the inheritance of two chromosomes of a chromosome pair from one parent and results from defects in chromosome segregation in meiosis. Isodisomy is more specific, and results in the inheritance of the identical chromosome pair. A defect in meiosis 1 (M1) results in disomy and a defect in meiosis 2 (M2) results in isodisomy. If a. isodisomy occurs in maternal meiosis 2 (M2) for the X chromosome, and b. the mother has a recessive X-linked trait then both Xs would be identical and the female would manifest the disorder. If the phenotypic female has a single X chromosome or a X and Y chromosome: Turner syndrome (45, X). (See question 1 above) An intersex situation where a genotypic male (46,XY) with androgen insensitivity or an inactivating mutation or crossover in the SRY region will develop as a phenotypic female. If the phenotypic female has either an XX or XY chromosomal complement: Fully penetrant dominant trait, inherited either from the maternal or paternal lineage.

In addition to the omphalocele, the ultrasound also showed a fetus that was large for gestational age, had an enlarged tongue (macroglossia). After birth, the infant's growth rate was greater than normal, and he had hemihyperplasia (asymmetric overgrowth). Genetic testing revealed that the infant was mosaic for a duplicate paternal segment of 11p15 (without a maternal copy), confirming a diagnosis of Beckwith-Wiedeman syndrome. Define mosaicism and speculate how genetic mosaicism may be related to hemihyperplasia.

Mosaicism is defined as the occurrence of two or more cell lines with different genetic or chromosomal constitutions in a single individual or tissue. Thus, for this infant, one cell line should have a normal male karyotype (46,XY), and the other cell line showing 11p15 duplication. The body part which exhibit hyperplasia or enlargement might have more cells with the 11p15 duplication than tissues that appear to have normal growth. For example, the cell line with the 11p15 duplication might be in the growth plates of the bones in an enlarged limb, and the normal limb might have normal karyotype.

Jimmy is a 3 year old boy with severe muscle weakness. He was described as "floppy" right after birth and was late in sitting and walking compared to other children his age. His pediatrician told his parents that he had "myopathic" facial features, with ptosis (drooping eyelids), downturned mouth, and little facial expression. A test for congenital myotonic dystrophy (DM) was positive with over 800 repeating units in the DMR1 gene. Jimmy's mother also had minor muscle weakness, ptosis, and unilateral cataract and his maternal grandmother did not have muscle weakness although she did have bilateral cataracts. Describe the inheritance pattern in myotonic dystrophy and explain how the mutation in the DMPK gene may have caused this patient's severe muscle weakness.

Myotonic dystrophy 1 (DM1) is an autosomal dominant form of muscular dystrophy that is caused by triplet repeat expansion in the DMPK gene. In contrast to classic autosomal dominant inheritance, DM1 mutations display both reduced penetrance and anticipation in pedigrees. In addition, individuals with full mutations display pleiotrophy and variable expressivity. This family shows anticipation with each generation developing a more severe phenotype at an earlier age than the previous generation. The Grandmother is the oldest and has the mildest phenotype (cataracts only), her daughter has more features (ptosis and mild muscle weakness) at a younger age, and the young son has the most severe features. DM is a triplet repeat disease, where a CTG repeating unit in the 3' untranslated region of the DMPK gene expands in meiosis of the parent. A genotype-phenotype correlation exists between the number of repeating units and phenotype, where unaffected parents can carry a pre-mutation that places their offspring at risk. Most forms of congenital DM1, like this patient, are caused by expansion of a smaller number of repeats to a larger number of repeats in maternal meiosis.

Name and describe the meiotic mechanism by which the fetus inherited the extra chromosome 21. In your answer, explain whether the error occurred in MI or in MII.

Nondisjunction of the two chromosomes 21 in the meiotic divisions. In Meiosis I nondisjunction, instead of segregating normally to opposite poles, homologous chromosome pairs move together (nondisjoin) to the same pole. The second meiotic division is normal. In Meiosis II nondisjunction, nondisjunction occurs in the second meiotic division. Two sister chromatids of one chromosome do not separate normally to opposite poles but instead, move together (nondisjoin) to the same pole. Interpretation of the microsatellite markers Figure 2 reveals that the nondisjunction event which resulted in Joan and Jeremy's fetus having trisomy 21 occurred in meiosis II (maternal meiosis II nondisjunction error).

Describe the general process by which a karyotype is obtained. Why was the microduplication missed using metaphase FISH?

Obtaining a karyotype is an attempt to arrest chromosome condensation during the process of mitosis, thus it requires living, dividing cells. The mitotic arrest is generally achieved by using chemicals such as colchicine, which destabilizes the microtubules and inhibits mitotic spindle formation. Other techniques have been developed, however, to ensure that the chromosomes do not condense too much. If condensation proceeds too far, the resulting chromosomes are too short, limiting the resolution of the banding pattern and only allowing for a count of the chromosome number. The arrested cells are "dropped" onto a microscope slide in such a way that allows the nuclei to rupture, releasing the chromosomes and allowing them to spread out and be viewed individually. The microscope slide is then stained, usually with Giemsa stain, which allows resolution between heterochromatin and euchromatin. A technician examines the slide, photographs cells that have ruptured in a fortuitous pattern, and then cuts and pastes the chromosomes onto a template such that each chromosome is placed next to its homologous partner. When a FISH probe is used, the process is the same, but the staining utilizes a fluorescently labeled probe that hybridizes to a specific chromosomal region. The presence or absence of the probe is directly assessed under a fluorescent microscope. A good, high resolution karyotype can produce a resolution of approximately 650-850 bands. That number corresponds to the number of alternating light and dark bands that are visible on the chromosomes. A band count of 400 means that the chromosomes are relatively short, and thus, is a lower resolution karyotype. One dark band on a 400 band resolution may resolve into three bands (dark-light-dark) on a 650 band resolution. The microduplication could have been missed in the initial karyotype and FISH because of chromosome condensation. If the chromosomes were very condensed, the two signals present on the duplicated chromosome were physically too close to each other and blended into one cytogenetic band or one FISH signal. When evaluated by other methods that did not condense the chromosomes (interphase FISH) or directly evaluated copy number (CGH microarray), the duplication was found.

Early descriptions of the interstitial microduplications of 22q11 suggested that learning deficits, dysmorphic features, urogenital malformations, and hearing loss were common, but colobomas were never present. Melanie, however, has the same microduplication, but shows a phenotype completely opposite of that description. What kind of study bias might account for this discrepancy?

One of the problems that plagues genetic descriptions is ascertainment bias. Ascertainment bias reflects the process of making generalizations based on a non-representative sample. Most studies of cytogenetic abnormalities are based on patients with obvious anomalies. This can shift or bias the analysis because the group of patients was highly abnormal to start with-otherwise no one would have bothered to study their chromosomes. Patients such as Melanie, who have very few to no findings, are often overlooked and do not have karyotypes drawn. Indeed, Melanie was not evaluated for nine years of her life.

EGFR1 is expressed, and even overexpressed, frequently in lung cancer. Unfortunately, administration of erlotinib is an effective treatment in only a minority of patients with lung cancer. Referring to the multistep model of cancer progression, propose a hypothesis that explains why erlotinib has not been found to be more effective as a sole therapy for lung cancer. Propose a second hypothesis that is not dependent on the multistep model.

