Information

Chance of child inheriting at least one copy of an allele - heterozygote mother, unknown father


Our genetics professor has posted up working for previous examination answers, but I am not convinced that his answer is correct. My answer is close but may just be due to co-incidence.

Question: There is a dominant "risk" allele with a frequency of 5%; what is the chance that a woman with one copy of this allele, and her husband will have a child with at least one copy of the allele.

Heterozygote frequency: 0.095
Non-risk allele homozygote frequency: 0.9025
Risk allele homozygote frequencyy: 0.0025

Professor's answer:
0.50 (woman's chance of passing allele) + 0.095/2 (heterozygote frequency in population divided by chance of passing allele) = 0.5475

My answer:
There are three possible situations, that the husband is homozygous recessive, homozygous dominant and heterozygous. The probability that any child will have the allele in each situation is 50%, 75% and 100% respectively.
Hence my answer = (50% x 0.9025) + (75% x 0.095) + (100% x 0.0025) = 0.525


It seems that what you are trying to do is calculate the probability of the child having at least one copy of the "risk allele" which I will denote as "A" (and it's counterpart "a").

First of all you know that the mother is Aa, and the child will get one allele at random. The probability that the mother passes on the A allele is then 0.5.

The tricky bit is the seemingly unknown genotype of the father. However, you do know genotype frequencies, which are the probability for the fathers three possible genotypes.

  • If the father is AA then it is certain the child will get the risk allele, and the probability of the father being AA is 0.0025.
  • If the father is heterozygous then there is a probability of 0.5 he will pass on the A allele, and the probability of the father being Aa is 0.095.
  • If the father is aa there is a probability if 0 that he will pass on the A allele.

Combined these give the probability that the father carries at least one A and will pass it to his child.

$$(1 imes 0.0025) + (0.5 imes 0.095) + (0 imes 0.9025) = 0.05$$

The probability of the child getting at least one A is then the probability of getting it from the mother plus the probability of getting it from the father.

$$0.5 + 0.05 = 0.55$$


I wrote out every possibility, and I agree with yours.

Mom passes good allele 50.00% Dad is homo wt 90.25% Odds of 0 bad alleles 45.13%

Mom passes good allele 50.00% Dad is het 9.50% Dad passes bad allele 50.00% Odds of 1 bad allele 2.375000%

Mom passes good allele 50.00% Dat is het 9.50% Dad passes good allele 50.00% Odds of 0 bad allele 2.375000%

Mom passes good allele 50.00% Dad is diseased 0.25% Odds of 1 bad allele 0.12500%

Mom passes bad allele 50.00% Dad is homo wt 90.25% Odds of 1 bad alleles 45.13%

Mom passes bad allele 50.00% Dat is het 9.50% Dad passes bad allele 50.00% Odds of 2 bad alleles 2.375000%

Mom passes bad allele 50.00% Dat is het 9.50% Dad passes good allele 50.00% Odds of 1 bad allele 2.375000%

Mom passes bad allele 50.00% Dad is diseased 0.25% Odds of 2 bad alleles 0.12500%

Odds of 0 bad alleles 47.50% Odds of 1 bad allele 50.00% Odds of 2 bad alleles 2.50%


Duchenne Muscular Dystrophy (DMD)

Until the 1980s, little was known about the cause of any of the forms of muscular dystrophy. In 1986, MDA-supported researchers identified a gene on the X chromosome that, when flawed (mutated), causes Duchenne, Becker, and an intermediate form of muscular dystrophies.

Genes contain codes, or recipes, for proteins, which are important biological components in all forms of life. In 1987, the protein associated with the DMD gene was identified and named dystrophin. The dystrophin gene is the largest gene yet identified in humans and is located in the short arm of the X chromosome, in the Xp21.2 locus (a locus is the position of a gene on a chromosome). The majority of mutations of the dystrophin gene are deletions of one or more parts of it. 1

DMD occurs because the mutated DMD gene fails to produce virtually any functional dystrophin. Individuals with BMD genetic mutations make dystrophin that is partially functional, which protects their muscles from degenerating as badly or as quickly as in DMD.

The dystrophin protein transfers the force of muscle contraction from the inside of the muscle cell outward to the cell membrane. Because it connects the center of the muscle cell to the edge of the cell, the dystrophin protein is extremely long. One end is specialized for linking to the muscle cell interior and the other end is specialized for linking to a variety of proteins at the cell membrane. The long middle section, called the rod domain, is taken up by a series of repeating units called spectrin repeats.

