File: ch14, Chapter 14, Human Genetics

Essay

1. Brianna and Sean have a child with cystic fibrosis, a serious genetic disease which affects lung cells. It is caused by the presence of a double homozygous recessive allele in the genotype (cc). How could this happen if neither of the parents has the disease?

Ans: Each parent may have the recessive allele, c, but it does not show up because each also has a “normal” wild type allele, C, to produce the correct gene product. There is a 1 in 4 chance that BOTH of them will contribute the recessive mutant allele to the child.

Cc X Cc à probability of 75% that offspring will be phenotypically normal (CC or Cc) and 25% CF (cc)

2. Brianna and Sean have a child with cystic fibrosis, a serious genetic disease which affects lung cells. It is caused by the presence of a double homozygous recessive allele in the genotype (cc). What are the chances that any child of these parents will be a carrier for the gene for cystic fibrosis?

Ans: An offspring would have a 50% or 2 out of 4 probability of being a heterozygous carrier (Cc). The odds apply to each separate birth event. This is the classic heterozygous cross.

3. Downs Syndrome (Trisomy 21) is a medical conditions caused by nondisjunction of chromosomes during meiosis in either parent, although it most frequently occurs in the mother. Can you explain why women after age 40 (as opposed to men of the same age) have a higher incidence of problems during meiosis in the egg cells?

Ans: Women are born with all the eggs they will ever have, and these eggs are arrested in an early stage of meiosis. They do not complete mitosis until released from the ovary. A woman may carry these eggs for up to 30-40 years before becoming pregnant. During that time mutations may accumulate as they sit in the ovary. This may affect their ability to continue through the last stages of meiosis (chromatid separation) correctly. Males have a high turnover rate for their developing sperm in the testes, (approximately every three months or so) and therefore have a renewable “fresh supply” of sperm, less subject to accumulation of mutations than eggs.

4. A man who is heterozygous for the Huntington’s disease allele (D) marries a woman who has two wild type alleles for this gene (dd). Use the Punnet square to determine the probability that any of their children would contract the disease

Ans: The probability is 50% (1 in 2) that the child will receive the dominant D allele from the father . This is equivalent to a classic Mendelian “backcross” or “testcross”.

d d

D Dd (disease Dd(disease)

d dd dd

5. In the question above, would the gender of the child affect the probability?

Ans: No, it would not, because the trait is carried on an autosome, not on the X chromosome and therefore shows no gender association.

6. In certain individuals with familial hypercholesterolemia, there may be one of three clinical situations occurring: 1) Extremely high cholesterol, >600mg/dl; 2) still dangerously high but only ~400 mg/dl; and 3) "normal" levels, ~200mg/dl. What type of single gene transmission pattern does this clinical situation follow? (Hint: see Chapter 13 for the plant version of this).

Ans: This suggests incomplete dominance because a full, or intermediate, or low level of gene product is produced depending on whether one (Rr) or two wild type genes (RR) for the LDL cholesterol receptor are inherited. If both genes for the LDL cholesterol receptor are present (RR) and working correctly, there are adequate quantities of the receptor (which removes LDL from the blood) on cell surfaces, and the levels of LDL are lowest (200 mg/dl); in the heterozygous situation one wild type gene alone (Rr) can only produce half the normal amount of receptor (the other allele is inactive) so there are fewer receptor molecules produced and the LDL in the blood is correspondingly higher (400mg/dl). If neither of the alleles functions, as in the double homozygous recessive state (rr) , there is no LDL receptor on cell surfaces and the level of LDL in the blood will be highest (≥ 600mg/dl).

7. If nondisjunction occurs during separation of the sex chromosomes a variety of syndromes may result. Diagram the chromosomal changes which must occur during meiosis to produce a female with Turner's syndrome.

