Important Sex Linked Disorder That Affects Humans - Assignment Example

?Question 2. One important sex linked disorder that affects humans is that of hemophilia. The disease is an X-linked recessive trait that results in a decreased ability to clot. Thus, if a hemophilic is cut they bleed for a much longer amount of time than a non hemophilic. Because the disorder is present on the X chromosome and is recessive, it is much easier for males to be afflicted than females. Males only inherit a single copy of the X chromosome, thus if they inherit a chromosome with the disease then they show the phenotypic characteristics. In contrast a female must inherit two copies of the allele, one from each parent to exhibit the disease. A cross of a female carrier and a male with the disease will mean that the children will have a 50% chance of inheriting the disease regardless of gender. It is the only case where a female is able to inherit the disorder. Crossing a female with the disease with a male that doesn’t have it will result in all male offspring having the disease and all females being carriers. Finally, crossing a female and a male both that have the disease will result in all offspring having it. Many genetic disorders are inherited recessively. As such the child must have two copies of the allele for it to show phenotypically. One such disorder is cystic fibrosis. This condition results in the excess build up of mucus in many organs including the lungs and liver, as well as an increased susceptibility to developing infections. Without treatment the disease often results in death at early childhood. A cross between a female carrier (Cc) and a male with the disease (CC) would result in 50% of the offspring having the disease, and the other 50% being carriers. A cross between two carriers (Cc) would result in a quarter of the offspring having the disease, half being carriers and the final quarter not inheriting the allele. Lastly, a cross between two individuals with the disease would result in all of the offspring having the disease. Although dominantly inherited disorders are rare they do exist. One such disorder is Huntington’s disease. This disease results in irreversible mental deterioration and instability. Because of the disease being dominant, there are no individuals that ‘carry’ alleles. Instead, in order to answer the questions, individuals that carry one of each allele will be used. Thus, a cross between two carriers (heterozygotes) would result in three quarters of the offspring having the disease and one not. A cross between a female that did not have the disease and a male carrier (heterozygote) would result in half of the offspring having the disease and the other half not. Finally, a cross between two individuals with the disease depends on the genotypes of each. If both have only one copy of the disease allele, then one quarter of their offspring will not have the disease. However, if even one of them is homozygous, then all offspring will have the disease. Question 3. Mendel’s laws predict the manner in which traits are passed from one generation to the next. His theories propose that two alleles exist for each trait, one which is dominant over the other. If an individual has one copy of the dominant allele then this trait is visible even if a copy of the recessive allele is present. If there are two copies of the recessive allele, then the recessive trait is visible. The inheritance of these traits follows the simple laws of probability. However, not all inheritance follows these patterns. Four variations to these laws are incomplete dominance, multiple alleles Incomplete dominance is a form of inheritance where an individual that has dominant and one recessive allele shows an intermediate phenotype between the two. For example, if a flower with RR is red, and rr is white, then one with the genotype Rr may be pink. However, it is clear that the two alleles do not blend with one another, as they are visible individually in later generations. As a consequence of this, a cross of a homozygous dominant and homozygous recessive individual will result in three phenotypes not two. Thus the pattern of inheritance does not follow Mendel’s laws. Another form of inheritance that does not follow Mendel’s laws in when there is more than two alleles present for a single gene. This occurs in human blood type, where alleles A and B are co-dominant and O is recessive. Because of this, inheritance no longer follows simple probabilities, and it the genotypes of second or