The repeated spectrin units in the middle of the protein play an important role in linking the two ends, but studies have shown that the exact number of these units is not critical for the function of the protein as a whole. Many cases of DMD are caused by mutations in the part of the gene that encodes this middle section. Production of the entire protein stops when the mutation is encountered.

The absence of dystrophin sets in motion a cascade of harmful effects. Fibrous tissue begins to form in the muscle, and the body’s immune system increases inflammation. In addition to its force-transfer role, dystrophin provides the scaffold for holding numerous molecules in place near the cell membrane. Loss of dystrophin displaces these molecules, with consequent disruptions in their functions. Lack of dystrophin causes muscle damage and progressive weakness, beginning in early childhood.

Inheritance in DMD

DMD is inherited in an X-linked pattern because the gene that can carry a DMD-causing mutation is on the X chromosome. Every boy inherits an X chromosome from his mother and a Y chromosome from his father, which is what makes him male. Girls get two X chromosomes, one from each parent.

Each son born to a woman with a dystrophin mutation on one of her two X chromosomes has a 50 percent chance of inheriting the flawed gene and having DMD. Each of her daughters has a 50 percent chance of inheriting the mutation and being a carrier. Carriers may not have any disease symptoms but can have a child with the mutation or the disease. DMD carriers are at risk for cardiomyopathy.

Although DMD often runs in a family, it is possible for a family with no history of DMD to suddenly have a son with the disease. There are two possible explanations. The first is that the genetic mutation leading to DMD may have existed in the females of a family for some generations without anyone knowing. Perhaps no male children were born with the disease, or, even if a boy in an earlier generation was affected, relatives may not have known what disease he had.

The second possibility is that a child with DMD has a new genetic mutation that arose in one of his mother’s egg cells. Because this mutation is not in the mother’s blood cells, it is impossible to detect by standard carrier testing.

A man with DMD cannot pass the flawed gene to his sons because he gives a son a Y chromosome, not an X. But he will certainly pass it to his daughters, because each daughter inherits her father’s only X chromosome. They will then be carriers, and each of their sons will have a 50 percent chance of developing the disease and so on.

A good way to find out more about the inheritance pattern in your family is to talk to your MDA Care Center physician or a genetic counselor. More information also is included in MDA’s booklet Facts About Genetics and Neuromuscular Diseases.

Females and DMD

Why don’t girls usually get DMD?

When a girl inherits a flawed dystrophin gene from one parent, she usually also gets a healthy dystrophin gene from her other parent, giving her enough of the protein to protect her from the disease. Males who inherit the mutation get the disease because they have no second dystrophin gene to make up for the faulty one.

Early in the embryonic development of a female, either the X chromosome from the mother (maternal X) or the one from the father (paternal X) is inactivated in each cell. Chromosomes become inactivated at random. In each cell, there is a 50 percent chance that either the maternal or paternal X chromosome will be inactivated, with the other left active.

Usually, girls do not experience the full effects of DMD the way boys do, although they still have symptoms of muscle weakness. A minority of females with the mutation, called manifesting carriers, have some signs and symptoms of DMD.

For these women, the dystrophin deficiency may result in weaker muscles in the back, legs, and arms that fatigue easily. Manifesting carriers may have heart problems, which can show up as shortness of breath or an inability to do moderate exercise. The heart problems, if untreated, can be quite serious, even life-threatening.

In very rare instances, a girl may lack a second X chromosome entirely, or her second X may have sustained serious damage. In these cases, she makes little or no dystrophin (depending on the type of dystrophin mutation), and she develops a dystrophinopathy just as a boy would.

A female relative of a boy with DMD can get a full range of diagnostic tests to determine her carrier status. If she is found to be a DMD carrier, regular strength evaluations and close cardiac monitoring can help her manage any symptoms that may arise. For more on DMD in females, see Debatable Destinies: Duchenne muscular dystrophy carriers carry on, despite uncertainty.