One possibility (multistep model) is that the cancer may have been dependent on constitutive activation of EGFR at some point in its development, but the accumulation of additional mutations allows the cancer to grow and survive even in the absence of activated EGFR1 by the time treatment is administered. Another possibility (also multistep model) is that there are redundant signaling pathways, and loss of any one of them does not have a major impact on cell proliferation. A third possibility (also multistep model) is that additional mutations may occur within the EGFR1 gene itself, allowing the receptor to become "immune" to the drug (i.e. mutation in the drug-binding site). A fourth possibility (not dependent on multistep model) is that the drug never gets to the target; for example, it is not absorbed in high enough concentrations, it is metabolized to inactive metabolites, and/or eliminated.

Describe the histopathologic changes in the pancreas seen in patients with type 2 diabetes mellitus and compare this to the pathology seen in a patient with type 1 diabetes. Please identify the images in Figure 1, 2 and 3 as a pancreas tissue section from a normal patient, a type 1 diabetic patient, or a type 2 diabetic patient.

Pancreatic lesions in Type 1 and Type 2 diabetes depend on the stage of disease. Type 1: -Insulitis early, -marked atrophy & fibrosis, -Beta cell depletion Type 2: -No insulitis, -focal atrophy & amyloid deposits, mild Beta cell depletion.

T3 (tri-iodothyronine) is available commercially under the brand name "Cytomel." Occasionally a patient will present to an endocrinologist with a puzzling form of thyrotoxicosis that turns out to be due to the patient's secret ingestion of excessive amounts of thyroid hormone pills. This is a psychiatric illness. The patient takes the excessive thyroid hormone for the secondary gain of having people worry over them and take care of them because they are sick. It is called "thyrotoxicosis factitia." Suppose a patient were secretly taking excessive amounts of Cytomel. What would the patient's blood levels of free T4, free T3, and TSH be?

Patient would have elevated T3, low T4, and low TSH in serum. The patient would still be clinically thyrotoxic because the body would be exposed to an excessive amount of T3, which is an active form of thyroid hormone.

The 22q11 region can be duplicated or triplicated, and can sometimes be found as a freestanding extra chromosome containing two copies of the 22q11 region (47,XX,inv dup pter22->22q11.2). The phenotypic spectrum is highly variable. Severe presentations include the cat-eye syndrome, which features mental retardation, bilateral coloboma, cleft palate, kidney defects, heart defects, and anal atresia. What might be the mediators of these phenotypic differences?

Penetrance is defined as the proportion of individuals who are known to be carriers for a specific genetic mutation that are affected. For example, if 70% of individuals who harbor gene mutations show any signs or symptoms of the condition, then the gene mutation is 70% penetrant. Although penetrance is a population based term, it is sometimes applied to an individual who either does or does not have findings of the condition. If a mutation carrier is asymptomatic, the person is incompletely penetrant or non-penetrant. In this context, penetrance is an all or nothing concept; the patient either has signs or symptoms (penetrant) or doesn't have signs or symptoms (non-penetrant). Given that Melanie's is mildly affected without other more severe manifestations, one can conclude that there is variable expression of the phenotype and not reduced penetrance. Variable expression may relate to either the severity of the defect or to the presence or absence of various features of the condition. Variable expression is only relevant for persons who are affected, displaying any feature of the disorder. The 22q duplication disorder is also pleiotropic, since the same molecular defect can cause anomalies in multiple different and unrelated systems. The mediators of the phenotypic differences may include: 1. The number of copies of the duplicated segment. Higher copy numbers correlate with a higher frequency of heart defects. 2. The length of the duplicated segment and how many additional genes it contains. Although the suggested paper states that this does not seem to be the case for interstitial duplications, the Cat-Eye syndrome does contain a larger duplicated region and is more prone to cardiac defects. 3. Genetic variation of genes within and outside of the region. Genes outside of the region may be involved in the phenotypic expression. However, one can also consider that differences in expression patterns or functional polymorphisms may have their effects amplified by being duplicated.

As indicated in this case, there was metastasis to a lymph node. Compare and contrast the ways that primary cancers can disseminate throughout the body. Is it relevant that only one lymph node is involved?

Primary cancers are either carcinomas or sarcomas. In general (and there are exceptions to the rule) carcinomas spread via lymphatic routes and sarcomas spread via blood vessels. In this case, the melanoma behaves in a manner similar to a carcinoma and the spread is to a local lymph node (N1). Since it is in one lymph node this is a much better scenario (lower stage) than in multiple lymph nodes although it would be better if there was no nodal involvement (N0).

Alan is a currently healthy man, age 29, who has four children ages 8 months to 6 years old. Because Alan was treated for retinoblastoma as an infant with enucleation of his right eye, his children's pediatrician recommended careful ophthalmologic screening of each of them from early infancy. His two oldest children appear not to have been afflicted, however, his two youngest did develop bilateral retinoblastoma. Each of their tumors was treated with conservative measures, and all are doing well with preserved vision in both eyes. Of note, Alan's siblings (three), parents, and parents siblings (seven in all) were unaffected by this disease. Discuss the pattern of inheritance of retinoblastoma. Can a founder be identified? Comment on the penetrance of the mutation.

Retinoblastoma occurs in family clusters, or sporadically. Cases can also be separated into bilateral and/or multifocal disease and unilateral disease. Bilateral or multifocal disease essentially always occurs in the setting of a germline mutation which inactivates a single allele of RB. This can either be inherited from a parent who carries a germline mutation (and would almost certainly have been affected by the disease) or can occur de novo in the index case. In the latter situation, the de novo mutation would most likey have occurred during meiosis of either the paternal or maternal sex cell. Once an individual receives the de novo mutation from a parent, all constitutional and germline tissue would harbor the mutation(so in some respects, is still inherited). Although mutation of this allele results in a loss of function of Rb, the disease is inherited in an autosomal dominant pattern with a high degree of penetrance. Sporadic retinoblastoma is not heritable. Inactivation of both alleles of the RB gene occurs only in the somatic cell giving rise to the tumor. In the case presented here, the disease is inherited from the father, Alan. As usual, the pattern of disease in Alan's children occurs with a frequency consistent with its autosomal dominant nature, and as is typical, is highly penetrant.

Spina bifida is associated with damage to the spinal cord, most commonly at lumbar and sacral levels. In what ways might this impact the pelvis and lower extremities?

Spina bifida can result in paraplegia, incontinence, and sexual dysfunction. Pelvic viscera receive autonomic innervation, both sympathetic and parasympathetic, that originates in lumbar and sacral levels of the spinal cord. The sympathetic innervation is derived from preganglionic neurons residing in the lateral horn of the gray matter at caudal thoracic and upper lumbar cord levels. Parasympathetic innervation originates in the lateral horns of the gray matter at the 2nd, 3rd, and 4th sacral cord levels. The lower extremity is innervated by nerves that arise from roughly the 2nd lumbar to 3rd sacral spinal cord levels.

T3 (tri-iodothyronine) is available commercially under the brand name "Cytomel." Occasionally a patient will present to an endocrinologist with a puzzling form of thyrotoxicosis that turns out to be due to the patient's secret ingestion of excessive amounts of thyroid hormone pills. This is a psychiatric illness. The patient takes the excessive thyroid hormone for the secondary gain of having people worry over them and take care of them because they are sick. It is called "thyrotoxicosis factitia." What would a radioactive iodine uptake and radioactive iodine scan reveal?

