Sunday, September 27, 2020
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Galactosemia Punnett: recessive and fatal

Classic galactosemia is an inherited disease: The enzyme galactose-1-phosphate uridyltransferase has a defective gene. This enzyme is involved in the pathway that converts galactose, a simple sugar found in milk products, to glucose. Lactose is the main sugar in milk, and that sugar is made up of two simple sugars: galactose and glucose.

As you know, glucose is used by our cells to produce ATP as a product of cellular respiration, which is the usable form of energy for the human body.

In the normal metabolic pathway, we convert 1-phosphate-galactose into glucose, which we can use to make ATP for energy. In galactosemia, galactose builds up in the blood. When 1-phosphate-galactose builds up, severe liver, kidney, central nervous system, and other damage can occur. The disease is fatal if left untreated.

Scientists know the allele for normal digestion (G) is dominant, and the allele for galactosemia (g) is recessive. If a male who is heterozygous for the galactosemia trait and a female who has galactosemia have a child, describe how the disorder could be passed down. What are all the possible genotypes and phenotypes for the male’s parents.

A response to this question is found in the comments to this post.

Paul Katula is the executive editor of the Voxitatis Research Foundation, which publishes this blog. For more information, see the About page.

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4 COMMENTS

  1. Galactosemia involves a single gene, for which the recessive allele, g, carries the malfunction of the enzyme listed above (G1P-UTase). Since the trait is recessive, a person with one or two dominant alleles, G, will have normal digestion. A person with two recessive alleles will have galactosemia.

    The possible genotypes and the associated phenotypes are shown here:

    • Genotype: Homozygous dominant (GG), Phenotype: normal
    • Genotype: Heterozygous (Gg), Phenotype: normal
    • Genotype: Homozygous recessive (gg), Phenotype: galactosemia

    In the text of the problem, we are told the male parent is heterozygous for the trait. This means his genotype is Gg and his phenotype is normal. He does not have galactosemia, but he carries one allele for the disorder. In some textbooks, this is known as being a carrier for the disease, but it is more proper to say he is heterozygous for the disorder.

    How can an individual inherit one dominant allele and one recessive allele for a disorder? A fundamental bit of knowledge you have to understand is that one allele comes from the mom and one allele comes from his dad. That is, he either inherited a recessive allele from his mom and a dominant from his dad, or he inherited a dominant allele from his mom and a recessive one from his dad.

    There are a few possibilities for the genotypes/phenotypes of his parents. The first is that one parent (either the mom or dad, placed on the top row) was homozygous dominant (GG), while the other parent was homozygous recessive (gg):

          G G
    g Gg Gg
    g Gg Gg

    If these were the genotypes of his parents, he couldn’t be anything but heterozygous, since no genotype besides the heterozygous (Gg) genotype is shown in the Punnett square above. A heterozygous offspring could also result from one homozygous dominant parent and one heterozygous parent:

          G G
    G GG GG
    g Gg Gg

    Here, there is a 50 percent likelihood that these parents’ offspring will be heterozygous, since two of the four squares in the above Punnett square contain a heterozygous offspring (Gg). Finally, a heterozygous offspring can result from two heterozygous parents:

          G g
    G GG Gg
    g Gg gg

    As you can see from the Punnett square above, these parents’ would also have a 50 percent chance of producing a heterozygous offspring.

  2. Now, if this heterozygous man and a woman who has galactosemia have a child, what are the chances that the child will have galactosemia? To determine odds in an inheritance problem, complete a Punnett square. The man’s genotype is Gg (heterozygous), and we will put his along the left side. The woman has galactosemia, and since the disorder is recessive, there is only one possibility for her genotype: gg (homozygous recessive).

          g g
    G Gg Gg
    g gg gg

    The Punnett square above shows a 50 percent probability that the couple’s offspring will have galactosemia, a phenotype that is only seen in individuals who are homozygous recessive (gg).

  3. Hospitals sometimes employ genetic counselors to help patients determine the chances that their children will be born with certain genetic disorders, such as galactosemia.

    Let’s suppose you are a genetic counselor and a couple walks into your office, seeking advice about how likely it might be that their next child will have galactosemia. First you would advise them that galactosemia can be controlled if detected early, so it was a good idea for them to come in and see you. Then you advise them that amniocentesis can detect galactosemia in the fetus. But they still want to put you to work.

    He has normal digestion, but she has galactosemia. Since she has galactosemia, the only possibility for her genotype is homozygous recessive (gg). If she had even one dominant allele (G), she would have normal digestion, since the dominant allele would mask the effects of the recessive allele, g, and she would not express the disorder physically.

    The man, however, presents more of a challenge. Let’s just pretend this is real life, where a test question wouldn’t give you the answer that he is heterozygous for the condition. You’re going to have to figure this one out all for yourself.

    Since you can’t just look at his DNA under a microscope, you have to use your reasoning powers. Since he has normal digestion, there are two possibilities for his genotype: it could be GG (homozygous dominant) or Gg (heterozygous). Since either of these genotypes could be found in an individual who does not express the disorder, how could these genotypes have originated?

