Lab 4 Mendel and Pedigrees Answer

1 Example 1: a) Autosomal Recessive Trait. For this example we will use the previously mentioned case of albinism. Albinism results from the homozygous occurrence of the autosomal recessive allele a (genotype aa). Consider two phenotypically normal parents, who have some albino children. What would be the possible genotypes of both the parents and the offspring? (use a Punnett square to show your work). The genotypes for both parents would have to be heterozygous, Aa, while the albino offspring would, like all expressing the albino phenotype, have genotype aa. As this Punnett square shows, the only way for two phenotypically normal parents to produce an albino child is for both parents to have one copy of the recessive albinism allele – thus, they are heterozygotes. (If even one parent was homozygous for the non-albinism allele, all of the children would have at least one A allele and thus would not be albino.) b) What genotypes would we expect from a family consisting of a normal man and an albino woman who have two albino and two normal children? We know that the woman’s genotype must be aa, since she expresses the albino phenotype. The albino children would also be aa. The normal father, and the normal children, must be heterozygotes, genotype Aa. As above, in order for a normal parent to produce an albino child, that parent must be a heterozygote; in this case, because mom is homozygous albino, the non-albino kids will be heterozygotes as well – they will certainly get an albinism allele from her, and they must have inherited a normal allele from dad, as seen in the Punnett square above. c) What genotypes would we expect for a family consisting of two albino parents who have only albino children? In this family, because everyone is an albino, everyone must be homozygous for the albinism allele, genotype aa. 2 Example 2: Codominance: Sickle-Cell Anemia (Autosomal Trait, i.e a trait that is not found on one of the sex chromosomes, X and Y) Let A = normal allele S = sickling allele Fill in the genotypes for the members of the pedigree. In grading this, I simply looked to see if you had labeled each individual in the pedigree with their genotype, or simply labeled the key. I’ve filled in the genotypes in the key. Generation Number Normal, AA I Sickle-Cell Trait, AS II III Sickle-Cell Anemia, SS IV If a heterozygote were to marry a normal homozygote, what proportion of the offspring would be homozygous for sickle trait? Fill in the grid. MALE A FEMALE A A AA AA S AS AS Proportion of offspring with the following genotypes: __1/2 AA __1/2 AS ___0 SS As the Punnett square shows, none of the offspring of this couple would be homozygous for sickle trait. 3 Example 3: Dominant or Recessive? (Autosomal Trait) By examining the pedigrees below, determine if the following traits are dominant or recessive: Let F = freckled, Let A = attached earlobe N = not freckled U = unattached earlobe FRECKLES ATTACHED EARLOBES Observed Phenotypes Observed Phenotypes Freckled Attached Not Freckled Generation Number I Unattached Generation Number I II II III III IV IV V V Dominant Allele __F__ Dominant Allele ___U___ Hints: Look for a case where parents with similar phenotypes (i.e. both have the trait) have at least one offspring with the opposite condition (i.e. does NOT have the trait). The child must be homozygous for the recessive allele and each parent must be heterozygous. (Try doing a few “test” Punnett squares to see that this is so.) Often, but not always, there will be more individuals with the dominant phenotype. Generally, but not necessarily, dominant traits won’t skip generations. When you think you’ve got it, test that you’re right by plotting the genotypes for individuals on the pedigree. For the first pedigree, note that there are more individuals with the freckled phenotype, which is one clue that the F allele is dominant. Also, note that those who have freckled always have at least one freckled parent, which also hints that F is dominant; the trait also does not skip generations. For the second, the give-away that U is the dominant allele is in generation II, where there’s an attached-earlobe daughter born to unattached-earlobe parents. When you have both parents expressing one variant and a child expressing the other, it means that the child is homozygous for the recessive allele (in this case, the A allele) and, incidentally, that both parents must be heterozygous. Since the A is recessive, U must be dominant. 