Since the thyroid gland is inactive when it is not exposed to thyroid stimulating hormone, and iodine uptake by thyroid gland is TSH dependent, the radioactive iodine uptake will be very low. The thyroid scan would show homogeneous but barely detectable uptake in the thyroid gland.

Kelli is a 25 year old female who presented for a first prenatal visit at 16 weeks gestation. She says she has not experienced any problems with the pregnancy up to this point. She is not taking any medications, including prenatal vitamins. Her history is significant only for migraines, which she was taking valproic acid for up until 10 weeks, when her doctor told her to stop all medications. She decides to undergo First Quad screening, which was negative for trisomy, but showed an elevated msAFP. She was told that her baby has spina bifida. Kelli asks several questions. Please formulate answers to her following questions: What is spina bifida?

Spina bifida is a common neural tube defect (NTD) of the spinal cord, often occurring in the lumbar region. The vertebrae do not completely cover the spinal column and tissue protrudes from the back. There is damage to the nerves. It can be associated with partial or complete lower limb paralysis, loss of bladder and bowel control and hydrocephalus.

What stage is Ms. Monroe's colon cancer (base this on her more-advanced tumor)? Is this stage curable? What is the rationale for chemotherapy for Ms. Monroe's cancer?

Stage III is enough (stage IIIb: T3N1M0 to be exact). Ms. Monroe has curable disease. She may have been cured with surgery but alternatively could have microscopic disease left behind. Chemotherapy in this setting is "adjuvant" i.e. it is being administered to help surgery cure cancer. Ideal answer should include this concept: the goal of adjuvant chemotherapy is to kill microscopic disease and increase the cancer cure rate.

Describe the mechanism by which the 22q11 region is duplicated or deleted.

The 22q11.2 region is flanked by highly homologous low-copy repeat (LCR) regions. A low copy repeat region is a stretch of DNA that has been duplicated elsewhere in the genome, often on the same chromosome. The chromosomal region lying between the LCRs may be several megabases in length and can contain genes within it. It is "low copy" because typically there are only a small number of that specific LCR within the genome. During meiosis I, homologous pairing between chromosomes occurs. During this process, homologous LCR regions that are located in different chromosomal regions can associate though misalignment and pair homologously. If a crossover event occurs across a misaligned, homologously paired region, two derivative chromosomes 22 will occur: one will have a duplication of the 22q11.2 region and the other will have a deletion of 22q11.2.

Is the gene involved a proto-oncogene, tumor suppressor gene, or caretaker gene? Explain why you came to your conclusion.

The APC gene is a tumor suppressor gene. While its normal function is not fully understood, mutations that decrease its function, such as truncation mutations seen in FAP, lead to an increased risk of cancer. APC forms a complex with GSK-3beta that binds beta-catenin. If APC is mutated, it cannot bind beta-catenin, which then is free to move to the nucleus, where it acts as a transcription factor, resulting in cellular proliferation. Both alleles of a tumor suppressor gene must be lost or inactivated to completely lose its function. Inheriting one mutant APC gene is not sufficient to initiate tumor formation. The second APC allele must be deleted or inactivated through somatic events in order to develop FAP. Students who answered MYH polyposis for question 1 should answer "caretaker gene" for question 2. The MYH gene is a base excision repair gene and is involved in repairing DNA that has suffered oxidative damage. As it maintains genomic integrity, it is a caretaker gene.

In your reading on the interplay between glucose and insulin you also find studies that have examined the genes that control insulin secretion. The myotrophin (MTPN) gene is involved in the positive regulation of insulin secretion. Propose a molecular mechanism based on miRNA regulation of MTPN that might lead to an elevated serum insulin level.

The MTPN gene is regulated in response to glucose levels. This regulation is mediated by miRNAs (specifically, miR-375). If the negative, miRNA-mediated regulation of MTPN translation was altered, insulin secretion would probably be effected. If miRNA synthesis was decreased there would be less miR-375 available to downregulate MTPN, leading to an increase in insulin secretion.

Another appealing target in the treatment of solid tumor malignancies is the vasculature. Blood vessel recruitment (angiogenesis) may be less heterogeneous in mechanism than progression of the cancer cells themselves, as cells that make up the vasculature are not derived from a malignant clone with countless genetic aberrations. Identify an FDA-approved therapeutic that targets angiogenesis. (Hint: it is approved for treatment of both colorectal cancer and non-small cell lung cancer.) Is this drug a large molecule (monoclonal antibody) or a small molecule inhibitor? On this basis, how must it be administered? What is its target, and how would inhibiting its target abrogate angiogenesis? How is this target different in principle from the targets of trastuzumab and erlotinib?

The angiogenesis-targeted therapeutic approved for use in colon cancer and lung cancer is bevacizumab. This agent is a monoclonal antibody, and therefore must be administered intravenously. Bevacizumab is an antibody that binds to circulating vascular endothelial growth factor (VEGF). VEGF is the ligand that activates a cell-surface receptor on proliferating endothelial cells (the cells that comprise capillaries, and are the first cells to appear and direct growth of new blood vessels). This is different from the targets of trastuzumab and erlotinib, which bind and inhibit receptors, not the ligands that bind to the receptors.

Two of the symptoms that can differentiate hyper- and hypothyroidism are amenorrhea and menorrhagia, respectively. What is the blood supply to the uterus? To the ovary? Describe the route taken by the blood vessels involved from their origins to the structures they supply. Include anastomotic connections.

The blood supply of the uterus is derived mainly from the uterine artery, a branch of the internal iliac artery. Branches of the uterine artery that supply the upper uterus and uterine tubes anastomose with branches of the ovarian artery, a branch of the abdominal aorta. The uterine artery is a branch of the anterior division of the internal iliac artery. It travels to the uterus in the base of the broad ligament, crossing the ureter superiorly ("water under the bridge") to reach the junction of the body of the uterus and the cervix. It then ascends along the lateral margin of the uterus to reach the uterine tube; here, it turns laterally and travels along the uterine tube to anastomose with the ipsilateral ovarian artery. Through its anastomotic connections, the uterine artery participates in the arterial supply of the ovary and vagina. The ovarian artery originates from the abdominal aorta and descends in the posterior body wall to cross the pelvic brim and supply the ovary. The ovarian vessels raise a "bump" in the peritoneum that is called, inaccurately, the suspensory ligament of the ovary. The ovarian artery continues medially, beyond the ovary, along the uterine tube to anastomose with the uterine artery.

How does spina bifida occur? (Please include a description of the normal embryologic process, as well as a description of what went wrong.)

The spinal cord forms from the posterior neural tube. The neural tube arises from the neural plate, which folds and seals to make the neural tube and cover itself with surface ectoderm. When the neural tube fails to close, the vertebrae cannot form around the spinal column, and the spinal cord is exposed to an abnormal environment, and nervous tissue degenerates.

A 64 year old woman presents to your clinic with a small, hard mass in the upper lateral quadrant of her left breast. After biopsy, you diagnose her with Her2/neu positive invasive ductal carcinoma and recommend a mastectomy and treatment with trastuzumab. Describe the anatomy of the breast.

