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1.- The character represented in black is dominant. We can see this in the first couple of generation I. From the crossing of two homozygous recessive individuals, individuals with the dominant character could not emerge.

2.- It is an autosomal character. If it were linked to the Y chromosome, only men would have that character. If it were linked to the X chromosome, being dominant, all the descendants of a father with that character should present it. If it were linked to the recessive X chromosome, all male children of a mother with that trait should manifest it.

They do not ask for it, but as we are practicing we will see the possible genotypes of this family:

I                Aa           x             Aa                                       aa            x          Aa

II           A_          aa            aa                    x               Aa           aa             Aa         Aa

III                                                    Aa          Aa         aa

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The first thing we notice is that this disease affects men and women in the same way, although we must check if it is linked to a sex chromosome or it is an autosomal chromosome.

• If it were linked to the Y chromosome, only men would have that character, and a man who had that character should be the son of another man who manifested that character. This is not the case.
• If it were linked to the X chromosome, being dominant, all the descendants (women) of a father with that character should present it. This occurs in the AB couple, whose daughters should be X^AX^A, since that allele would come to them from the father and the mother, but one of them has offspring with a man without that character but no descendant shows that character. .
• If it were linked to the recessive X chromosome, all male children of a mother with that trait should manifest it. This is not the case either.

For all this, it is shown that it is an autosomal character.

In the AB couple, both manifest that character and all their sons and daughters, too. This would indicate that they are homozygous individuals for that character, dominant (AA) or recessive (aa), since all their offspring are the same as their parents.

In the case of the couple CD, C would manifest the character D would not manifest it, having children with that character and without that character. If C is aa and D is Aa he could have such offspring. But if C were AA and D were aa, all children would be with C. Therefore, it appears that the genotype of A, B, and C is aa.

The other couple of the second generation would also corroborate this statement. The mother (aa) would have children with a father (AA) and all the children would be like the father (Aa) heterozygous.

The genotypes of the family would be:

I                                                       aa                      x                  aa

II                       aa              AA        x       aa          aa           aa        x        Aa

III                                  Aa          Aa          Aa                   Aa       aa       Aa      aa

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The character represented by the color black is a character due to a recessive allele, since individuals 6 and 7 (both Aa) have offspring represented in black (aa) and white (AA or Aa). If the allele that determines the character represented in black were dominant, to have children with that character, parents should also have it.

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To know if the allele that determines the character represented in black is dominant or recessive, we must look at the generation II couple, since pairs 1-2 and 3-4 would not provide the necessary information. The couple formed by woman 3 and man 4, both represented in black, have three children, two with the black character and one with the white character. If it were a recessive character, all the descendants should have the black character. Therefore, it is a dominant allele.

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The first thing we notice is that this disease affects both men and women, although it seems more common in it. We will have to check if it is linked to a sex chromosome or it is an autosomal chromosome.

• If it were linked to the Y chromosome, only men would have that character, and a man who had that character would transmit that Y chromosome to his male children, who would also manifest that character. This is not the case.
• If it were linked to the X chromosome, being dominant, all the descendants (women) of a father with that character should present it. This is not the case in the problem family.
• If it were linked to the recessive X chromosome, all male children of a mother with that trait should manifest it. This is not the case either. In addition, the AB pair should have all the offspring with that character.

Therefore, it appears that this disease is linked to an autosomal (non-sexual) chromosome.

A and B have the disease but there are descendants who do not. This seems to indicate that the disease is due to a dominant allele, and that A (Aa) and B (Aa) would be heterozygous for this trait. When two individuals who do not have a certain character, it does appear in the offspring it is because it was hidden, it was recessive, in the parents but they transmit it to their children and it appears if it is homozygous. This explains why healthy children (AA and Aa) and sick children are born from A (Aa) and B (Aa), as in the case of C (aa).

Although he does not ask for it in the problem, the children of the second generation who have children among them (nobody has said that they were people, and if they were, it would be their problem, well, now ours) would probably be AA and AA or Aa, because all his offspring are healthy.

C (aa) is homozygous recessive for this trait and has children with D. Part of the offspring is healthy but one daughter is sick (aa), so the father, D, should be heterozygous (Aa).

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The first thing we notice is that this disease affects men and women in the same way, although we must check if it is linked to a sex chromosome or it is an autosomal chromosome.

• If it were linked to the Y chromosome, only men would have that character, and a man who had that character should be the son of another man who manifested that character. Furthermore, all men descended from a sick man would be sick. This is not the case in the problem
• If it were linked to the X chromosome, being dominant, all the descendants (women) of a father with that character should present it. This is not the case.
• If we had not been warned in the problem that C is homozygous for this character, we might have thought that it was a recessive X-linked character if C had been heterozygous for this character.

Therefore, it is an autosomal character, not linked to a sex chromosome.

At first glance, couple B and A, sick, have all the offspring with this character, so they could appear homozygous (dominant or recessive), but they have a son E, who together with C (homozygous woman) have sick children and healthy, so E has to be heterozygous for this character (Aa).

Thus, B and A would be Aa and AA (it is not exclusive that both were Aa), E would be Aa and C, aa. D would be aa.