Genes and Chromosomes

In humans, each baby inherits half of its chromosomes from its mother and half from its father. Each sperm and egg carries half of the full human chromosome set thus, when they join during fertilization, the product is a new organism with a complete chromosome set. This equals two matched sets of 22 chromosomes from each parent, called autosomal chromosomes, plus a sex chromosome from each. The egg from the mother always contributes an X chromosome, while the sperm from the father contributes either a Y or an X chromosome. Each autosomal chromosome carries many genes, and these genes each have a match or allele on the corresponding chromosome from the other parent -- although of course the traits these genes code for can vary. The X chromosome is larger than the Y chromosome and carries more genes, however, so some genes on the X chromosome do not have corresponding alleles on the Y chromosome.


Inheritance Inheritance

Hereditary hemochromatosis is inherited in an autosomal recessive or autosomal dominant manner, depending on which type a person has. [2]

Hemochromatosis types 1, 2, and 3 are inherited in an autosomal recessive manner. [2] This means that people with these types of hemochromatosis have a genetic change ( mutation or pathogenic variant) in both copies of a gene causing hemochromatosis in each cell of the body. We inherit one copy of every gene from our mother and the other from our father. The parents of a person with hemochromatosis types 1, 2, or 3 are each expected to have one changed copy of the gene causing hemochromatosis. People with one changed copy of a gene are known as carriers . Carriers typically do not have signs or symptoms of hemochromatosis.

  • 25% chance to have hemochromatosis
  • 50% chance to be a carrier like each parent
  • 25% chance to have two working copies of the genes causing hemochromatosis, meaning the child is unaffected and is not a carrier
  • 50% to inherit hemochromatosis
  • 50% chance to be unaffected

Hemochromatosis is a disease that shows reduced penetrance. This means that some people with pathogenic variants causing hemochromatosis never show symptoms of the disease. However, children or family members who have pathogenic variants causing hemochromatosis may show symptoms of the disease. [1] [2]


Dominant Inheritance

When a trait is dominant, only one allele is required for the trait to be observed. A dominant allele will mask a recessive allele, if present. A dominant allele is denoted by a capital letter (A versus a). Since each parent provides one allele, the possible combinations are: AA, Aa, and aa. Offspring whose genotype is either AA or Aa will have the dominant trait expressed phenotypically, while aa individuals express the recessive trait.

One example of a dominantly inherited trait is the presence of a widow’s peak (a V-shape) at the hairline. Let (W) represent the dominant allele, and (w) represent the recessive allele. An individual with a (WW) or (Ww) genotype will have a V-shaped peak at the hairline. Only ww individuals will have a straight hairline. To determine the probability of inheritance of a widow’s peak (or any other dominant trait), the genotypes of the parents must be considered. For example, if one parent is homozygous dominant (WW) and the other is homozygous recessive (ww), then all their offspring will be heterozygous (Ww) and possess a widow’s peak. If both parents are heterozygous (Ww), there is a 75% chance that any one of their offspring will have a widow’s peak (see figure). A Punnett square can be used to determine all possible genotypic combinations in the parents.

A pedigree that depicts a dominantly inherited trait has a few key distinctions. Every affected individual must have an affected parent. Dominantly inherited traits do not skip generations. Lastly, males and females are equally likely to receive a dominant allele and express the trait. In this pedigree both heterozygous and homozygous individuals are affected since the trait is dominant.

Image courtesy of Michael A. Kahn, DDS

CLICK HERE to learn more about patterns of inheritance
CLICK HERE to learn more about recessive inheritance
CLICK HERE to learn more about X-linked inheritance


Imprinted gene mutations

Some genetic disorders are now known to result from mutations in imprinted genes. Genetic imprinting involves a sex-specific process of chemical modification to the imprinted genes, so that they are expressed unequally, depending on the sex of the parent of origin. So-called maternally imprinted genes are generally expressed only when inherited from the father, and so-called paternally imprinted genes are generally expressed only when inherited from the mother. The disease gene associated with Prader-Willi syndrome is maternally imprinted, so that although every child inherits two copies of the gene (one maternal, one paternal), only the paternal copy is expressed. If the paternally inherited copy carries a mutation, the child will be left with no functional copies of the gene expressed, and the clinical traits of Prader-Willi syndrome will result. Similarly, the disease gene associated with Angelman syndrome is paternally imprinted, so that although every child inherits two copies of the gene, only the maternal copy is expressed. If the maternally inherited copy carries a mutation, the child again will be left with no functional copies of the gene expressed, and the clinical traits of Angelman syndrome will result. Individuals who carry the mutation but received it from the “wrong” parent can certainly pass it on to their children, although they will not exhibit clinical features of the disorder.