4 Example 4: Sex-linked Inheritance (e.g. genes found on the X-chromosome) Genes are said to be sex-linked when they occur on those chromosomes (in humans, the 23rd pair of chromosomes) which determine the sex of the individual, the X and Y chromosomes. A female offspring receives an X from her mother and an X from her father. A male offspring receives an X from his mother and a Y from his father. Most of the sex-linked genes are carried on the X chromosome since the Y chromosome contains very few of the loci found on the X chromosome. This means that while a female will have two alleles for the sex-linked trait, a male will only have one allele. Hence, the male will always phenotypically show the allele carried on his X chromosome (whether it is dominant or recessive) because he does not have a second allele to potentially mask the first one. In humans, there are about 70 traits, most of them pathological, that are certainly X-linked (ex. color blindness, G-6-PD deficiency, and hemophilia). Most X-linked traits are recessive, although some are dominant. Color blindness is a sex-linked recessive mutation. In the following pedigree, fill in the genotypes. Let XB = normal phenotype Xb = color blind Where XBXB and XBXb are normal females, XBY is a normal male, XbXb is a colorblind female, and XbY is a colorblind male. Hints: Since this is a sex-linked trait, the male carries only one allele and therefore phenotypically shows his genotype. First go through and fill in the genotypes of the males. If a son shows a trait and his mother doesn’t, then the trait must be recessive and mom is heterozygous. You may find it helpful to keep track of whom each X chromosome comes from. = color blind trait is expressed Generation Number I II X BX b X BX b = normal trait is expressed XB Y XbY X bY XB X B XB Y III X BY XBXb XbXb XBY XBXb X BY XBXb XBY In the case of an X-linked recessive gene, males, having only one X chromosome, will always express the recessive, whereas women will only express the recessive phenotype if they have copies of the recessive allele on both of their X chromosomes. The place to start here is by indicating the genotypes of those who are colorblind, and then trace the history of the X chromosomes from parent to offspring – for instance, the woman on the far left of generation II must be heterozygous because she has produced a colorblind daughter in addition to her normal son. A couple of problem areas – the woman on the far right of generation II is almost certainly homozygous 5 normal, because she has 5 normal children (to whom she passed an XB chromosome); however, it’s still possible that she’s a heterozygote who just happened not to pass on her Xb chromosome to any of her children. The two unlabeled women in the middle of generation II could be either XBXB or XBXb ; we know that each had to inherit an XB from her normal father, but each could have inherited either of heterozygous mom’s chromosomes; since they have no offspring, we can’t tell for sure. Statistically we would expect one of each of the normal female genotypes; in grading, I looked for some sort of acknowledgment of this. Example 5: Determining Mode of Inheritance. Examine the pedigrees below. Using the information and hints presented previously, determine the most likely mode of inheritance for the traits illustrated in each pedigree. (As before, black circles and squares represent individuals that express the trait.) For each pedigree, indicate the probable mode of inheritance, and briefly explain how you made that determination. Pedigree (a) represents autosomal recessive inheritance. We know that it’s recessive because the affected individuals are born to unaffected parents (thus, the parents must be heterozygous, and the affected offspring homozygous for the recessive allele). We kow that it’s autosomal (not sex-linked) because both women and men are affected; also, if it were sex-linked, it wouldn’t be possible to have an affected daughter of an un-affected father (as we see in the third generation here). Pedigree (b) represents autosomal dominant inheritance. Clues that this trait is dominant include the fact that affected individuals are numerous (the trait doesn’t skip generations) and that affected offspring have affected parents; clues that it’s autosomal include the fact that both men and women are roughly equally affected, and you’ve got the trait passing from affected fathers to affected sons, even with an unaffected mother. 6 Pedigree (c) represents sex-linked recessive inheritance. We know that it’s recessive because there’s an affected child born to two unaffected parents; we know that it’s sex-linked because it preferentially affects males. Also notice that the affected male in the third generation has inherited the trait from his affected grandfather, though his mother who is a carrier.