I                            AA           x           Aa

II                       A_      A_       A_         Aa         x          aa

III                                                      Aa      aa       Aa        aa

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a) Indicate whether the allele that determines the alternative represented in black is dominant or recessive. Reason for the answer using the possible genotypes of all the individuals.

It is dominant. In the case of pairs 2 and 3, the parents have a phenotype in which the alternative represented by the color black is shown, and yet both phenotypes (white and black alternative) are present among their descendants. In the case of a recessive allele, the alternative represented by white would not be present. To get the maximum score, it must be well justified.

* In the lower part of the individuals, the genotype that the offspring would have is shown, based on how the genotype of the parents would be in each hypothesis of allele dominance.

b) Given this detected dominance relationship, could it be a sex-linked gene? Give reasons for the answer.

* In the lower part of the individuals, the genotype that would explain the phenotype of each individual of the offspring is shown.
Given the statement and the result of section a) it would result that XD would be the dominant allele for the character and Xd the recessive allele. Therefore, the black color would represent the phenotype that the trait shows, while the white color does not show that trait.
During development, the genotypes that would explain the phenotype indicated in the tree have been put. It will be seen that in several cases (framed and crossed out) it is not possible to obtain this genotype from the cross between its parents.
If we write the genotypes of the parents (pair 1 and pair 2) assuming that the gene is linked to the X chromosome, we will see that it would not be possible to explain the phenotypes of the offspring (II.2, II.3 and II.7). The same occurs if we start from pair 3, the phenotypes of their offspring cannot be explained by an X-linked gene whose phenotype represented in black was determined by a dominant allele (XD). In conclusion, we can say that it is not a sex-linked gene.

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In this problem we are asked how the type of transmission (dominant, recessive or linked to sex) can be compatible with the genealogical tree of the image (affected in black, healthy in white).

We will analyze each of the possibilities:

• Autosomal dominant transmission . It seems likely that the transmission is autosomal dominant since affected individuals emerge from affected individuals. The parents could be AA x AA and all the F1 offspring would be AA. One of these descendants has a healthy child with another affected one. Therefore, the cross of the parents would be AA x Aa, and all the descendants of F1 would be AA or Aa, phenotypically affected. One of them, Aa, had a male child aa with another affected individual Aa. Therefore, the autosomal dominant transmission is compatible with the genealogical tree in the figure.
• Autosomal recessive transmission . From what can be seen in the left part of the image, it could be thought that from the crossing between individuals aa x aa an F1 aa arose, all of them affected. But if we look at the crossing on the right, of two homozygous recessive individuals (aa x ​​aa), a heterozygous (Aa) or homozygous dominant (AA) cannot emerge. Therefore, autosomal recessive transmission is not consistent with the pedigree in the figure.
• Sex-linked transmission .
  • Linked to the Y chromosome . It would only affect males. This possibility is not supported .
  • Linked to the X chromosome . We will look at the crossroads on the right. 
    • Dominant. At the last crossing, an XDXd heterozygous affected woman, together with an XDY man, may have a healthy XdY child. It is compatible with the presented family tree.
    • Recessive  At the last crossing, an XdXd affected woman, along with an XdY man, could not have a healthy XDY child. This possibility is incompatible.

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Section A

To see if the transmission of inheritance is dominant or recessive, we will look at the cross between individuals II2 and II4. They both have the character that is discussed in the problem and have a son without character and another with a daughter with him. The inheritance has to be dominant since, if it were recessive, in the third generation an individual III2 could not have appeared that did not manifest this character.

Section B

Genotypes of generation I

                                   Aa (male)     x      aa (female)

Section C

Generation I            Aa       x            aa

Generation II      aa      aa      Aa      x       Aa

Generation II             A_ (AA or Aa)     aa  

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As seen in the family tree, only some men manifest the disease, so we can think that it is a sex-linked genetic transmission disease.

If it were linked to the Y chromosome, all male children of those affected would suffer from the disease, and this is not the case.

If it were linked to the X chromosome and were dominant, the daughters of those affected would manifest the disease, and this is not the case.

Therefore, it is most likely an X-linked recessive inheritance transmission.

We will try to make the crossing:

I                                       XeY          x        XEXE

II     XEY       x         XEXe          XEY         XEXe

III XEX_      XeY      XEY

The family tree supports an X-linked recessive inheritance.

There is another possibility:

The disease could also have a non-sex-linked transmission. It could be an autosomal recessive disease.

Let's try the crossover

I                        aa                x          AA

II     Aa      x       Aa             Aa            Aa

III    A_         aa       A_     

Although the first option may seem more likely, the second option is also possible.

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In the statement of the problem it is said that it is a monogenic and autosomal disease, so it is not on the sex chromosomes.

a) Indicate whether the allele that determines the presence of the disease is dominant or recessive. Give reasons for the answer (0.5 points).

If we look at the crossing 7-8, so that a diseased offspring comes out of two healthy individuals, that character has been hidden, they being carriers. Therefore, the disease is recessive, manifesting itself in homozygosity (aa).

b) Indicate the genotypes of individuals I.1, I.2, II.7 and III.11, using “A” for the dominant allele and “a” for the recessive one.

                        aa         x       Aa                                     aa         x         Aa

               aa            aa                   Aa            x            Aa           Aa               aa

                                                      A_         aa           A_

c) Define what a test cross is and what it is used for.