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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 is not the case.
• 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 this character and some of their children, not all, also. This seems to indicate that the disease would be determined by a dominant allele and that both A and B would be heterozygous for that trait (Aa).

Similarly, in the case of the couple CD, C would manifest the character (Aa) and D would not manifest it (aa), having children with that character (Aa) and without that character (aa).

The genotypes of the family would be:

I                                            Aa               x                  Aa

II                 A_           aa       x         aa           aa              Aa           x           aa

II                    aa          aa            aa                            aa    Aa     aa      Aa

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a) Indicate if the character presents dominant or  recessive inheritance. Give reasons for the answer.

From the crosses observed in the diagram, it can be deduced that the studied character is dominant. This is evidenced in the cross between II4 and II5, that if they were homozygous recessive, all their children should be homozygous recessive for this trait.

At the cross between I1 (represented in black) and I2 (represented in white), individuals with this character (black) and without it (white) originate, so that individual I1 would be heterozygous (Aa), and I2 would be homozygous recessive. (aa).

This would also be confirmed between crosses II4 and II5, both heterozygous (Aa), which would give rise to individuals with that character (black, A_) and without that character (white, aa). 

b) Indicate the genotypes of individuals from generation I and  individuals II.4 and II.5, using “A” for the dominant allele  and “a” for the recessive allele.

I                            Aa             x                 aa

II                     aa     Aa         Aa         Aa      x          Aa

III                                                              A_    aa  

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The first thing we notice is that curly hair appears in the same way in males as in females, and although there is no male with straight hair, 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 males would have that character, and a male who had that character should be the son of another male who manifested that character. This is not the case.
• If it were linked to the X chromosome, being dominant, all the descendants (females) of a father with that trait should present it. This is not the case.
• If it were linked to the recessive X chromosome, all male children of a mother with that trait should manifest it. This does occur in the pair on the right, but from the crossing between a male (X^aY) and a female (X^aX^a) with curly hair, all the offspring should have curly hair, and it does not happen.

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

In the crossing indicated, it can be seen that from the crossing of two curly-haired guinea pigs, two curly-haired guinea pigs and one with straight hair are born. If the curly-haired character were recessive, all children should have had curly hair.

Therefore, it seems clear that the allele that determines curly hair is dominant over that that determines straight hair.

In this genealogical tree, no male guinea pig with straight hair appears, but this inheritance is not linked to sex since that offspring would not appear that way if the character were linked to one of the sex chromosomes.

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a) Only two individuals, two men, are those who manifest this disease. That can make us think that it is possibly a character linked to sex.

In the crossing of generation I, between a healthy woman and a sick man, all their offspring are healthy. This suggests that the gene that determines the disease is on the X chromosome man (X^aY) and, that being recessive, not manifested in daughters (X^AX^a). In the children's father also happens because the Y chromosome (X^AY).

Generation II women are carriers (X^AX^a) and when they have children with a normal man (X^AY) they have healthy daughters (X^AX^A) and the probability of having 50% healthy children (X^AY) and 50% of having them sick (X^aY).

b) Genotypes:

I                                              X^aY           x         X^AX^A

II         X^AY     x    X^AX^a                 X^AY           X^AX^a

III        X^AX_          X^aY          X^AY

Although this is the most likely solution, since the only two affected are men, it may also be possible that it is a trait linked to an autosomal recessive gene, not sexual. The genotypes of the family would be:

I                                              aa                 x                AA

II                  Aa           x         Aa              Aa               Aa

III                A_          aa         A_

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a) Indicate if the character presents dominant or recessive inheritance. Give reasons for the answer.

It is a character determined by a dominant allele. If the trait were determined with a recessive allele (aa), all the offspring of I1 (aa) and I2 (aa) would have that trait (aa).

b) Indicate the genotypes of the individuals of generation I and their descendants (II.2 to II.5), using “A” for the dominant allele and “a” for the recessive allele.

Given that inheritance is dominant and there are descendants of I1 and I2 that have and others that do not have that trait, it seems logical to think that both I1 and I2 are heterozygous for that trait.

Therefore, the genotypes would be as follows:

I              1 (Aa)                          I2 (Aa)

II       II1 (aa) II2 (Aa, since III2 is aa) II3 (aa) II4 (aa) II5 (Aa, since III2 is aa) II6 (aa)

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We see that the character of the attached earlobe appears in similar proportions in men and women.

Both in the first couple of generation I and in the couple of generation II we see that individuals who do not have this character appear individuals who do and others who do not. 