Upon rare occasion, persons are identified with an imprinted gene disorder who show no family history and do not appear to carry any mutation in the expected gene. These cases are now known to result from uniparental disomy, a phenomenon whereby a child is conceived who carries the normal complement of chromosomes but who has inherited both copies of a given chromosome from the same parent, rather than one from each parent, as is the normal fashion. If any key genes on that chromosome are imprinted in the parent of origin, the child may end up with no expressed copies, and a genetic disorder may result. Similarly, other genes may be overexpressed in cases of uniparental disomy, perhaps also leading to clinical complications. Finally, uniparental disomy can account for very rare instances whereby two parents, only one of whom is a carrier of an autosomal recessive mutation, can nonetheless have an affected child, in the circumstance that the child inherits two mutant copies from the carrier parent.


What APOE Means for Your Health


Genes are one of many risk factors for dementia. While a quarter of Alzheimer's patients have a strong family history of the disease, only 1% directly inherit a gene mutation that causes early-onset Alzheimer's, also known as familial Alzheimer's disease (FAD) [1]. But another gene called APOE can influence your risk for the more common late-onset type of Alzheimer's.

There are three types of the APOE gene, called alleles: APOE2, E3 and E4. Everyone has two copies of the gene and the combination determines your APOE "genotype"&mdashE2/E2, E2/E3, E2/E4, E3/E3, E3/E4, or E4/E4. The E2 allele is the rarest form of APOE and carrying even one copy appears to reduce the risk of developing Alzheimer's by up to 40%. APOE3 is the most common allele and doesn't seem to influence risk. The APOE4 allele, present in approximately 10-15% of people, increases the risk for Alzheimer's and lowers the age of onset. Having one copy of E4 (E3/E4) can increase your risk by 2 to 3 times while two copies (E4/E4) can increase the risk by 12 times [2].

Despite this association, the National Institutes of Health only recommends genetic testing for APOE status to advance drug research in clinical trials. APOE4 is just one of many risk factors for dementia and its influence can vary across age, gender, race, and nationality [3][4]. For example, having one copy of the E4 allele may pose more risk to women while having two copies seems to affect men and women similarly [5].

To learn more about the genetics of Alzheimer's disease and the contribution of the APOE genes, check out the National Institute on Aging's Alzheimer's Disease Genetics Fact Sheet.

THE BIOLOGY OF APOE

The APOE protein plays many important roles, including the transport of cholesterol across different tissues and cells. The proteins made by varying APOE alleles handle this transport function differently.

Outside the brain, APOE4 can increase the risk of atherosclerosis (i.e., hardening of the arteries) and stroke [4], which may explain why APOE4 is a risk factor for vascular causes of cognitive impairment and dementia [6][7]. Inside the brain, APOE helps to clear beta-amyloid, a component of plaques. APOE2 appears to perform this function more effectively than APOE4, with APOE3 in the middle. This difference in beta-amyloid transport represents what scientists call "loss-of-function" toxicity. However, researchers suspect that APOE4 proteins may also have toxic "gain-of-function" activities, such as increased response to stress or injury [4].


APOE4 may increase the risk of dementia through toxic gain of function and through the loss of normal healthy function. Figure adapted from [4]

APOE4 AND ALZHEIMER'S DRUG DISCOVERY

Some drugs in development (called "structure correctors") may change the physical structure of the APOE4 protein so that it behaves more like the APOE2 protein [8]. Another approach is gene therapy, which attempts to insert APOE2 genes into the brains of people with APOE4 genes [9]. To learn more about these programs and other APOE-related drug discovery programs supported by the Alzheimer's Drug Discovery Foundation, review our research portfolio with a filter for "APOE4."

DOES APOE AFFECT HOW THERAPIES WORK?

Researchers are exploring whether the APOE genotype influences the effects of drugs and other therapies in development for Alzheimer's disease and general cognitive health. Highlights of the scientific research in which a differential effect is possible follows, with links to reports.

Estrogen: Several studies suggest that the side effects of estrogen-containing hormone replacement therapy may be worse in people who carry the APOE4 allele, at least in terms of brain aging and dementia risk. However, the evidence is inconsistent.

