Monohybrid Cross Experiments - Essay Example

Genetics By [Presented to] of a Monohybrid cross: A monohybrid cross is a breeding experiment between parental generation organisms that differ in one trait. Monohybrid inheritance involves inheritance of a single trait. A monohybrid cross between two pure breeding plants, one that is a tall plant (dominant trait) and the other that is a short plant (recessive trait) is expected to produce an F1 (first) generation of plants that are al tall because the allele of tallness would dominant to shortness in the offspring produced. However, all plants produced in F1 generation would be heterozygous for their trait unlike their parents that were homozygous for their respective trait. If the F1 generation is further paired among itself by a yet another monohybrid cross than an F2 generation of plants will be produced. The expected phenotype of F2 generation would be 75% tall plants and 25% short plants. In F2 generation 25% of plants would be tall but would be homozygous for their trait of tallness. Another 50% of plants would also be tall but would be heterozygous for their trait of tallness while the remaining 25% of would be short and would be recessive for their trait. Dihybrid cross: A dihybrid cross is a cross between two individuals that involves two pairs of contrasting characteristics. If two plants that are pure bred (homozygous) for two contrasting traits for example one plant that is tall and produce yellow seeds (both dominant traits) and the other plant that is short and produces green seeds (both recessive traits) are crossed with other, than the F1 generation produced would be all heterozygous for both characteristics. But only dominant trait would appear in the phenotype of F1 generation and all plants of this generation would appear tall and would produce yellow seeds. If the members of this F1 generation are further crossed with each other than an F2 generation would produced which would have different types of plants: tall plants producing yellow seeds, tall plants producing green seeds, short plants producing yellow seeds, short plants producing green seeds in a ratio of 9:3:3:1 b) If a boar that is homozygous for lop ears and spotted skin is sow that is homozygous for pricked ears and unspotted skin than the F1 generation produced will all have pricked ear and spotted skin, as these two traits are dominant. The genotype of F1 generation would be PpSs. (Key: P= pricked ears, p=lop ears, S=spotted skin, s-unspotted skin). However the phenotype and genotype of F2 generation produced is shown as follows Heterozygous F1 generation Pricked ear spotted skin mating * Different gametes produced byF1 generation The Punnett Square below shows the F2 generation produced by random mating PS Ps pS ps PS PPSS PPSs PpSS PpSs Ps PPSs PPss PpSs Ppss pS PpSS PpSs ppSS ppSs ps PpSs Ppss ppSs ppss The Punnett Square shows the possible genotype results of F2 generation. The genotype results explain the phenotype results such that nine offspring are pricked ear and spotted, three offspring are pricked ear and unspotted, three other offspring are unpricked (lop) ear and spotted while one offspring is lop eared and unspotted 2.a)Homogametic sex: It refers to that gender of specie that has both sex chromosomes identical. For instance in humans females have two X sex chromosome thus females in humans are referred to as homogametic sex. But in birds males of two identical Z chromosomes so in birds, males are homogametic sex Heterogametic sex: It refers to that gender of a specie that has two heteromorphic (differently shaped) sex chromosome. In humans it refers to males because they have two different sex chromosomes X and Y. However, in birds it refers to females who have two different sex chromosomes Z and W. b) Of all the kittens produced in the F1 generation there is a chance of 50% of them born as male kittens and 50% of them are born as female kittens. Since the female parent has tortoiseshell coat and tortoiseshell coat only appears in the phenotype when the creature is heterozygous for B1B2 allele, it becomes clear that B1 and B2 are sex linked genes that can only be present on X chromosomes and not Y chromosomes. This is because a female cat has both sex chromosome as X and since here a female parent has two sex linked genes B1 and B2 it confirms that the female parent has one X chromosome containing B1 allele and the other X chromosome containing B2 allele. However since the parent father has only one X chromosome he has only B2 allele and it expresses in his phenotype as black coat. The possibility of coat colour of the F1 generation however can be predicted through the following punnett square. X B2 allele from the father Y with no allele X with B1 allele from mother XX female with B1B2 XY male with B1 only X with B2 allele from mother XX female with B2B2 XY male with B2 only The Punnett Square shows that of all the female kittens produced in F1 generation there is 50% chance that any particular kitten has tortoise shell coat and is heterozygote B1B2 and 50% chance that any particular female kitten has black coat and is homozygous for B2 allele. Of all the male kittens produced there is 50% chance that any particular male kitten has ginger colour coat and possess one X chromosome with B1 allele and 50% chance that any particular male kitten has black coat and possess one X chromosome with B2 allele. Since the genes for the length of hair is present on autosomal chromosome and short hair is dominant over long hair and the female parent is homozygous for short hair, all offspring of F1 generation will be short haired because all kittens will inherit one chromosome with allele for short hair from their mother which will be dominant