If it were a character linked to a sex chromosome:

• If it were linked to the Y chromosome, no man would have it without his father, so this is not the case.
• If it were linked to the X chromosome and had a dominant character, no daughter could have that character if her mother did not have it.
• If it were linked to the X chromosome and had a recessive character, a woman with that character should have received it through her mother (who manifested it or was a carrier) and from her father. In the generation II couple we see that a woman is born with this trait from a couple that does not have it, so it is also shown that it cannot be a recessive trait linked to the X chromosome.

Therefore, it is an autosomal character. In the first couple of generation I, from two individuals who do not have this character, there appear individuals who do and others who do not. This indicates that it is a recessive character that only manifests itself in homozygosity.

The genotypes of the family would be:

I                       Aa               x               Aa                        aa           x         Aa

II           A_          aa               aa           Aa          x           Aa                  aa

III                                                             aa               A_

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a) The first thing we look at is that this disease affects both men and women, although it is more common in men and 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 is not the case.
If it were linked to the recessive X chromosome, all male children of a mother with that trait should manifest it. It could be the case, but the left part of the graph, in generation III, all the presumed descendants of XhXh and XhY should manifest the disease, and it is not the case.
Therefore, the disease is due to an allele on an autosomal dominant chromosome. It can be seen in the generation II couple on the left who have children with and without the disease. If it were due to a recessive allele, it would only manifest itself in homozygosity.

b) People affected by the disease will be those who have the dominant H allele, homozygous HH, or heterozygous Hh. Healthy individuals will be homozygous recessive for this hh trait.

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a) Indicate if this anomaly is dominant or recessive. Give reasons for the answer.

The problem statement says that the abnormality is caused by a single sex-linked gene located on the X chromosome. You have to remember that men only have one X chromosome and women, two, XX. Therefore, as no woman has the anomaly, we must think that it is a recessive anomaly linked to the X chromosome. If it were dominant, all mothers of children with this characteristic should manifest it.

b) List the possible genotypes of all individuals. Use the letter A (uppercase) for the dominant allele and the letter a (lowercase) for the recessive.

                                               X^DY     x    X^DX^d                        X^DY     x       X^DX^d

  X^dY       X^DY     X^DX_       X^DY                x              X^DX^d             X^DX_

X^dY     X^DY     X^DX^d    x        X^DY

   X^dY

c) Taking into account, exclusively, the data of this genealogy, could the possibility that the character was autosomal be totally excluded? Give reasons for the answer. 

It cannot be excluded, taking into account exclusively the data of this genealogy, that it is an autosomal character. The three individuals that present the anomaly could be homozygous recessive for the allele that determines said anomaly, and their parents, heterozygous, who, being recessive, would not manifest the anomaly.

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The first thing that strikes us is that of a couple, he is sick and she is healthy, all their male descendants are healthy (at least the two that exist) and all women have the disease. This seems to indicate that it is a genetic inheritance linked to sex, since it is linked to the X chromosome that the father has passed to them, so it seems that it is a dominant character.

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 recessive X chromosome, all male children of a mother with that trait should manifest it. This is not the case either.

The genotypes of the family members are:

I                                                         X^AY                     x                     X^aX^a

II             X^AX^a       X^aY      X^aY        X^AX^a          X^AX^a        x        X^aY

III                                                                             X_X^a     X_X^a     X_Y     X_Y

The granddaughters could be either X^AX^a or X^aX^a, at 50% probability. That is, the 50% probability of being sick (X^AX^a) and the other 50% of being healthy (X^aX^a).

Grandchildren, equal, 50% healthy (X^aY) and 50 patients% (X^AY).

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a) Indicate if this type of cataract is dominant or recessive. Give reasons for the answer. 

In the statement it is said that this disease is produced by a single sex-linked gene located on the X chromosome, and in the genealogical tree we see that it appears equally in men and women, alternating in the generations.

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 true in this family.

If it were linked to the recessive X chromosome, all male children of a mother with that trait should manifest it. In generation III we see that this case does not occur.

Therefore, it is a dominant character. 

b) List the possible genotypes of all individuals. Use the letter A (uppercase) for the dominant allele and the letter a (lowercase) for the recessive.

X^aX^a              x             X^AY

X^aY     x          X^AXa        X^aY        X^AX^a         X^AX^a          x            X^aY

X^aX^a   x     X^AY       X^aY       X^aX^a                                     X^AY

X^aY    X^aY     X^AX^a      X^AX^a      x       X^aY

c) Indicate the probability that the woman indicated with an arrow has a descendant with this type of cataract: (1) if it is a child; (2) if it is a girl.

 X^AX^a      x       X^aY

Gametes                X^A     X^a           X^a     Y

F1                     X^AX^a     X^aX^a          X^AY    X^aY

50% of girls will have the disease.

50% of children will have the disease.