Hypertension Management: Effective management of mid-life hypertension is likely to reduce the risk of dementia and cognitive decline in most people. Observational studies suggest that APOE4 carriers might be particularly likely to reap the benefits of effective hypertension management. However, the complex relationships between cardiovascular health, APOE status, and cognition are not well understood.

DHA: Although DHA may be part of a healthy diet for APOE4 carriers, evidence from observational studies, clinical trials, and some preclinical research suggests that it is less likely to protect against dementia or cognitive decline in APOE4 carriers. Some researchers are testing the idea that APOE4 carriers simply need higher doses of DHA because it does not reach their brains as effectively [10].

Statins: Evidence is mixed on whether statins have different effects on brain health in people who carry at least one APOE4 allele. Several observational studies found that APOE4 allele status had no effect while another suggested different effects on cognition in patients with at least one APOE4 allele.

Nicotine: Although there is no evidence suggesting different Alzheimer's disease benefits from nicotine between APOE4 carriers and non-carriers, some evidence suggests nicotine may be a stronger acute cognitive enhancer in APOE4 carriers than non-carriers.

Cerebrolysin: One clinical trial comparing the Exelon&trade patch with cerebrolysin found no difference in response rates in patients with at least one APOE4 allele but a 3-fold higher response rate in patients without an APOE4 allele.

  1. Bird, T.D., Alzheimer Disease Overview, in GeneReviews(R), R.A. Pagon, et al., Editors. 1993: Seattle (WA).
  2. Michaelson, D.M., APOE epsilon4: the most prevalent yet understudied risk factor for Alzheimer's disease. Alzheimers Dement, 2014. 10(6): p. 861-8.
  3. Farrer, L.A., et al., Effects of age, sex, and ethnicity on the association between apolipoprotein E genotype and Alzheimer disease. A meta-analysis. APOE and Alzheimer Disease Meta Analysis Consortium. JAMA, 1997. 278(16): p. 1349-56.
  4. Liu, C.C., et al., Apolipoprotein E and Alzheimer disease: risk, mechanisms and therapy. Nat Rev Neurol, 2013. 9(2): p. 106-18.
  5. Riedel, B.C., P.M. Thompson, and R.D. Brinton, Age, APOE and sex: Triad of risk of Alzheimer's disease. J Steroid Biochem Mol Biol, 2016. 160: p. 134-47.
  6. Dwyer, R., et al., Using Alzgene-like approaches to investigate susceptibility genes for vascular cognitive impairment. J Alzheimers Dis, 2013. 34(1): p. 145-54.
  7. Sun, J.H., et al., Genetics of Vascular Dementia: Systematic Review and Meta-Analysis. J Alzheimers Dis, 2015. 46(3): p. 611-29.
  8. Mahley, R.W. and Y. Huang, Small-molecule structure correctors target abnormal protein structure and function: structure corrector rescue of apolipoprotein E4-associated neuropathology. J Med Chem, 2012. 55(21): p. 8997-9008.
  9. Zhao, L., et al., Intracerebral adeno-associated virus gene delivery of apolipoprotein E2 markedly reduces brain amyloid pathology in Alzheimer's disease mouse models. Neurobiol Aging, 2016. 44: p. 159-72.
  10. Yassine, H.N., et al., The effect of APOE genotype on the delivery of DHA to cerebrospinal fluid in Alzheimer's disease. Alzheimers Res Ther, 2016. 8: p. 25.

Dr. Penny Dacks was previously the Director of Aging and Alzheimer’s Disease Prevention at the Alzheimer's Drug Discovery Foundation. She was trained in neuroscience at the Mount Sinai School of Medicine, the University of Arizona, and Queen's University (Canada) with individual fellowships from the National Institute of Health, the Evelyn F. McKnight Brain Research Foundation, the ARCS Foundation and the Hilda and Preston Davis Foundation. She has authored over 18 peer-reviewed scientific articles and is a member of the Society for Neuroscience, the Gerontological Society of America, the Endocrine Society and the Association for Women in Science.


