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Scheda 4. SEX LINKAGE Human Female Karyotype Human Male 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. 1. Determine for each cross if the character is sex-linked or not and assign the genotype to the crossed individuals. PROGENY PHENOTIPES PARENTS PHENOTYPES FEMALE pale MALE X b Xb dark FEMALES XB Y dark 45 pale 0 MALES XB Xb dark 0 pale 48 Xb Y 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 XBe)The males are emizygote then the phenotype corresponds to the unique allele that they carried. The dark parent will be B then XB Y, while the pale sons will be b (recessive allele inherited from the mother) then Xb Y f) The mother is homozygous recessive Xb Xb while the dark daughters will be heterozygous XB (father’s allele) Xb (mother’s allele) 1. Determine for each cross if the character is sex-linked or not and assign the genotype to the crossed individuals. PROGENY PHENOTIPE PARENTS PHENOTYPES Female rough Male Rr rough Females Rr Rough 87 Smooth 33 Males RR/Rr Rough 92 RR/Rr Smooth 27 r r rr 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. d) Rough is dominant (R), the recessive (r) determine the smooth phenotype: The parents are both heterozygous. (Rr). e) From Rr X Rr, we expect ¾ rough (1/4 RR + 2/4 Rr) and ¼ smooth (1/4 rr) If the gene was x-linked? All the females will be rough because they inherit the father’s allele R. 1. Determine for each cross if the character is sex-linked or not and assign the genotype to the crossed individuals. PROGENY PHENOTIPE PARENTS PHENOTYPES Female red Male X R Xr red Females XR Y Red 102 White 0 Males XR - Red 49 XR Y White 52 Xr Y 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 d) Dominant allele? = females phenotypes (all red individuals then Red is dominant (XR) e) The males are red and white. The sons will be: red XR Y and white Xr Y f) The mother’s genotype must be XR Xr: the sons have two different genotypes (XR Y, Xr Y) g) The daughters have two different genotypes (XR XR e XR Xr ) 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. PROGENY PHENOTIPE PARENT PHENOTYPES Female black Male NN white Females nn black Males 98 Nn withe 0 black 103 N n white 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 xlinked d) The mother is homozygous dominant: N N or XN XN e) The father is homozygous recessive: n n or Xn Y PROGENY PHENOTIPE PARENT PHENOTYPES Female black Male XN XN white Females Xn Y black 98 white 0 Males XN Xn black 103 XN Y white 0 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. XA Y XA Xa Xa Xa The daughter must be homozygous recessive. Then the parents contribute with gametes Xa The mother must be heterozygote because she is health. Yes he is right he can not be the father of the girl. A recessive sex-linked gene is responsible for the most common kind of haemophilia. Considering the family tree shown below, answer the following questions: XE Y XE Xe I II Xe Y The mother gave the mutant allele and she is heterozygous XE Xe, while the father will be XE Y The son II,3 is Xe Y. The daughter II,2 inherited the allele Xe with probability 1/2 (½ XE XE, ½ XE Xe). a) If II.2 marries a normal man (XE Y), which is the probability to have children affected by haemophilia? The probability to produce sick child will ¼ x ½ = 1/8 (haemophilic males). b) If II.2 already has an affected child, which is the probability to have a second affected child? If the II.2 has already a sick child, the mother is certainly heterozygous then the probability is ¼ x 1= ¼ . Determine if the character shown in the following family tree can be due to: a) the dominant allele of an autosomic gene aa Aa Aa Aa aa Aa aa aa Aa aa aa aa aa aa Try to complete the scheme with the genotypes. 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. Hypothesis could be right 4. b)the recessive allele of an autosomic gene Aa aa aa aa Aa aa Aa Aa aa Aa A- A- A- A- If the allele is recessive, all the sick indivduals are homozygous recessive 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 Hypothesis could not be right c) the dominant allele of a sex-linked gene; XA Y Xa Y XA Y Xa X a XA Xa Xa Y XA Xa XA Y Xa Xa Xa Y Xa Y Xa Xa Xa Y Xa Xa If the allele is x-linked and dominant, all the health females are homozygous recessive Xa Xa and the health males are Xa Y. The sick men will be XA Y. The sick females will be heterozygous XA Xa. Hypothesis could be right d) the recessive allele of a sex- linked gene Xa Y XA Y Xa Y XA Xa Xa Xa XA Y Xa Xa Xa Y XA Xa XA Y XA Y XA Y XA- XA - If the allele is x-linked and recessive, all the sick males are emizygote Xa Y All the health males will be emizygote XA Y . The sick females of the second generations must be homozygous Xa Xa then the original mother must be heterozygous XA Xa Hypothesis could be right, but if the character is rare this configuration is not probable.