Ans: A female with Turner’s syndrome has only one functional X chromosome so her genotype will be XO. She inherited only one X chromosome and is missing one. One of her parents produced gametes which did not complete meiosis properly so 2 X chromosomes were present in one of the gametes and no X chromosomes were present in the other. The most likely situation is incomplete separation of the twin chromatids for the X chromosome during meiosis II, although it is possible that incomplete separation of the tetrad of X chromosomes during meiosis I might have occurred

8.Hemophilia is caused by an X-linked recessive gene (X ^h). A man who is not hemophiliac fathers a child by a woman with a normal phenotype (X/_?) but unknown genotype. Their first son is normal. Use the Punnett to determine whether it is still possible for them to have a child with hemophilia. Why/Why not?

Ans: If the man shows no signs of hemophilia then his X chromosomes must be normal wild type. However, the woman may be heterozygous for the allele,

(X ^h), and not be aware that she is a carrier. Their first son would have received a wild type X chromosome from the mother. However, there is a 50% (1 in 2) chance that a subsequent son might receive the X ^h allele from the mother. A female child would be a carrier but would have a normally functioning X chromosome from the father.

X X ^h

X XX XX ^h

Y XY X ^hY

9. If an individual with normal alleles for hemoglobin (HbHb) has a child with a carrier of the sickle cell hemoglobin gene (Hb Hb^s) , what are the chances that the child will be a carrier?

Ans: Do the Punnett square here too. There is a 50% probability that any child will be a carrier of the gene Hb^s

Hb Hb

Hb HbHb HbHb

Hb^s HbHb^s HbHb^s

10. Assume that the allele for blue eyes (c) is autosomal and recessive to the allele for brown eyes (C). How might a blue-eyed man and a brown eyed woman have a blue-eyed child?

Ans: Although eye color in humans is more complicated than this, the simplified version is as follows:

The brown eyed woman might be a heterozygote, and carry the recessive allele, c, for blue eyes (so her genotype is Cc). If her egg contains a recessive allele, and his sperm does too (as it must if they have a blue-eyed child) then there is a 50 % chance that their child will have blue eyes even though brown color is dominant in the mother.

11.Name at least two of the medical considerations in testing fetuses by amniocentesis.

Ans: No invasive medical procedure is ever totally without risk. There is danger of infection (at the very least), or miscarriage / spontaneous abortion of the fetus if something goes wrong. However, the risk is quite low relative to the benefits of the procedure.

12. There is now a test for the detection of the defective part of the DNA sequence seen in Huntington's Disease. What are the advantages and disadvantages of such a genetic test in this case?

Ans: Knowing that you carry the allele means knowing that you will get the disease at some point in the future because we do not yet know enough about Huntington’s disease to cure or prevent it (although that is the hope for the future). This is an advantage for some people want to know the future, because then they can prepare themselves both economically and in terms of not becoming parents and passing on the allele to their offspring. The knowledge is at the same time a disadvantage to others who may find the truth painful and choose not to be tested or not to learn the results.

13. Is the ability to diagnose a genetic disorder the same thing as the ability to treat or cure the same disorder? Justify your answer.

Ans: Several genetic disorders can now be diagnosed using biochemical or molecular tests that demonstrate the presence or activity of a given gene (e.g., Huntington’s disease, etc) but these same diseases may not be treatable or cured because our understanding of the disorders has not advanced as rapidly as our ability to predict or detect their occurrence

14. What is the evolutionary advantage to a eukaryotic organism of having a pair of chromosomes (i.e., two alleles for each gene) as opposed to just one chromosome?

Ans: An organism that has two alleles of a critical gene (on two chromosomes) has a better chance of surviving any mutations or chromosomal damage that may arise in that gene. In an organism with only a single chromosome (e.g., bacterial cells) what you see is what you get. Damage to a critical gene is often lethal.

15. Assume that you are blood group O, and your mother is blood group A. Your father could be any blood group except which one?? Use a Punnett square to work out the various possibilities for the genes for blood groups.