Chapter 14 - Mendel and the Gene Idea Flashcards Preview

Fill in this diagram of a cross of round- and wrinkled-seeded pea plants. The round allele (R) is dominant and the wrinkled allel (r) is recessive.

c. F1 generation i. Rr (round)

e. F2 generation k. 3 round:1 wrinkled

A tall pea plant is crossed with a recessive dwarf pea plant. What will the phenotypic and genotypic ratio of offspring be

a. if the tall plant was TT ?

b. if the tall plant was Tt ?

b. 11 tall (Tt ) to dwarf (tt )

You can use a Punnett square to determine the expected outcome of a test cross, but a shortcut is to use only 1 column for the recessive individual's gametes, since they produce only one type of gamete.

A true-breeding tall, purple-flowered pea plant (TTPP ) is crossed with a true-breeding dwarf, white-flowered plant (ttpp ).

a. What is the phenotype of the F1 generation?

b. What is the genotype of the F1 generation?

c. What four types of gametes are formed by F1 plants?

d. Use the picture of the Punnett square to show the offspring of the F2 generation. Shade each phenotype a different color so yuo can see the ratio of offspring.

e. List the phenotypes and ratios found in the F2 generation.

f. What is the ratio of tall to dwarf plants?

Of purple- to white-flowered plants?

(Note that the alleles for each individual character segregate as in a monohybrid cross.)

e. 9 tall purple:3 tall white:3 dwarf purple:1 dwarf white

f. 12:4 or 3:1 tall to dwarf

12:4 or 3:1 purple to white

Apply the multiplication rule to a dihybrid cross. How would you determine the probability of geting an F2 offspring that is homozygous recessive for both traits?

Dihybrids produce four types of gametes. The probability of getting both recessive alleles in a gamete is 1/4, and for two such gametes to join is ¼ x ¼, or ⅟ 16.

a. In the following cross, what is the probability of obtaining offspring that show all three dominant traits?

probability of offspring that are A_B_C_ = _____

( _ indicates that the second allele can be either dominant or recessive without affecting the phenotype determined by the first dominant allele.)

b. What is the probability that the offspring of this cross will show at least two dominant traits?

c. What is the probability of offspring that show only one dominant trait?

a. Consider the outcome for each gene as amonohybrid cross. The probability that a cross of Aa x Aa will produce an A_ offspring is ¾. The probability that a cross of Bb x bb will produce a B_ offspring is ½. The probability that a cross of cc x CC will produce a C_ offspring is 1.

To have all of these events occur simultaneously, multiply their probabilities: ¾ x ½ x 1 = ⅜.

b. Offspring could be A_bbC_, aaB_C_ , or A_B_C_. The genotype A_B_cc is not possible. Can you see why?

Probability of A_bbC_ = ¾ x ½ x 1 = ⅜

Probability of aaB_C_ = ¼ x ½ x 1 = ⅛

Probability of A_B_C_ = ¾ x ½ x 1 = ⅜

Probability of offspring showing at least two dominant traits is the sum of these independent probabilities, or ⅞.

c. There is only one type of offspring that can show only one dominant trait (aabbCc ). Can you see why? Its probability is ¼ x ½ x 1 = ⅛.

List the possible genotypes for the following blood groups.

A dominant allele M is necessary for the production of the black pigment melanin mm individuals are white. A dominant allele B results in the deposition of a lot of pigment in an animal's hair, producing a black color. The genotype bb produces brown hair. Two black animals heterzygous for both genes are bred. Fill in the table for the offspring of this cross.

The ratio of offspring from this MmBb x MmBb cross would be 9:3:4, a common ratio when one gene is epistatic to another. All epistatic ratios are modified versons of 9:3:3:1.

The height of spike week is a result of polygenic inheritance involving three genes, each of which can contribute an additional 5 cm to the base height of the plant, which is 10 cm. The tallest plant (AABBCC ) can reach a height of 40 cm.

a. If a tall plant (AABBCC ) is crossed with a base-height plant (aabbcc ), what is the height of the F1 plants?

b. How many phenotypic classes will there be in the F2?

a. The parental cross produced 25-cm tall F1 plants, all AaBbCc plants with 3 units of 5 cm added to the base height of 10 cm.

b. As a general rule in the polygenic inheritance of a quantitative character, the number of phenotypic classes resulting from a cross of heterozygotes equals the number of alleles involved plus one. In this case, 6 alleles (AaBbCc ) + 1 = 7. So, there will be 7 different phenotypic classes in the F2 among the 64 possible combinations of the 8 types of F1 gametes. These 7 classes will go from 6 dominant alleles (40 cm), 5 dominant (35 cm), 4 dominant (30 cm), and so on, to all 6 recessive alleles (10 cm).

