Side 2 - Arvegang, genotype m.m.

Nej, hun kan kun være heterozygot.

Tak, Samuel. Har du også taget fejl i nogle af dine første bemærkninger i # 5?

Du kan jo efterprøve argumentationen.

Hvilken argumentation? Giver du mig ret i, at hun kun kan være heterozygot?

Jesus...

Jeg forstår simpelthen ikke, hvad du skriver, når du skriver så indirekte. Hvordan kan hun være homozygot, når farens genotype hedder AA og morens hedder Aa/aa?

Jeg beklager. Da jeg skrev homozygot, mente jeg heterozygot.

Tak, Samuel. Har du også taget fejl i nogle af dine første bemærkninger i # 5?

Jeg må bede dig specificere.

Er resten af det, jeg har skrevet, hvis jeg har korrigieret for dine skarpe indvendinger, korrekt?

Ja.

Mange tak. Din hjælp har været stor. Jeg håber, at jeg må spørge dig til råds en anden god gang.

Naturligvis. Jeg beklager, hvis jeg undervejs har givet forkerte/forvirrende svar.

Jeg klarer den.

Thanks for using AllExperts. There are two questions here, both of which are fairly straightforward once you have an understanding of how to approach pedigree diagrams.

1. What is the mode of inheritance? Consider the possibilities. Autosomal dominant diseases appear in every generation, and anyone getting the disease must have a parent with the disease. We can see in F2 on the right that two unaffected parents have an affected child, so the inheritance is not autosomal dominant. Autosomal recessive inheritance is next. On multiple occasions, unaffected parents produce affected children, which is characteristic of recessive disease. We cannot rule out autosomal recessive inheritance.

Now consider X-linked inheritance. We know it's not X-linked dominant inheritance because in F2, a healthy woman produces a diseased son in F3. Had the disease been X-linked dominant inheritance, the mother in F2 would have been affected by her dominant X chromosome and would have been sick. This is not the case, so it is not X-linked dominant inheritance. X-linked recessive inheritance usually prefers males, which is the case in this pedigree. In F2, two unaffected parents produce an affected son in F3, which supports the pedigree being X-linked recessive inheritance.

This pedigree most likely represents X-linked recessive inheritance, but the possibility of autosomal inheritance cannot be ruled out.

2. Assume that the trait is inherited in an X-linked recessive fashion. In my diagram, Xr represents an affected X chromosome while X is the normal X chromosome. Y represents the Y chromosome. In that case, the genotypes work out like this:

F1: XrY X? XY XrX
F2: XrX XY XrX XY XrY
F3: XrY

Thanks for using AllExperts.

3. To determine if the trait is sex-linked or not, assume first that the disease is indeed sex-linked dominant and draw the pedigree accordingly. You will see quickly that the disease cannot be sex-linked; I would encourage you to pay attention to F3, individuals 2 and 6.

4. After question 1, you've determined that that the trait is autosomal dominant. Affected individuals will therefore be AA or Aa, where A is the dominant trait; unaffected individuals will have genotype aa. Fill in all the aa's first. Then determine if any of the individuals must have the AA genotype. You'll be able to tell this because all of the children of an affected homozygote (AA) and an unaffected homozygote (aa) would have the disease: Aa is the only possible genotype from that pairing. The only possible way to obtain a homozygous dominant genotype is for two affected parents--who could be either AA or Aa--to have children. The affected children of one affected parent and one unaffected parent must be heterozygotes; they will necessarily inherit one recessive allele from the unaffected parent. Look at the pedigree--is there anywhere where two affected parents have children? No. All the affected offspring must be heterozygotes! The only uncertainty is the father in F1; he could be homozygous dominant or heterozygous. There simply isn't enough information to tell.

5. This question takes into account what you determined in question 2. III-1 is unaffected, so call her aa. III-2 is affected, so he is either Aa or AA. We determined in question 2 that the affected parent in F2 would be a heterozygote (Aa) because he had one affected parent and one unaffected parent. So we have one parent who is aa and one who is Aa. The question then becomes a simple cross: what proportion of individuals in an Aa x aa cross would have the disease?

6. This is similar to question 3. Both individuals are affected, so they must be AA or Aa. Since they had one affected and one unaffected parent, they must both have the Aa genotype. This also becomes a simple cross: what proportion of individuals in a Aa x Aa cross will have the disease?