Ans: This gene system is an example of co-dominance of multiple alleles. If you are blood group O, you have no surface markers for either group A or group B (if you had, they would show up) so your mother must be heterozygous AO (I^A or i genes). You inherited the i gene for blood type O from her. You must have also inherited the same gene from your father since your genotype must be ii. Your father could NOT be blood group AB (I^A I^B ) because any gene products of the alleles for I^A or I^B would show up against the genetic background of I^A genes and i genes from the mother. See the Punnett square below for the genotypes if the father were type AB.

I^A i

I^A I^A I^A iI^A

I^B I^A I^B i I^B

16. It is generally thought that the antibodies present in your blood that are reactive against type A and type B blood groups are induced by similar sugar structures on the surfaces of bacteria we encounter early in life. It would certainly not be a good idea to make antibodies against your own blood group but what is the possible evolutionary advantage of the cross-reactive antibodies against foreign blood groups?

Ans: The immune system’s job is to react against and remove any foreign molecules. Antibodies against foreign blood groups will react with and remove small amounts of foreign blood cells which might stimulate an inflammatory and/or autoimmune reaction, or might, in the case of malaria and other diseases, carry infectious parasites which can cause disease. However there is a disadvantage to this process because transfusion of larger amounts of the foreign blood cells can cause a massive reaction which may be dangerous or even fatal.

17. If an individual has two different alleles at the same genetic locus will they always make equal amounts of each protein coded by the respective alleles? Explain why or why not.

Ans: They may or may not depending on the allele itself and/or the expression system for each gene (see genetic regulation in Ch.15) . Some alleles may make different amounts of mRNA when transcribed, others may make similar amounts but translate it at different rates or not at all. In some genetic systems, one of the alleles may be a mutation which does not code for active protein at all (see Ch13) so only half the amount of protein may be made.

18. Can a dominant trait like ability to roll a tongue have arisen in the human population even without any selective pressure? Explain your answer.

Ans: Yes, because traits arise as a result of random events such as mutation in the DNA their initial appearance is not related to usefulness to the organism. Some mutations are evolutionarily neutral, i.e., do not have any advantage or disadvantage to the organism, and may be retained in the absence of selective pressure.

19. Based on the Mendelian 3:1 ratio for inheritance of a single gene trait, if a couple has a child with cystic fibrosis does this mean their next 3 children will be normal? Explain your answer.

Ans: No it doesn’t. Each event (birth of a child) is independent of all others and the 3:1 probability exists for each child, regardless of whether a child has been born with the disease already. In unfortunate and rare cases, more than one child with the disease may be born into the same family.

20. Although ultrasound is becoming more and more accurate in predicting the sex of a baby based on visual observations of differences in genital appearance between male and female fetuses, it is still not perfect unless the baby is in the correct position. What technique can be used unequivocally to determine the sex of a child, and how is it performed?

Ans: Amniocentesis removes small amounts of cells (shed normally by the fetus) from the amniotic fluid and they can be analyzed for karyotype (XX or XY) or specific DNA analysis to determine the gender of the child. Chorionic villus sampling of fetal cells can also be used in this way.

21. What are the chances that two individuals with wavy hair will have a curly-haired child? Curly-hair and straight hair exhibit incomplete dominance, with wavy hair as an intermediate state. (Curly = CC; straight = cc)

Ans: Wavy hair results from the heterozygous state (Cc). If two heterozygotes with wavy hair produce children, there is a 1 in 4 chance that any child will have curly hair (CC) , a 1 in 2 chance that a child will have wavy hair (Cc), and a 1 in 4 chance for straight hair (cc). This is the classic ratio for heterozygote crosses in this situation. (see snapdragons in Ch13).

22. Assume that the allele for blond hair color (b) is autosomal recessive to the allele for brown hair (B). If blond-haired Susan has a blond-haired child could the child's father, Mark, have brown hair? Explain your answer.