third generations cannot be predicted as simply. Mendel’s laws assume that different alleles are inherited independently of one another. Consequently, a 9:3:3:1 ratio is predicted among genotypes for a cross of AaBb with AaBb. However, if the two alleles are present on the same chromosome, then there is a much higher chance that they will be inherited together. Thus, if one chromosome on has AB on it and the other ab, gametes are more likely to have one of these two combinations than Ab or aB. As a consequence, the outcome of a cross cannot be predicted through simple probability. Finally, some genes occur on the sex chromosomes, and as a consequence their inheritance does not match the predictions of Mendel’s laws. For example, if a gene is only present on the X allele and not on the Y, then whatever allele a male has it will always be visible, while the dominant and recessive relationship will still be evident for females. Question 4a. The child’s genotype is A, thus phenotype must be either IAIA or IAi. The mother’s phenotype is B, thus she must have passed on the i allele to the child, and thus have the genotype of IBi. Consequently, the child cannot have the genotype IAIA and must be IAi. The father is of phenotype A, and because of the son’s genotype he could be either IAi or IAIA. Child: A – I­­Ai Mother: B – IBi Father: A – IAi or IAIA If the father is of genotype IAi, then the following punnett square predicts the distribution of the offspring between the father and the mother. It predicts a 25% probability of each of the four possible blood types (AB; A; B; O) Father Mother IA i IB IAIB (AB) IBi (B) i IAi (A) Ii (O) If the father is of genotype IAIA, then the following punnett square predicts the distribution of the offspring between the father and the mother. It predicts a 50% distribution of blood type A and blood type AB. Father Mother IA IA IB IAIB (AB) IAIB (AB) i IAi (A) IAi (A) If the son has blood type O, then this indicates that the genotype of the father is IAi. b. For the purposes of this analysis L can be used to signify the dominant allele, which results in arms of the same length (consequently, l represents the recessive allele). A cactus with two arms of different length would thus have the genotype ll. For the cactus to have the genotype ll (phenotype: different length) it must have inherited one l allele from each of its parents. Thus, each parent could have the genotype Ll or ll. If the parent had the genotype Ll then their phenotype would be arms of the same length. Consequently either one or both of the cactus’s parents could have had the phenotype of arms of the same length. The initial cactus had the genotype ll, thus could only pass the l allele to offspring. The neighbor’s cactus had the phenotype same length, thus the genotype could either be LL or Ll. If the genotype was LL then all offspring would be Ll (LL x ll cross) and thus show the same length phenotype. As the half of the offspring was reported to have arms of the same length and the other half to have arms of different lengths, the genotype of the neighbor’s cactus must have been Ll. The punnett square for a self cross of a heterozygote cactus with the same arm length is shown below. It predicts a 1:2:1 genotypic ratio for the offspring (LL:Ll:ll), and a 3:1 phenotypic relationship (same length: different length). L l L LL (same length) Ll (same length) l Ll (same length) ll (different length) c. The fruit fly (Drosophilia melanogasta) is commonly used in genetic experiments due to their fast generation time and low expense. They produce a new generation within a couple of weeks, and can be kept and bread both easily and inexpensively. Many early genetic experiments used fruit flies, and as a consequence much of their inheritance patterns are known. Red eyed, female (heterozygous): XRXr Red eyed, male: XRY White eyed, male: XrY White eyed, female: XrXr Crosses I would do would be: 1. XRXr (red eyed female) x XRY (red eyed male) 2. XrXr (white eyed female) x XRY (red eyed male) 3. XRXR (red eyed female) x XrY (white eyed male) 1. The punnett square for the first cross is examined below. Thus, from a cross of a heterozygote red eyed female with a red eyed male will result in three quarters of the offspring having red eyes and the final quarter having white eyes. In addition all of the white eyed offspring will be male. If the cross produces 500 files as offspring, then each cross then approximately 125 offspring of each phenotype (as listed in the punnett square) would occur. As