The breast, or mammary gland, is a modified sweat gland specialized for milk production. It consists of glandular and connective tissue elements, and both of these tissue types are subject to endocrine regulation. The breast lies in the superficial fascia overlying the pectoralis major muscle; a portion of its superolateral quadrant, the axillary tail, extends into the axilla. The retromammary space lies between the breast and the deep fascia overlying the pectoralis major muscle. Connective tissue thickenings, or condensations, that extend from the dermis of the skin deeply to the pectoral fascia are known as suspensory ligaments (known clinically as Cooper's ligaments). The spaces demarcated by the suspensory ligaments are filled with variable amounts of adipose tissue, in which is embedded the parenchyma of the gland -- a series of lobules of glandular epithelium and ducts. The ducts converge to form lactiferous ducts (about 15-20 per breast), which travel superficially in the breast and dilate to form lactiferous sinuses before opening onto the nipple. The portion of pigmented skin surrounding the nipple is called the areola and contains sebaceous glands that aid in nursing (note that the lactiferous ducts open on the nipple itself, not the areola).

Is there anything I did or didn't do that caused this?

The causes of spina bifida include genetic and environmental factors. Among the environmental factors, valproic acid early during pregnancy is associated with an increase in the incidence of spina bifida. High doses of valproic acid cause neural tube defects in mice. Supplementation with dietary folate reduces the incidence of neural tube defects in the population, and can suppress neural tube defects in some mouse models. There are likely to be many different genes which contribute to neural tube defects, as there are a large number of gene mutations in mice which have partially penetrant neural tube defects.

After finding out the result of the radioactive iodine uptake and scan, what features in the patient's history, physical examination, and laboratory testing might help narrow the differential diagnosis (i.e. What are the possible diagnoses?) to enable you to figure out why the patient was thyrotoxic? Explain your answers.

The differential diagnosis of thyrotoxicosis with a low radioactive iodine uptake in the thyroid includes: 1) painless thyroiditis, 2) subacute thyroiditis, 3) iodine-induced thyrotoxicosis, 4) thyrotoxicosis factitia, 5) struma ovarii, and 6) widely metastatic well-differentiated thyroid cancer. Struma ovarii is an ovarian dermoid tumor containing thyroid tissue. Metastatic differentiated tumors would produce T3. You can rule these out during the radioactive iodine scan by looking to see if there is any abnormal area of radioactive iodine uptake in the patient's pelvis or bones/lungs. Iodine-induced thyrotoxicosis could be ruled out by eliciting a negative history for ingestion of iodine containing drugs like amiodarone or recent exposure to x-ray contrast dye (which contains iodine). 24-hour urine iodine level that was not elevated would also rule out iodine-induced thyrotoxicosis. Painless thyroditis and thyrotoxicosis factitia cannot be ruled out without further testing. In subacute thyroiditis the thyroid gland is typically enlarged, painful, and tender. The absence of an enlarged, painful, tender thyroid gland would make subacute thyroiditis unlikely.

Tammy ultimately developed a malignancy of the cervix. Please name four groups of lymph nodes in the pelvis; describe their locations and the areas of the pelvis they drain.

The four lymph node groups in the pelvis are: (1) external iliac lymph nodes, (2) internal iliac lymph nodes, (3) sacral lymph nodes, and (4) common iliac lymph nodes. (1) The external iliac lymph nodes receive lymph mainly from the inguinal lymph nodes as well as lymph from areas relatively anterior and superior in the pelvis, e.g., lower ureter, urinary bladder, lower uterus. They lie along the pelvic brim along the external iliac vessels. (2) The internal iliac lymph nodes drain somewhat inferior areas in the pelvis, e.g., prostate gland, vagina, lower rectum and anal canal; these nodes are located posteriorly in the pelvic cavity, around the anterior and posterior divisions of the internal iliac artery. (3) The sacral lymph nodes, which lie in the concavity of the sacrum, drain the posteroinferior pelvic viscera. All the lymph nodes mentioned so far drain into the (4) common iliac lymph nodes. The nodes are located along the common iliac blood vessels, and provide a gateway for drainage from the lower extremity and pelvis to the abdominal lymph nodes.

The patient is concerned about his shaking and jittery spells. You explain to him the normal response to low glucose levels and how it can cause his symptoms. In your answer, list the normal hormonal response to hypoglycemia and discuss how this response causes his symptoms.

The hormones secreted in response to hypoglycemia are called "counter-regulatory regulatory" hormones. Goal is to increase blood glucose. -The first or immediate hormone response is an increase in epinephrine and glucagon levels. -A delayed response is seen in increased level of serum cortisol and serum growth hormone. The adrenergic symptoms occur as a result of increase in adrenergic hormone epinephrine. Symptoms include palpitation, hunger, increased sweating and shakiness. The neurogenic symptoms or neuroglycopenic symptoms can result if the hypoglycemia persists. These symptoms include headache, slow response, drowsiness, confusion and coma.

One common feature of Turner syndrome is primary amenorrhea due to underdeveloped, or streak, gonads. Describe the normal internal female genitalia, including the important features of each structure.

The internal female genital organs include the uterus, the fallopian tubes, and the ovaries. The uterus has thick walls composed of smooth muscle; the lining of the uterus, the endometrium, is where a fertilized oocyte will embed itself and develop until birth. Grossly, the uterus is described as having four parts: 1) the fundus - the rounded superior portion of the organ that is situated above the level of the entrances to the uterine tubes; 2) the body - the majority of the organ between the fundus and the isthmus; 3) the isthmus - the narrowed portion connecting the body to the cervix; and 4) the cervix - the inferior portion of the uterus that projects into the vagina. The cervix contains the slit-like cervical canal that opens superiorly into the main cavity of the uterus via the internal os and inferiorly into the superior part of the vagina via the external cervical os. The uterus is normally in an anteverted, anteflexed position, such that its anterior surface rests on the posterosuperior surface of the bladder. The fallopian tubes (uterine tubes) extend bilaterally from the uterus to the ovaries and connect the uterine cavity to the peritoneal cavity. A uterine tube is described as having four parts. From proximal to distal, these are the uterine part (or intramural part, the part that traverses the wall of the uterus), the isthmus (a narrowing in the tube), the ampulla (the widest portion and the usual site of fertilization), and the infundibulum (the funnel-shaped, terminal portion of the tube that presents numerous, finger-like projections, the fimbriae that help draw the ovulated oocyte into the tube). The ovary lies posterior and lateral to the fimbriae of the uterine tube. It is a firm, almond-shaped organ usually about 3.5 cm long, 1.5 cm wide and 1 cm in thickness.

The loss of a tumor suppressor gene is a common cause of many types of cancers. Imagine a 32-year old patient coming into your clinic complaining of testicular pain. You find that he has a painful scrotal mass and an elevated AFP. Suspecting a possible mixed germ cell tumor (the most common adult testicular cancer), you decide to do a biopsy. During the biopsy, what are all the layers of the scrotum that the needle must pierce? From what are each of these layers derived?

The layers of the scrotum from superficial to deep are skin, tunica dartos (subcutaneous tissue and the dartos muscle), external spermatic fascia, cremaster muscle and fascia, internal spermatic fascia, and the parietal and visceral layers of the tunica vaginalis testis. The dartos fascia is a continuation of Scarpa's fascia in the abdomen. The smooth muscle (dartos muscle) functions to wrinkle the scrotal skin when cold, bringing the testes closer to the body. The external spermatic fascia is derived from the aponeurosis of the external oblique muscle. The cremaster muscle and fascia are derived from the internal oblique muscle and its corresponding aponeurosis. This muscle participates in the cremasteric reflex during which sensation on the superomedial portion of the thigh results in contraction of the muscle, pulling the scrotum and testis closer to the body. An absence of this reflex can indicate lower motor neuron disorders or testicular torsion. The internal spermatic fascia is derived from the transversalis fascia (note that the transverses abdominus muscle does not contribute to layers of the scrotum). The tunica vaginalis is derived from the peritoneum. Like the peritoneum, it has both a parietal and visceral layer with a small fluid-filled space in between (this is also the site of a hydrocele - an accumulation of fluid that can result from infection or injury and can be diagnosed using ultrasound). The layers of the scrotum can be remembered using the mnemonic, "Some Damn Englishman Called It The Testes," -> Skin, Dartos, External spermatic fascia, Cremaster, Internal spermatic fascia, Tunica vaginalis, Tunica Albuginea * * The tunica albuginea is the tough fibrous covering of the testes that has fibrous septa projecting inward between lobules of the seminiferous tubules. It is NOT considered a "layer of the scrotum", but the needle would necessarily have to pass through it in order to obtain a sample of the above tumor.