Consider this pedigree for the trait of albinism (lack of skin pigmentation) in three generations of a family. (Solid symbols represent individuals who are albinos.) From your knowledge of Mendelian inheritance, answer the following questions.

a. Is this trait caused by a dominant or recessive allele? How can you tell?

b. Determine the genotypes of the parents in the first generation. (Let A and a represent the alleles.) Genotype of father _____ of mother _____.

c. Determine the probable genotypes of the mates of the albino offspring in the second generation and the grandson 4 in the third generation. Genotypes: mate 1 _____ mate 2 _____ grandson 4 _____.

d. Can you determine the genotype of son 3 in the second generation? Why or why not?

a. This trait is recessive. If it were dominant, then albinism would be present in every generation, and it would be impossible to have albino children with two nonalbino (homozygous recessive) parents.

b. father Aa mother Aa, because neither parent is albino and they have albino offspring (aa )

c. mate 1 AA (probably) mate 2 Aa grandson 4 Aa

d. The genotype of son 3 could be AA or Aa. If his wife is AA, then he could be Aa (both his parents are carriers) and the recessive allele never would be epressed in his offspring. Even if he and his wife were both carriers (heterozygotes), there would be a 243 /1024 (¾ x ¾ x ¾ x ¾ x ¾) or 24% chance that all five children would be normally pigmented.

a. What is the probability that a mating between two carriers will produce an offspring with a recessively inherited disorder?

b. What is the probability that a phenotypically normal child produced by a mating of two heterozygotes will be a carrier?

b. 2 /3. Of offspring with a normal phenotype, 2 /3 would be predicted to be heterozygotes and, thus, carriers of the recessive allele.

If two prospective parents both have siblings who have a recessive genetic disorder, what is the chance that they would have a child who inherits the disorder?

Both sets of prospective grandparents must have been carriers. The prospective parents do not have the disorder so they are not homozygous recessive. Thus, each has a 2 /3 chance of being a heterozygote carrier.

The probability that both parents are carriers is 2 /3 x 2 /3 = 4 /9 the chance that two heterozygotes will have a recessive homozygous child is ¼.

The overall chance that a child will inherit the disease is 4 /9 x 1 /4 = 1 /9.

Should this couple have a baby that has the disease, this would establish that they are both carriers, and the chance that a subsequent child would have the disease is ¼.

After obtaining two heads from two tosses of a coin, the probability of tossing the coin and obtaining a head is


Example: Hemophilia A

Hemophilia A is an X-linked recessive disease caused by a lack of a coagulant, or blood clotting agent, called factor VIII (factor 8). This is caused by a mutation in a gene on the X chromosome called F8. If a father is affected, his daughters will be carriers of hemophilia A and his sons will be unaffected. If a mother is an unaffected carrier, each daughter has a 1 in 2 chance (i.e., 50%) of being an unaffected carrier and each son has a 1 in 2 chance (i.e., 50%) of being affected with hemophilia A.


Click here to order our latest book, A Handy Guide to Ancestry and Relationship DNA Tests

A friend of mine just told me his great uncle was diagnosed with Huntington’s disease? What are his chances for having the disease? And what about his son?

-A curious adult from Iowa

This is a trickier question than you might think. With dominant diseases like Huntington’s Disease (HD), it is usually pretty easy to figure out risks. Generally if one parent has it then each child has a 50% chance of having it too.

And if neither parent has the disease, then odds are that none of the kids will either. With these diseases, you are almost never a carrier like you can be with recessive genetic diseases like albinism or cystic fibrosis. You usually can’t pass on a gene that causes the disease because you don’t have it.

So if the great uncle had HD but your friend’s grandparents didn’t, then we’d say he couldn’t have it. After all, if the grandparents don’t have the disease, then they don’t have the gene. Which of course means they can’t pass it to your friend’s parents!

But HD is complicated by the fact that it often doesn’t kick in until someone is in their 50’s. If a parent with the condition happened to pass away from other causes before the symptoms appeared, then they might have passed the disease on. In these cases you can have a chance to have the disease even if neither of your parents showed any symptoms. Here is what that sort of family tree might look like:

Here it looks like the mother and father on the far right in the second generation have two kids with the disease even though neither parent had it. Most likely this is because the mom passed away before her symptoms became apparent (the slash indicates she has passed away). Most likely her kids got HD from her.