Ans: Yes he could. Susan should be double homozygous recessive and can only contribute a “b” allele. To have a blond-haired child, Mark would have to also contribute a “b” allele. Mark would therefore have to be heterozygous at this locus (Bb) and would have brown hair himself but carry the “b” allele for blond hair. (NOTE: hair color in humans is more complex than this, requiring the actions of several genes. This situation has been simplified for the question ).

23. Prenatal testing reveals that a fetus has Tay-Sachs disease, a neurological disorder caused by a homozygous recessive allele where an important brain cell protein is not made. The parents have not had the disease and, in fact, they appear healthy. What are the genotypes of the parents (t = Tay Sachs disease, T = normal)?

Ans: The parents must be heterozygous carriers of the “t” allele for Tay-Sachs and their genotype must be Tt. Each contributed the “t” allele to their child. They are phenotypically normal because each has the dominant wild type allele, “T” which can code for normal amounts of the necessary brain cell protein.

24. If your father has a normal phenotype but your mother is heterozygous (Dd) for Huntington's disease, a dominant autosomal disorder, what are your chances of escaping the disease? Use a Punnet square to explain your answer.

Ans: This is the classic profile for dominant autosomal disorders. Dd x dd à 1 in 2 chance of Dd, the dominant allele appearing in any given child. Similarly there is a 1 in 2 chance of the normal “d” allele being donated by the affected parent so the child will be dd and escape the disease. The odds are 50:50 for each situation.

25. What would be the advantage to having many potential alleles at a given gene locus? (Remember: this is different from having many different loci control a single trait.) Be sure that your response considers survival of the species as a whole, not just for an individual member of a species.

Ans: Each allele might code for a protein with slightly different characteristics, some of which might be better than others (or, unfortunately, worse). Having multiple alleles available to a species means that some members of the species have a selective advantage for survival in a given situation which insures the survival of the species, but not necessarily every individual in it. Consider gene X with 5 possible alleles—X1,X2, etc. If the X4 allele gene product (protein) somehow protects against an infection by a lethal virus, then it would offer a survival advantage to any individual who has it by protecting them from the viral infection. Individuals with X1, X2, X3, and X5 might not survive the infection and die, but those with X4 would survive and carry on the species.

26. Some disorders are suspected to have a genetic component but have not been conclusively demonstrated to follow a pattern of inheritance. If you were a geneticist, how would you investigate a new disease that you suspect has a familial component to its inheritance before you subjected it to expensive DNA analysis?

Ans: Most human geneticists start with a family pedigree to trace the appearance of the trait through as many generations of extended family as they can. Pedigree analysis can give clues to whether or not the disorder is dominant or recessive, autosomal or X-linked and even whether or not it is controlled by more than one locus (a frequent occurrence). Later, once the genetics have been established by the pedigree analysis, the search for the actual gene in the DNA of many family members can begin.

27. For many years, the comedian Jerry Lewis hosted a telethon every Labor Day Weekend to raise money for children with muscular dystrophy, a chronic disease causing wasting of the muscles. One form (of the many that this disease can take) is seen only in male children of apparently disease-free parents. Explain whether you think this is a sex-linked trait and whether it is dominant or recessive.

Ans: If only males are affected this is strong evidence that it is an X-linked disorder transmitted by female carriers as a recessive gene. Women would normally have a wild type version of the gene on their other X chromosome so would be spared the disorder.

28. Some disorders are inherited as X-linked dominant traits. If a man manifesting such a trait was to have children, would the trait be seen in his sons or daughters? Why?

Ans: If a man has a dominant gene on his only X chromosome, it will be transmitted to each of his daughters who will have the gene and the disorder. His sons will get their X chromosome from their (presumably wild type gene) mother and will not have the disorder. The daughters can transmit this disorder to their own sons.

29. Sickle cell disease is a group of disorders caused by a single mutation in the gene for hemoglobin (Hb) causing the mutant Hb to become "sticky" and the molecules to aggregate in the red cell, leading to an abnormal, sickle-shaped cell which can plug up small blood vessels. Can you explain why such a molecular alteration would have pleiotropic effects on organs and organ systems, including enlargement of the spleen, anemia, brain damage, heart, liver and kidney problems? HINT: Think about the function of various organs mentioned.