the calculation considers probability and ratios actual numbers are likely to differ, but should fall roughly within the 1:1:1:1 phenotypic ratio. Mother Father XR Xr XR XR XR (red eyes) XR Xr (red eyes) Y XRY (red eyes) XrY (white eyes) 2. The punnett square for the second cross is examined below. As can be seen, for this cross all of the female offspring will be of the genotype XRXr (red eyed), while all of the male offspring while be of the genotype XrY (white eyed). Thus the ratio of offspring of each phenotype or genotype should be 1:1, and in the case of 500 offspring, approximately 250 of each variation. Mother Father Xr Xr XR XR Xr (red eyes) XR Xr (red eyes) Y XrY (white eyes) XrY (white eyes) 3. The third cross is examined in the punnett square below. In this cross all of the offspring will have red eyes. The females will have the genotype XRXr while the males will have the genotype XRY. If 500 offspring are produced, then there should be roughly 250 of each type. Mother Father XR XR Xr XR Xr (red eyes) XR Xr (red eyes) Y XRY (red eyes) XRY (red eyes) Question 5a. Assuming each of the genes are independently assorted, then the probability of certain combinations can be determined using the product rule. Thus for any gene, for a heterozygous cross the probabilities of each outcome is as follows: A a A AA Aa a Aa aa Thus, 50% for Aa, 25% for AA and 25% for aa. Therefore, the probabilities for certain combinations are as follows: aabbccdd = ? x ? x ? x ? = 1/256 = 0.004 AaBbCcDd = ? x ? x ? x ? = 1/16 = 0.063 AaBBccDd = ? x ? x ? x ? = 1/64 = 0.016 AABBCCDD OR aabbccdd = (? x ? x ? x ?) + (? x ? x ? x ?) = 1/128 = 0.008 b. In this example the allele striped fur can be allocated the symbol F (and white fur f), while normal eyes can be assigned N (and cross eyes n). Thus, a tiger that is heterozygous for each gene would have the phenotype FfNn. Each tiger will produce four types of gametes: FN, Fn, fN and fn. The phenotypes that could result would be: striped fur normal eyes, striped fur crossed eyes, white fur normal eyes and white fur crossed eyes. The actual genotypes could be as follows: Striped fur normal eyes: FFNN, FFNn, FfNN, FfNn Striped fur crossed eyes: FFnn, Ffnn White fur normal eyes: ffNN, ffNn White fur crossed eyes: ffnn The genotype of the striped fur normal eyes tiger could be one of the four listed above, while the genotype of the white fur crossed eyes tiger must be ffnn. If some of the offspring have striped fur and crossed eyes, then the striped fur normal eyed tiger must have carried the n allele, and thus his genotype was either FFNn or FfNn. c. For this example, the ability to run normally can be coded N (while the waltzing allele can be coded n). Furthermore, the dominant black hair color can be coded B while the recessive brown can be coded b. Thus, a mouse that is heterozygous for both conditions would have the genotype NnBb while one that is homozygous for both (and is running and black) would be NNBB. A cross would yield the following offspring: NB Nb nB nb NB NNBB NNBb NnBB NnBb NB NNBB NNBb NnBB NnBb nb NnBb Nnbb nnBb nnbb nb NnBb Nnbb nnBb nnbb The ratios of genotypes produced from this cross are: 2 NNBB: 2 NNBb: 2NnBB: 4NnBb: 2 Nnbb: 2nnBb: 2 nnbb or NNBB: NNBb: NnBB: 2NnBb: Nnbb: nnBb: nnbb. In terms of phenotype this can be expressed as 10 running black: 2 running brown: 2 waltzing black: 2 waltzing brown, 5 running black: 1 running brown: 1 waltzing black: 1 waltzing brown. A mouse that is homozygous for running and is brown would have the genotype NNbb, while one that is waltzing and is black (heterozygous) would have the genotype nnBb. A cross would yield the following offspring. Nb Nb Nb Nb nB NnBb NnBb NnBb NnBb nb Nnbb Nnbb Nnbb Nnbb nB NnBb NnBb NnBb NnBb nb Nnbb Nnbb Nnbb Nnbb Thus, of the offspring produced by this cross 50% would have the genotype NnBb (running, black) and 50% would have the genotype Nnbb (running, brown). The genotypic ratio would then be NnBb:Nnbb, and the phenotypic ratio would be running black: running brown. NB Nb nB nb NB NNBB NNBb NnBB NnBb Nb NNBb NNbb NnBb Nnbb nB NnBB NnBb nnBB nnBn nb NnBb Nnbb nnBb nnbb Thus the ratio for this cross would be: 1 NNBB: 2 NNBb: 2 NnBB: 1 NNbb: 4NnBn: 2 Nnbb: 1 nnBB: 2 nnBb: 1 nnbb 9 running black: 3 waltzing black: 3 running brown: 1waltzing brown Consequently, if 250 offspring were produced, then approximately: 15 mice would be homozygous dominant 50 mice would be running and brown 15 mice would be waltzing and brown 140 mice would be running and black Read More