As a 3rd year medical student, you are assigned to interview and examine two patients in the neonatal intensive care unit (NICU). The first infant is a six hour old 3200 gram (between 25-50%tile) boy born at 37 weeks gestational age to a 24 year old otherwise healthy mother. An abdominal wall defect identified as a gastroschisis was seen on a prenatal ultrasound done in the second trimester. Your exam shows a non-dysmorphic infant with intestines extruding through a defect on one side of the umbilicus. The second infant is day of life 4 boy born to a 28 year old mother at 34 weeks gestation. The infant was large for dates and had an omphalocele repaired soon after delivery. He had a large tongue and creases on his ear lobe. An ultrasound of his abdomen showed large liver and kidneys. Other than his surgery, you find that he has been treated for hypoglycemia and that a karyotype is pending. Describe normal mid-gut herniation, rotation and retraction during embryonic development. In your description, include the germ layer origins of the cells of the mid-gut, and the gestational age at the major steps.

The midgut lining is derived from the endoderm, the midgut smooth muscle and connective tissue derive from the splanchnic mesoderm and the enteric nervous system derives from the neural crest. At 6 weeks, a midgut loop herniates into the umbilicus. From 7 to 9 weeks the midgut undergoes a ninety degree rotation and at 10 weeks the midgut retracts into the abdomen.

Shirley Monroe is a 52 year old woman who was well except for hypertension (high blood pressure) until she developed fatigue over the past three months. Her primary care physician found her to have anemia (low red blood cell count) and also to be iron-deficient (suggesting blood loss as the cause of anemia). She had never been screened for colon cancer. He referred Ms. Monroe for a colonoscopy to look for a source of blood loss. The colonoscopy revealed a large mass in the right colon arising from a polyp as well as at least 40 other polyps. Biopsies of the mass revealed invasive adenocarcinoma. Ms. Monroe was referred to a colorectal surgeon. The surgeon was concerned that Ms. Monroe may have a genetic-type of colon cancer, which should be treated with a total colectomy (removal of the whole colon) rather than the usual right-sided colectomy that would be performed for a sporadic-type of colon cancer. However, Ms. Monroe's family history was notable only for her father dying of lung cancer at age 80. As there was not enough time for genetic testing before surgery, the surgeon refered Ms. Monroe for an upper endoscopy to evaluate her stomach and duodenum. She was found to have numerous fundic gland polyps in the stomach and a polyp in her duodenum, which was removed and was an adenoma (a premalignant lesion). The surgeon performed a total colectomy. The pathologic findings from surgery were 1) the large mass was an invasive adenocarcinoma arising in a polyp and extending through the muscularis propria into the pericolonic tissue. Three of 88 lymph nodes identified contained the same cancer 2) a second adenocarcinoma that also arose from a polyp but invaded only into the submucosa and 3) at least 40 polyps that were adenomas. A CT scan of the chest through pelvis revealed no evidence of metastases. Ms. Monroe's oncologist recommended that she be treated with chemotherapy for 6 months. He also referred her to Cancer Genetics for counseling and testing. 1 What features of this case suggest an inherited type of colon cancer and what type of inherited colon cancer do you suspect Ms. Monroe has? Explain how you came to your conclusion.

The patient is a little younger than the average age of a person with sporadic colon cancer (median 71 years of age per SEER database). The cancer is associated with many more polyps than typically seen with sporadic colon cancers (usually < 10). The patient has two colon cancers. (I would give credit for having 2 of these 3 points) This patient actually had attenuated Familial Adenomatous Polyposis (FAP). Her excessive number of polyps makes this a polyposis syndrome but she does not manifest 100's-1000's of polyps as seen in classic FAP. The presence of fundic gland polyps and duodenal polyps is associated with FAP (attenuated or classic). Another correct answer for this case would be MYH-associated polyposis, which is a more recently recognized polyposis. Patients with MYH-associated polyposis typically have 10-100 colon polyps and can have fundic gland and adenomatous polyps in the stomach. MYH -associated polyposis demonstrates a recessive pattern of transmission. Lynch Syndrome (or HNPCC), however, would be incorrect, as it is not a polyposis syndrome.

Daniel, a 54 year old ex-smoker, has been taking oral glucocorticoids (i.e. steroids) for 6 months to treat his chronic obstructive airway disease (COPD). Now that his pulmonary symptoms have improved, his physician wants to stop the glucocorticoids. However, every time he goes off these medications, Daniel feels extremely weak, dizzy and has a drop in his blood pressure. You take a complete history and perform a physical exam, noting that Daniel does not have skin hyperpigmentation. What other signs and symptoms might he have during the history and physical? Why is skin hyperpigmentation, which is a finding in patients with primary adrenal insufficiency, not found in Daniel?

The patient may have symptoms such as lethargy, easy fatigability, anorexia, nausea, vomiting, abdominal pain, diarrhea, orthostatic dizziness, myalgias and arthralgias. The physical examination would reveal an orthostatic drop in the blood pressure and rise in the pulse, suggesting hypovolemia. The patient would not be hyperpigmented because he has hypothalamic or secondary form of adrenal insufficiency. The levels of ACTH would be low in hypothalamic adrenal insufficiency, because the hypothamalo-pituitary-adrenal (HAP) axis is suppressed. In primary adrenal insufficiency, which could be caused by destruction of the adrenal glands, the lack of glucocorticoid production leads to an absence of negative feedback on ACTH secretion and a rise in serum ACTH. The elevated ACTH can cause skin hyperpigmentation.

What other abnormal laboratory tests (related to hemoglobin, white cells and blood chemistry) might you expect in patient with isolated ACTH deficiency?

The patient would be expected to have many of the typical laboratory abnormalities seen in chronic adrenal insufficiency. This would include a low serum sodium concentration, a mild anemia, and an increase in certain white blood cells called "eosinophils". The blood sugars might also be low. However, the acidosis and high potassium levels seen in primary adrenal insufficiency are caused by loss of another adrenal hormone, aldosterone. ("Primary" adrenal insufficiency is caused by destruction of the adrenal glands themselves.) In secondary adrenal insufficiency, aldosterone secretion is normal. ("Secondary" adrenal insufficiency is caused by dysfunction of the corticotroph cells in the pituitary gland.) For this reason, acid and potassium levels are normal in secondary adrenal insufficiency.