    An offspring with a GG genotype would be impossible if either parent had the disorder. If even one of his parents had the disorder, that parent would have a homozygous recessive (gg) genotype, and that parent could not possibly give the man a dominant allele (G) because they would not have any dominant alleles in their genotype to give him. So we ask: Did either of your parents have galactosemia? Unfortunately for us (good for them), neither of his parents had galactosemia.

    Keep in mind, we need to determine his genotype in order to know which Punnett square applies to this couple’s children. If we knew his normal digestion phenotype were the result of a heterozygous genotype, we would construct the Punnett square below (his genotype being shown along the left side):

          g g
    G Gg Gg
    g gg gg

    This Punnett square shows a 50 percent chance that a child this couple has will have galactosemia (the two gg theoretical genotypes shown in red out of four squares total) — if only it were that easy, but it’s not.

    In order to know for sure what the man’s actual genotype is, we’ll have to consider his brothers and sisters. So we ask: Do any of your brothers or sisters have galactosemia? “No,” he replies, referring to his three siblings. This is a little bigger clue, because if any of his siblings had galactosemia, that would tell us that sibling got a recessive allele (g) from this man’s mom and a recessive allele (g) from this man’s dad. Since neither his mom or dad had galactosemia, we would know for sure that both were heterozygous, because the homozygous dominant genotype, GG, would not give them any recessive allele to pass down.

    See, if both of his parents were heterozygous, their children would have a 25 percent chance of producing a homozygous recessive child, who would have galactosemia. That Punnett square is shown in the previous comment (Gg x Gg). If none out of four children were homozygous recessive, that goes against the odds, but this sort of thing happens all the time. Punnett squares give us the theoretical probability of a certain genotype occurring. All the genotypes in the Punnett square are possible, so two heterozygous parents could produce a hundred homozygous dominant children and zero homozygous recessive children. The odds are against that happening, but it could easily happen.

    What this means is that we still don’t know for sure what the man’s genotype is. It’s starting to look like he’s homozygous dominant, because neither his parents nor any of his siblings had galactosemia.

  4. So, what are this couple’s chances at this point? Neither of his parents had galactosemia, so we can’t know for sure that he is heterozygous. If he is heterozygous, the Punnett square below (Gg x gg) would apply straight away to this couple’s children, and we tell them they have a 50 percent chance of having a child with galactosemia.

          G g
    g Gg gg
    g Gg gg

    On the other hand, if he has a homozygous dominant genotype, GG, the Punnett square below would apply to this couple’s children:

          G G
    g Gg Gg
    g Gg Gg

    This Punnett square shows no chance that this couple will produce any children with galactosemia, but they will all be heterozygous for the condition, meaning they could pass it onto their children.

    The problem for us now comes down to determining how likely it is that the first Punnett square applies and how likely it is that the second Punnett square applies.

    Now, if any of the man’s siblings has children and any of those children has galactosemia, that sibling must have a heterozygous genotype, and we can conclude that at least one of the man’s parents is heterozygous (remember, neither of his parents has galactosemia). In fact, this is what brought the couple into the genetic counseling office in the first place: his brother’s daughter has galactosemia, so he wondered what his chances were of passing it on, since his wife has it.

    Let’s do a quick summary at this point. Because the man’s brother is heterozygous, there is no chance that both of his parents were GG. If one of his parents was Gg and the other GG, there’s a 50 percent chance that he is Gg and 50 percent chance that he’s GG. If both of his parents were Gg, there’s a 50 percent chance that he’s Gg, based on the Punnett square in the earlier comment (Gg x Gg). However, since he does not have galactosemia, there is no chance that he is gg, the other possible outcome in the Gg x Gg Punnett square. Thus, there’s also a 50 percent chance that he’s GG if this is the case. The probabilities are 50-50, regardless of what genotype his parents are. Do you follow this logic?

    Then, if he’s GG, there’s no chance any of his kids will have galactosemia, since they would all be heterozygotes; if he’s Gg, there’s a 50 percent chance his kids will have galactosemia, since the woman will pass on a recessive allele no matter what and he will pass on a recessive allele half the time, according to the theoretical odds.

    In conclusion, we can’t give a definitive answer, because we don’t know the man’s genotype for sure. But we can apply the following steps. Let P(g|t) denote the probability the couple will have a child with galactosemia, given that the man is heterozygous, and P(g|m) denote the probability the couple will have a child with galactosemia, given that the man is homozygous dominant.

    • P(g|t) = .5
    • P(g|m) = 0
    • P(t) = .5, the chance the man is heterozygous
    • P(m) = .5, the chance the man is GG

    Our answer should probably be the average of the two probabilities, P(g|t) and P(g|m), since P(t) = P(m). If P(t) could be found to be greater than P(m), our answer should be adjusted accordingly, but since this couple has no children yet, we can’t get a more definitive estimate for these probabilities. Our answer would be .25: the couple has a one-in-four chance of producing a child with galactosemia. If this were the real world, we would of course include a footnote to our answer, explaining that it is based on incomplete information, since the genotype of the man was not determined with certainty.

Comments are closed.