1. Scheda 4. SEX LINKAGE

2. Human Male Karyotype Human Female Karyotype X-LINKED inheritance: If the trait is recessive and associated with X chromosome, the phenotype will be more frequent in males (XY) than in females (XX) since they are EMIZYGOTE.

3. 1. Determine for each cross if the character is sex-linked or not and assign the genotype to the crossed individuals. FEMALEMALEFEMALESMALES paledarkdark 45dark 0 pale 0pale 48 PARENTS PHENOTYPES PROGENY PHENOTIPES a)PARENTS HAVE DIFFERENT PHENOTYPES. b)IN THE PROGENY THE PHENOTYPEs DISTRIBUTION is NOT EQUAL BETWEEN MALES AND FEMALES. c)THE GENE IS X-LINKED d) In order to establish which is the dominant allele, i observe the females: they are dark then they are X B - f) The mother is homozygous recessive X b X b while the dark daughters will be heterozygous X B (father’s allele) X b (mother’s allele) X b X B X b e)The males are emizygote then the phenotype corresponds to the unique allele that they carried. The dark parent will be B then X B Y, while the pale sons will be b (recessive allele inherited from the mother) then X b Y X B Y X b Y

4. FemaleMaleFemalesMales rough Rough 87Rough 92 Smooth 33Smooth 27 a) The parents have the same phenotype b) In the progeny males and females have the same distribution.(1:3) c) The gene is autosomic. If the gene was x-linked? All the females will be rough because they inherit the father’s allele R. e) From Rr X Rr, we expect ¾ rough (1/4 RR + 2/4 Rr) and ¼ smooth (1/4 rr) r RR/Rr R r d) Rough is dominant (R), the recessive (r) determine the smooth phenotype: The parents are both heterozygous. (Rr). 1. Determine for each cross if the character is sex-linked or not and assign the genotype to the crossed individuals. PARENTS PHENOTYPES PROGENY PHENOTIPE

5. FemaleMaleFemalesMales red Red 102Red 49 White 0White 52 a) The parents have both the same phenotype b) In the progeny we have non equal distribution between males and females. c) The gene is x-linked X R X r X R Y X r Y X R -X R Y d) Dominant allele? = females phenotypes (all red individuals then Red is dominant (X R ) e) The males are red and white. The sons will be: red X R Y and white X r Y f) The mother’s genotype must be X R X r : the sons have two different genotypes (X R Y, X r Y) g) The daughters have two different genotypes (X R X R e X R X r ) but all of them are red. 1. Determine for each cross if the character is sex-linked or not and assign the genotype to the crossed individuals. PARENTS PHENOTYPES PROGENY PHENOTIPE

6. FemaleMaleFemalesMales blackwhiteblack 98black 103 withe 0white 0 a) The parents have different phenotypes. b) All the individuals of the progeny are black, black is the dominant character (N). c) Since the trait does not segregate, we cannot define if the gene is autosomic or x- linked N n N n d) The mother is homozygous dominant: N N or X N X N e) The father is homozygous recessive: n n or X n Y FemaleMaleFemalesMales blackwhiteblack 98black 103 white 0 X N X n YX N X n X N Y 1. Determine for each cross if the character is sex-linked or not and assign the genotype to the crossed individuals. PARENT PHENOTYPES PROGENY PHENOTIPE PARENT PHENOTYPES PROGENY PHENOTIPE

7. 2. In humans the presence of a fissure in the iris (coloboma iridis) is controlled by a recessive sex-linked gene. An affected daughter is born from a normal couple. The husband asks for divorce, accusing his wife of infidelity. Is he right? We can draw the family tree. X a The daughter must be homozygous recessive. Then the parents contribute with gametes X a X A X a The mother must be heterozygote because she is health. X A Y Yes he is right he can not be the father of the girl.

8. a) If II.2 marries a normal man (X E Y), which is the probability to have children affected by haemophilia? b) If II.2 already has an affected child, which is the probability to have a second affected child? I II X e Y The son II,3 is X e Y. The mother gave the mutant allele and she is heterozygous X E X e, while the father will be X E Y X E X e X E Y The daughter II,2 inherited the allele X e with probability 1/2 (½ X E X E, ½ X E X e ). The probability to produce sick child will ¼ x ½ = 1/8 (haemophilic males). If the II.2 has already a sick child, the mother is certainly heterozygous then the probability is ¼ x 1= ¼. A recessive sex-linked gene is responsible for the most common kind of haemophilia. Considering the family tree shown below, answer the following questions:

9. Determine if the character shown in the following family tree can be due to: a) the dominant allele of an autosomic gene Try to complete the scheme with the genotypes. Aa aa Aa Hypothesis could be right If the allele is dominant, all the health individuals will be homozygous recessive aa. The sick individuals will be homozygous dominant AA or heterozygous Aa. The individual I,1 is heterozygous because in the second generation not all the individuals are sick.

10. 4. b)the recessive allele of an autosomic gene Aa A- Aa A- Aa aa If the charachter is autosomic recessive we have to suppose that two individuals that come in to the family with marriage are heterozygous. If the character is rare this is not probable If the allele is recessive, all the sick indivduals are homozygous recessive aa. Hypothesis could not be right

11. c) the dominant allele of a sex-linked gene; X A X a X A Y X a Y X a If the allele is x-linked and dominant, all the health females are homozygous recessive X a X a and the health males are X a Y. The sick men will be X A Y. The sick females will be heterozygous X A X a. Hypothesis could be right

12. d) the recessive allele of a sex- linked gene X A X a X a X a Y X A Y X A X a XA-XA- X A - The sick females of the second generations must be homozygous X a X a then the original mother must be heterozygous X A X a If the allele is x-linked and recessive, all the sick males are emizygote X a Y All the health males will be emizygote X A Y. Hypothesis could be right, but if the character is rare this configuration is not probable.