Now consider the following: Tammy is a 21 year old premed student recently diagnosed with malignant melanoma of the skin. After having the diagnosis made by biopsy, she underwent a wide local excision and lymph node dissection. Although the primary skin lesion was tiny, there was evidence of a single metastatic focus of disease in one lymph node. At her oncologist's recommendation, she is now receiving adjuvant interferon treatments to try to improve her risk for recurrence. Familial melanoma has been associated with a mutation in the p16/INK4a gene, whose protein product is an inhibitor of cyclin dependent kinase 4 (CDK4). Tammy carries a mutation in this gene. Address the following 3 issues: 1) Based upon the above pedigree, postulate possible patterns of inheritance of the p16 mutation. Comment on the founder, penetrance, carriers. Are currently unaffected family members at risk? Which ones? 2) Based upon p16's putative cellular function, describe it in terms of a protooncogene, oncogene, or tumor suppressor gene. Discuss p16's influence on Rb activity. 3) Although p16 and Rb mutations cause related defects in cell cycle regulation, they lead to very different patterns of cancer. Provide a hypothesis to explain this observation.

The pedigree is not large enough to provide conclusive evidence as to the pattern of inheritance. The two possibilities, both consistent with the pedigree, are autosomal dominant inheritance with reduced penetrance, or autosomal recessive. It is not likely to be X-linked because Tammy's mother was not affected, and her cousin appears to have inherited the gene through her unaffected father. The founder mutation is either Tammy's grandmother, or more likely is an ancestor not shown on the pedigree. Because the most likely pattern of inheritance (and in fact the correct pattern for p16) is autosomal dominant with low penetrance, approximately half of Tammy's sibs, and her mother and sibs are likely to be carriers. Any of them are at higher risk of developing melanoma and pancreatic cancer than the general population. p16 protein, also called p16INK4a, is an inhibitor of CDK 4 and 6. Its tumorigenic function is most likely mediated via its inhibition of CDK4 activity. Presumably, loss of the inhibitory action of p16 on cyclinD1/CDK4 complexes leads to increased phosphorylation of Rb. This in turn lessens the inhibitory effect of Rb on E2F-stimulated transcription of genes involved in the G1/S transition. p16 is therefore a tumor suppressor gene. There is no set answer here. One hypothesis is that the key limiting steps or pathways for cell cycle checkpoint control are heterogeneous in different cell types and even in the same cell type at different levels of development. A case in point is Rb in retinoblastoma. Only in the developing cells of the retina is there extreme sensitivity to Rb dysfunction. Once those cells reach maturity, their sensitivity to loss of Rb apparently disappears. In addition, there is a great degree of redundancy of these control mechanisms. Other somatic cells are sensitive, but over a much longer timeframe. Another concept is that of "necessary and sufficient." Many tumors display aberrations, if not outright loss, of function of Rb and p16. This loss of function may be necessary for uncontrolled proliferation, but only in the context of additional mutations; i.e. loss of Rb or p16 is frequently necessary for proliferation, but rarely sufficient for full-blown malignancy. The degrees to which this holds true for specific combinations of defects is clearly cell-type specific.

Alan's wife is concerned about their future grandchildren. What are the risks for developing retinoblastoma for Alan's grandchildren?

The risks for developing cancer can be determined by the chance that the individual has a gene mutation multiplied by the penetrance for the condition. The risks for Alan's grandchildren to develop retinoblastoma will vary greatly, depending upon the gene mutation status and age of the grandchild. Each offspring of Alan's two affected children have a 50% chance of inheriting the RB mutation. If they do, they have between a 95-98% chance of developing retinoblastoma. The children of Alan's unaffected children will have a risk that is average for the population, except in the very rare event that this parent was a non-penetrant carrier.

Mytonic dystrophy is associated with proximal muscle weakness, which refers to muscles that are related to the trunk, shoulders, and hips, as opposed to distal limb muscles. A number of muscles that serve as lateral rotators of the thigh at the hip joint originate on pelvic structures and attach to the femur. Two of these muscles are the piriformis and obturator internus. They are innervated by branches of the sacral plexus (more in Block 5). The piriformis m. exits the pelvis through the greater sciatic foramen; the obturator inernus m. exits the pelvis through the lesser sciatic foramen. What are the boundaries of these exit pathways? What neurovascular structures pass through these exits? What is the course of the pudendal nerve, and what does it innervate?

The sacrotuberous ligament converts the greater sciatic notch into the greater sciatic foramen. The sacrospinous and sacrotuberous ligaments convert the lesser sciatic notch into the lesser sciatic foramen. The greater sciatic foramen transmits the superior gluteal vessels and nerve above the piriformis. Inferior to the piriformis pass the inferior gluteal vessels and nerve, sciatic nerve, pudendal nerve, and internal pudendal vessels. (...and a few other small nerves you will learn about later in the curriculum. ) The lesser sciatic foramen transmits the internal pudendal vessles, pudendal nerve, and the nerve to the obturator internus m. The internal pudendal vessels and pudendal nerve exit through the greater sciatic foramen, hook around the ischial spine, and enter the perineum through the lesser sciatic foramen. The pudendal nerve innervates structures of the perineum providing sensation to the external genitalia and motor branches to perineal muscles, including the external urethral sphincter and external anal sphincter.

Cystic fibrosis results in the majority of men (97%) who have it being infertile, but not sterile, due to congenital bilateral absence of the vas deferens (but fully functioning testes). These men can reproduce, but only with the help of assisted reproductive techniques. What structures comprise the reproductive tract through which sperm pass from the testis to the outside world? Include a description of the course of the reproductive tract of the male from testis to penile urethra.

The seminiferous tubules (which manufacture sperm) in the testis empty into the rete testis ("straightened" portions of the seminiferous tubules), which, in turn, empty into about 10 ductus efferentes (efferent ducts of the testes) that actually form the head of the epididymis. These coalesce to form the duct of the epididymis, a highly coiled structure that measures 4 to 6 meters in length, that serves as a reservoir for sperm, and that forms the body and tail of the epididymis. The tail of the epididymus continues as the ductus deferens (vas deferens). This muscular tube provides the peristaltic impulse that generates the force of ejaculation. It ascends in the scrotum, traverses the inguinal canal, emerges from this canal below the peritoneum and crosses the external iliac vessels and the pelvic brim to enter the pelvis. Its course is then inferior and medial to reach the posterior aspect of the urinary bladder. It crosses above and medial to the ureter (water under bridge),reaches the posterior aspect of the prostate gland, and enlarges to become the ampulla of the ductus deferens. (The ampullae of the two sides nearly touch one another in the midline.) The ampulla of the ductus deferens unites with the duct of the seminal vesicle to form the ejaculatory duct. The secretions of the glandular epithelium of the seminal vesicle are discharged into the ejaculate by the muscular walls of the seminal vesicle. (Spermatozoa are not stored in the seminal vesicle.) The ejaculatory ducts open into the prostatic urethra in its posterior wall as tiny slits in the seminal colliculus, a swelling beneath the posterior wall of the prostatic urethra. Numerous ducts of the prostatic glands empty into the prostatic urethra on either side of the prostatic colliculus. The prostatic urethra becomes the short membranous urethra immediately inferior to the prostate gland as the urethra passes through a sheet of skeletal muscle that constitutes the external urethral sphincter. Finally, sperm pass through the penile urethra.

Compare and contrast type 1 and type 2 diabetes mellitus with regard to the symptoms that are seen when the disease first develops.