Ans: In all these organs, proper blood flow is critical. Plugs in the small capillaries of major organs can reduce flow and therefore decrease function which can be dangerous and even lethal. Sickled red blood cells are removed by the spleen causing anemia as well as overtaxing the spleen which can become enlarged and fragile.

30. Several of the situations in this chapter document human gene disorders which occur predominantly in one or another population (For example Irish, Scandinavian, Asian, Jewish, Mediterranean, and/or people of African Descent.) If you are not a member of a particular group cited, could you possibly carry that group-associated allele for a given disease? Explain why certain alleles show up with increased frequency in certain groups.

Ans: Alleles are often seen with increased frequency in certain groups because until quite recently, humans married only members of their individual ethnic or regional group. You tend to marry people from your own “village”, thereby increasing the chances that if the allele is present it will be in the heterozygous state and will be retained in the population. Some alleles are introduced into new populations by migrations, intermarriage, and other situations. E.g., the gene for sickle cell hemoglobin is thought to have arisen in populations in West Africa but it is found now in many of the North African, Mediterranean, Middle Eastern and Hispanic population groups. If you are not a member of a group with a high incidence of the allele and/or disorder, you may still carry the allele for the above mentioned reasons.

31. Is the presence of an alternative allele of a gene a guarantee that it will be expressed, and the individual will exhibit a particular disease? Site examples of situations you may have heard in the news or on TV that might support your conclusion.

Ans: Individuals carrying recessive alleles in the heterozygous state may show little or no sign of the disorder. E.g., in sickle cell disease, the heterozygous individuals produce sufficient wild type hemoglobin from their wild type allele, that they may live completely normal lives. An allele’s presence does not guarantee its expression at all. Some alleles of certain genes can be expressed, and then inhibit the expression of the other allele in the genome.

32. Does sickle cell disease affect only those individuals who are of African descent? Keep in mind that current human genetic evidence suggests that all individuals on the planet came from a common ancestor, i.e., “out of Africa”, thousands of years ago and migrated to Europe, Asia, and the Americas. Explain your answer.

Ans: Many individuals of Mediterranean and Middle Eastern ancestries show the sickle cell trait since their ancestors probably migrated out of central and west Africa many thousands of years ago. Even populations in the Americas who derive from early European (Mediterranean) settlers and explorers, and slaves from West Africa may carry the allele.

33. In this country we generally do not marry our close relatives for social and genetic reasons. However, this was not always the case. In many cultures at different times in human history, it was considered appropriate for two closely related individuals to produce offspring (to preserve a dynasty, or to cement a powerful family’s wealth or influence). Using a pedigree for any double homozygous recessive disorder, show why two first cousins might produce an affected child more often than two unrelated individuals.

Ans: It is likely that recessive alleles might be present in the carrier state but go undetected if each carrier has a normal wild type allele to perform the specific function of that gene. Close intermarriage between, for example, first cousins might be a classic Heterozygote X Heterozygote cross, yielding a 25% chance that each offspring will be double homozygous for the recessive allele and therefore have the disorder. Unrelated individuals have a less likely chance of both having the same allele at that gene locus

Aa AA

I.

AA Aa Aa AA

II. Aa Aa

III.

aa

34The following is a pedigree for a family with two normal-appearing parents where two of the offspring have an inherited disorder. By analyzing the genders, ratios and appearance of the disorder in the offspring, can you tell whether it is autosomal or X-linked? Dominant or recessive?

Key: male, normal; male, affected;

female, normal ; female, affected

Ans: The affected individuals are equally male and female so it is probably an autosomal trait. There is a ratio of roughly 3 unaffected for every affected individual in the F1 generation, suggesting a heterozygote “cross”. This means that the parents are probably both heterozygous for the allele and it is therefore recessive.