What this all means is that if your friend’s grandparents lived to a ripe old age without the disease, then the odds are pretty good your friend doesn’t have the gene. The same is true if your friend’s parents are in their 60’s without the disease.

Still, in genetics nothing is 100% (click here to learn more). Fortunately for the people who want to know as for sure as possible, there is a genetic test that can look for the DNA differences that can lead to HD.

Where the 50% Number Comes From

You get HD if you have a broken version of a specific gene. Since you get your genes from your parents, a common way to get the disease is to inherit a broken copy of the gene from them.

As I mentioned, HD is a dominantly inherited disease. This means that if one of your parents has the disease, you have a 50% chance of getting it from them. And if your parents don’t have the disease, you probably won’t get it. Let’s look more closely at why.

Everyone has two copies of each gene. You get one of your mom’s copies and one of your dad’s copies. Sadly you don’t get to pick which one.

So let’s say your dad has HD and your mom doesn’t. Your mom has two good copies of the gene, so whichever one you get from her will work just fine.

But your dad has one broken copy of the gene and one good copy. You will get one of them randomly, so your chances of getting the broken copy are one in two or 50%. It’s just like the chance of getting tails if you flip a coin.

The good news is that if you don’t get the broken copy of the gene, then you can’t pass the disease on to your children. And even if you do have it, your kids only have a 50% chance of getting it from you.

If you do have a broken gene, there is a way to make sure your kids don’t have it. It’s called preimplantation genetic diagnosis, or PGD.

Scientists do PGD by fertilizing an egg outside of a mother just like in vitro fertilization (IVF). Once the embryo grows a little, they can use a small part of it to test for the broken gene. If the embryo has the broken gene, they won’t implant it back into the mother, so she won’t have a child with the disease.

A Tangled Problem

Not all diseases are dominant like HD. Many genetic diseases only happen when both copies of a gene are broken or lost.

But HD is different. One broken copy is enough to cause the disease.

Each gene has the instructions for making a specific protein. And each protein has a specific job to do in the cell. In this case, the huntingtin gene has the instructions for making the huntingtin protein.

Genetic diseases often happen when a DNA change messes up the instructions in a gene. Now the protein the gene codes for is broken and can’t do its job.

Sometimes the protein simply stops working and other times it gains some new function. In HD, the huntingtin protein does something it shouldn’t.

In HD, a part of the huntingtin gene repeats itself over and over. This causes a longer form of the huntingtin protein to get made. And this long protein gets tangled up with many of the other proteins in the cell.

This causes bunches of other proteins to stick together and stops all of them from doing their job in the cell. Eventually the cells die because so many things are not working. This is what causes the symptoms of HD.

So losing a gene doesn’t cause HD. Instead it is caused by gaining a longer protein that stops everything else from working.

But not everyone with HD has the same length of extra protein. And this makes HD an even more complicated disease.

The HD Gray Area

As I mentioned before, it is possible to get HD even if your parents don’t have it. If your parent passes away before showing symptoms of HD, you might not know whether that they had a broken HD gene.

Although this doesn’t apply to your case, there is another way that you can get HD even if both of your parents live long and healthy lives. It has to do with the fact that different people have different lengths of the huntingtin protein. And the same length protein can have different effects in different people.

There are certain lengths of the huntingtin protein that are safe for everyone and certain lengths that cause anyone who has them to end up with HD. But there are in between lengths that are tricky. Even the same length protein can cause HD in one person but not someone else.

So you can imagine a parent with one of these intermediate length proteins. She happens to have the right genetic background so that her cells can deal with this slightly longer protein.

Now imagine she passes it down to her son whose cells can’t deal with a protein that long. He ends up with HD even though his mom did not have any symptoms. And this isn’t the only way these slightly longer proteins can cause problems.

When healthy people with slightly longer proteins pass the gene on to their child, it sometimes gets a little bit longer. And this can sometimes make it long enough to cause the disease. So then the child gets HD even when neither parent did.

So even if you get a genetic test to know for sure what your HD gene is like, there can still be some uncertainty about whether you will get HD. Which is why such a seemingly simple question needed such a long answer!