The symptoms of type 1 and type 2 diabetes mellitus are the same. Symptoms of a high blood sugar: polyuria, polydipsia, blurred vision, and fatigue. Weight loss is a feature in the presentation of most type 1 diabetics and some type 2 diabetics, but this is simply a reflection of the severity of the insulin deficiency. -In general, symptoms tend to present more explosively in type 1 than in type 2 diabetes mellitus. The symptoms in type 2 diabetes mellitus may also have been present for a much longer period of time than in type 1 at the time of diagnosis. -Type 1 diabetes mellitus frequently present in a severe state of metabolic derangement called "diabetic ketoacidosis" due to the increased hepatic synthesis of ketoacids brought on by absolute insulin deficiency.

Compare and contrast type 1 and type 2 diabetes mellitus with regard to treatment.

The treatment of type 1 diabetes mellitus always includes insulin. Treat type 2 diabetes mellitus either by decreasing the insulin resistance or by increasing the amount of insulin in the body. -Insulin resistance can be decreased non-pharmacologically by a combination of weight loss and regular exercise. -Insulin resistance decreased: metformin, the glitazone class such as pioglitazone and rosiglitazone. -Insulin levels increased by: sulfonylurea and meglitinide class, exenatide, by DDP-4 inhibitor sitagliptin (increases the action of GLP1) or by injections of insulin itself.

Because Ms. Long's child has an abdominal wall defect, he or she will be at an increased risk for future abdominal wall injuries - namely, ventral hernias. What are the two major types of hernias that occur in the region of the groin? How are they differentiated? Which is most common in a previously healthy male and which in a female? What are the deep and superficial inguinal rings?

The two most common types of groin hernias are inguinal hernias and femoral hernias. Protrusion of abdominal contents through the anterior abdominal wall inferiorly may occur through the inguinal canal or through the femoral canal. The inguinal canal permits structures to pass between the abdominopelvic cavity and the scrotum/labium majorum; the femoral canal allows structures to pass between the abdominopelvic cavity and the thigh. Inguinal hernias occur above the inguinal ligament and femoral hernias occur below the inguinal ligament. Femoral hernias are always "acquired" hernias and constitute only about 4% of all groin hernias; they are more common in females. Inguinal hernias may be congenital (present at birth) or acquired and are more common in males, mainly because the descent of the testes creates a "larger" inguinal canal. The external iliac artery passes beneath the inguinal ligament to become the femoral artery and the femoral vein ascends behind the inguinal ligament to become the external iliac vein. As these vessels pass beneath the inguinal ligament they are invested in a somewhat tubular arrangement of fascia (derived from the transversalis fascia), only about 1.5 cm in length. This is known as the femoral sheath. Just medial to the sheath is the femoral canal, a potential space into which the femoral vein can expand when the venous return through it increases. The mouth (proximal end) of the femoral canal, just over 1 cm in diameter, represents a potential weakness through which a herniating portion of intestine/peritoneum can descend beneath the inguinal ligament and into the thigh. Femoral hernias occur through the femoral canal. A femoral hernia is particularly likely to obstruct and strangulate. The inguinal canal is a normal "defect" in the inferior portion of the anterolateral abdominal wall. It is created during development by the process of gonadal descent. It lies entirely above the inguinal ligament. The deep inguinal ring is at the lateral extremity of the canal and is the point of communication of the inguinal canal with the abdominopelvic cavity. The superficial inguinal ring is at the medial extent of the canal and is the point of communication of the inguinal canal with the scrotum/labium majorum. Inguinal hernias may protrude through either ring. Hernias can be extremely painful and carry risk for incarceration of the bowel. They are usually diagnosed with the help of imaging (CT, MRI, etc.) and most are ultimately treated with surgery.

You explain that the extra chromosome is usually derived from the mother, but could also have come from the father. Joan and Jeremy ask that the origin of the extra chromosome 21 in their fetus be determined, given Jeremy's positive family history. You talk to the laboratory's cytogeneticist and learn that the best way to track a specific chromosome is to use molecular markers unique to chromosome 21. To isolate the molecular markers from the rest of the genomic DNA a molecular biology method called polymerase chain reaction (PCR) is performed on a sample of the DNA extracted from an individual's cells. Joan and Jeremy agree to have their blood drawn for molecular analysis. What type of molecular markers would be generated by this study and how would it help in determining the origin of a specific chromosome? In your answer, briefly describe the polymerase chain reaction (PCR).

The type of marker used for Joan and Jeremy is most likely a repetitive DNA sequence that is found only on chromosome 21. One type is called short tandem repeat (STR) sequences (=microsatelites) consisting of repeating units of two or more nucleotides. Unique. A person will have two copies of a specific STR; one inherited from their mother and the other from their father. Thus, the STR analysis will identify both alleles (or both copies) of each STR in Joan and in Jeremy.

Some features of microduplications of 22q11 are genitourinary anomalies, including urethral stenosis. Describe the urethra in an unaffected male and an unaffected female.

The urethra of the male traverses the prostate gland, the urogenital hiatus of the pelvic diaphragm, the sphincter urethrae muscle, the perineal membrane and the corpus spongiosum of the penis. It is about 20 cm in length and is subdivided into four different regions. The first portion of the male urethra is at the base of the bladder and is known as the preprostatic portion. It is approximately 1 cm in length and is surrounded by the internal urethral sphincter (circularly arranged smooth muscle fibers). The second portion of the male urethra traverses the prostate and is known as the prostatic urethra (about 3.5 cm in length). It receives the ejaculatory ducts and the many ducts of the prostate gland. This portion of the urethra can become compressed in cases of benign prostatic hyperplasia, leading to the clinical symptoms associated with this pathology. The membranous urethra is the third part of the male urethra. It passes first through the external urethral sphincter muscle and, just below this muscle, the perineal membrane; its length is about 1 cm. The penile or spongy, urethra (about 15 cm in length) travels through the corpus spongiosum of the penis. The spongy urethra opens distally at the external urethral orifice. In females, the urethra, about 4 cm in length, travels through the urogenital hiatus of the pelvic diaphragm, the external urethral sphincter and perineal membrane, and opens into the vestibule of the vagina. Its posterior wall closely abuts the anterior wall of the lower part of the vagina.

What are the chances of this happening again in a future pregnancy, or one of my sisters having a child with spina bifida?

There is a slight increase in the risk for a subsequent fetus with any NTD, including spina bifida and anencephaly. There is less than a 10% risk of recurrence for a single child with a NTD, and about 10% risk of recurrence for two or more neural tube defects. Her sister's risk for future pregnancies is only slightly increased. However, spina bifida and other neural tube defects are a common congenital anomaly (1/2000), so all women attempting pregnancy should be placed on prenatal vitamins containing folate. All pregnant women should be offered screening for NTD using the ultrasound or msAFP. For women who have had a child with a NTD, ultrasound with careful examination of the spine is usually recommended.

Daniel, a 54 year old ex-smoker, has been taking oral glucocorticoids (i.e. steroids) for 6 months to treat his chronic obstructive airway disease (COPD). Now that his pulmonary symptoms have improved, his physician wants to stop the glucocorticoids. However, every time he goes off these medications, Daniel feels extremely weak, dizzy and has a drop in his blood pressure. What is causing his symptoms?

These symptoms suggest presence of secondary adrenal insufficiency, resulting from the suppression of hypothamalo-pituitary-adrenal (HAP) axis from chronic steroid use. Since there was exogenous steroid intake by the patient, it would suppress the secretion of corticotropin releasing hormone (CRH) from the hypothalamus and ACTH from the pituitary. Patient's own adrenal glands would become atrophic. So when the patient stops exogenous steroid, patient's own adrenal gland are not able to make glucocorticoid, and this results in adrenal insufficiency.

A 28 year-old otherwise healthy man was evaluated for possible infertility after he and his wife were unable to conceive after a year without using contraception. A sperm sample showed no viable sperm (azospermia) and a CF test was "positive". What type of azospermia does this patient have? Explain his infertility in terms of his own embryologic development.

This patient's infertility is due to a form of obstructive azospermia. Men with classic CF have a developmental defect called congenital bilateral absence of the vas deferens (CBAVD). In CF males, infertility is the rule and is due to abnormalities in the Wolffian duct derivatives. This includes agenesis of the vas deferens and incomplete development of the epididymis, though testicular structures are normal. These structures are formed in the 7th week of gestation. Non-classic CF males can have an isolated "genital" form of CF, consisting solely of bilateral or unilateral absence of the vas deferens (CAVD). In some individuals, unilateral renal agenesis has also been associated with non-classic CF, although the association with CFTR mutations is controversial.

You find a 2003 paper that describes 93 infants with an omphalocele and abnormal karyotypes all with a partial trisomy of chromsome 3q. The authors suggest a hypothesis for the genetics of the developmental defect. From your knowledge of chromosome structure and function, propose a general mechanism that explains the potential association of omphalocele to chromosome 3q. (Yatsenko SA, Mendoza-Londono R, Belmont JW, Shaffer LG. Omphalocele in trisomy 3q: further delineation of phenotype. Clin Genet. 2003 Nov;64(5):404-13)

This study discusses an association between omphaocele and a specific karyotype abnormality of the long arm of chromosome 3. This is a type of aneuploidy, which refers to the addition or loss of a part or an entire chromosome. Thus, the infants in the study all have a trisomic region of chromosome 3. If each of the three alleles in the trisomic region is normally expressed, the amount of protein product is increased, which may disrupt the normal regulatory mechanisms important during mid-gut formation. Alternatively, the trisomic region may harbor a single gene, that when over expressed may cause the phenotype. Further mapping and correlative studies will need to be done before any of these hypotheses are proved to be correct.

In just the past few years, some of the most exciting developments in cancer therapy have been the development of specific inhibitors of a class of cell surface receptors that function as tyrosine kinases. Herceptin (trastuzumab) targets the protein Her2/neu, also called ErbB2, which is overexpressed in many breast cancers. Tarceva (erlotinib) targets epidermal growth factor receptor 1 (EGFR1, ErbB1) which is overexpressed in certain lung cancers. Describe the type of molecules of which each of these targeted therapies is composed. How does their structure affect their routes of administration?

Trastuzumab (notice that it ends in -mab, read monoclonal Ab) is a monoclonal antibody directed against the extracellular portion of the Her2 protein, which is a receptor that spans the cell membrane. Because it is a large protein molecule, it must be administered intravenously, as it would not be absorbed from the gut. Erlotinib (notice that it ends in -inib, read inhibitor), on the other had is a small molecule drug that binds in the ATP-binding pocket of the EGFR1 receptor. It is orally bioavailable. Both function to inhibit activation of the respective receptor. (The key point is that these targeted therapies can be either monoclonal antibodies against, or small molecule inhibitors of, receptor tyrosine kinases).

Compare and contrast type 1 and type 2 diabetes mellitus with regard to etiology.

Type 1: -Autoimmune disorder in which the body's own immune system attacks and destroys the beta cells in the islets of Langerhans in the pancreas. -Heritable genetic predisposition toward developing this autoimmune attack on the beta cells, which is then triggered by something in the environment. -Trigger is thought to be a viral attack. Type 2: -Appears to begin with the development of insulin resistance that gradually gets worse and worse. Pancreas is able to overcome this insulin resistance initially by making more insulin. -As time goes on, the pancreas falls behind. Although it is often making supra-physiologic amounts of insulin, it still does not make enough insulin to overcome the insulin resistance. -The exact cause of the insulin resistance and the reason why the pancreas cannot make more insulin and overcome the insulin resistance indefinitely is not known. Thought to be due to cytokine interference in normal glucose metabolism.

CTG expansions in DMPK cause myotonic dystrophy 1 (DM1). Explain the molecular pathogenesis of this disease. In your answer, include your understanding of the mutation and how this mutation causes its effects. In addition, explain the molecular implications of the figure below.

Upon expansion of the CTG (CUG in the RNA) repeat within the 3' UTR of the DMPK gene, specific RNA binding proteins are sequestered. These include CELF proteins and (CUG-BP1 & ETR 3-like). These particular RNA binding proteins have been shown to play an important role in alternative pre-mRNA splicing in developing cardiac muscle, skeletal muscle and the brain. Specifically, these alternative splicing factors are important for the temporal switch from fetal to adult spliced variants. To date, 27 alternatively spliced targets have been shown to be affected in DM1. Not every CELF protein is altered in the same manner. For example, MBNL1( also known as muscleblind, an alternative pre-mRNA splicing factor) is sequestered by the repeat-containing RNA, which effectively results in a loss of its protein function. In addition, CUG-BP1 exhibits an increased phosphorylation with a concomitant gain of new protein functions. The figure illustrates that CUG repeat-containing RNA, and by extension, sequestered alternative pre-mRNA splicing factors,such as MBNL1, exist in foci within the nucleus.

X-inactivation can cause women and girls to be affected with a X-linked disorder. What is X-inactivation and explain how this mechanism can cause an X-linked phenotype in a female.

X-inactivation is the process by which one of the two copies of the X chromosome in females in inactivated. This process equilibrates the amount of genetic material in females and in males, as females have two X chromosomes and males have only one. In mammals, the imbalance between the sexes is solved by inactivating one X chromosome, early in female embroyogenesis, a process also referred to as Lyonization (after the scientist Mary Lyon). For humans, X-inactivation is complete by the first week of life. One X chromosome is selected for inactivatation through a process of methylation and chromatin condensation mediated by the DNA-organizing proteins known as histones. The condensed chromosome is recognizable in the nucleus as a dense clump, known as a Barr body (after the Candian cytologist Murray Barr). Either the maternal or paternal X chromosome can become inactivated in early embryogenesis, but in subsequent mitoses, that X chromosome remains inactivated and the inactivation pattern becomes "fixed" in a clonal fashion. That is to say, if a particular cell inactivates the paternal X, all of its daughter cells will also inactivate their paternal X. In cases where there are differences between the two X chromosomes in their DNA sequence, such as a genetic mutation exists on one X but not the other, the female becomes a mosaic. Mosaicism is defined as the presence of two cell lines that are genotypically different, but derived from the same source. Skewed inactivation can occur in some females, where either the paternal or maternal X is preferentially inactivated, resulting in a female displaying an X-linked disorder. If an individual carries a gene mutation on one of the X chromosomes and if the "normal" chromosome is inactivated more than the "abnormal" chromosome, there will be a reduction in number of cells expressing the normal allele. If, for example, a woman carrier of a fragile X full-mutation had a skewed inactivation pattern, there is a chance that the normal allele is silenced in more cells than the Fragile X allele resulting in inadequate normal FMR1 protein. Lack of FMR1 expression in the brain correlates with the severity of mental retardation and may be a cause for females to express some of the features of Fragile X syndrome. Alternatively, skewed X inactivation could result in the Fragile X allele being silenced in more cells than the normal allele. In this situation, the level of normal FMR1 protein could be increased and result in a milder phenotype.