to the recessive allele for long hair they inherit from their father. Though the F1 generation will all be shorthaired but unlike their mother all the kittens will be heterozygous for short hair. c) If F1 generation is allowed to mate freely than only tortoiseshell coat female cats be potential mother for ginger coat color males because at the time of mating they can contribute an X chromosome with B1 allele that can express ginger coat in the phenotype of the offspring. However any male cat in the F1 generation can be a potential father for a ginger coat colour male because father just has to contribute Y chromosome with no allele for coat colour at the time of mating. The tortoiseshell female cats and all male F1 cats can also be potential parents for producing longhaired offspring because all F1 generation is heterozygous for short hair and possess one recessive allele for long hair in one of their autosomal chromosomes. For long hair to be expressed in the phenotype of the offspring it should be homozygous for long hair recessive allele. Thus if recessive alleles for long hair are contributed from both the parents at the time of mating than a longhaired offspring can be produced. Thus to produce specifically a ginger coat longhaired male cat, tortoiseshell females and any male cat of F1 generation should be allowed to mate. 3. a) Mutation is a permanent change in the DNA sequence of a gene. Mutations can be caused by copying errors in the genetic material during cell division, by exposure of DNA to ionizing radiations, by viruses or by chemical mutagens or can occur deliberately under cellular control during processes such as hypermutation. b) i. Frameshift mutation: It s a point mutation involving either a deletion or insertion of a nucleotide in a gene. Due to frameshift mutation the normal reading frame used to decode the nucleotide triplets in the gene is altered to another reading frame. This altered reading frame thus results in altered protein product. Using the base sequence given frame shift mutation can be showed as follows. After the first triplet code if another base of adenine is added than the base sequence would change to CGA-AAU-UUU-GGC-UAG-AGG-C from the original sequence of CGA-AUU-UUG-GCU-AGA-GGC. This altered base sequence would produce an altered amino acid chain and thus an altered protein product. ii) Missense mutation: It is a mutation in which a nucleotide is substituted within a gene such that the substitution changes the codon so that it codes for a different amino acid in the protein chain. If in the given base sequence the second triplet code is changed from AUU to AGU i.e. the second base of uracil is changed to guanine than this altered triplet code will code for amino acid serine instead of amino acid isoleucine in the protein chain. This will result in an altered protein product. iii) Same-sense mutation: It is a mutation which changes the nucleotide sequence of a codon but does not change the amino acid the codon codes for due to the degeneracy of the genetic code. For example if in the given base sequence of CGA - AUU - UUG - GCU - AGA – GGC, the fifth codon is changed from AGA to AGG i.e. that if the last base of adenine of this codon is substituted to guanine than too the altered codon will code for amino acid arginine as the original codon had to code for amino acid arginine. This altered base sequence will not produce an altered protein product. 4 a) Hardy-Weinberg principle states that the gene frequencies and genotype ratios in a population remains the same from generation to generation unless disturbing influences like non-random mating, genetic drift, mutation, gene flow or natural selection are introduced. b) The Hardy Weinberg’s Equilibrium formula is p+q=1 where p is the frequency of one allele which usually dominant in a population while q is the frequency of the other allele which is usually recessive in a population. If both sides of the equation are squared than the equation will become as (p + q)2 =1. If this equation is expanded than it becomes p2 +2pq + q2 =1 .In this equation p2 give us the frequency of homozygous dominant individuals in a population, q^2 give us the frequency of homozygous recessive individuals in a population and 2pq give us the frequency of hetrozygote in a population. 5a) Terminal haemorraghic syndrome (TMS) is caused by a recessive gene and rats that are homozygous recessive for this gene suffer from this disease and eventually die of it. In a breeding population of 400 rats it has been seen that 48 have died of TMS. Since some of the rats in the population are homozygous recessive for TMS gene it confirms that the parents who were bred were either both or one of them was carrier of the recessive gene. If it is assumed that both the parents were heterozygous for the gene, i.e. both carrying one dominant and one recessive gene for TMS than the following diagram can see the expected genotype of the offspring. (Key T= dominant TMS gene, t= recessive TMS gene) Both heterozygous parents mating Genotype of * parents Gametes Produced By parents Random mating Genotype of offspring TT Tt Tt tt The diagram show that if both parents are carriers of TMS recessive gene but are not suffers of TMS than the offspring produced is non suffer and suffer of the disease in the ratio of 3:1. According to the figures given to us 48 out of 400 rats are suffers of TMS and the ratio of non-suffers to suffers in this population is also roughly 3:1. Through the diagram we can predict that parents of the given population of rats were heterozygous for TMS gene. The diagram above also shows that of all the rats who are non-sufferer of TMS the ratio of number of rats that are pure breeding for dominant allele to the number of rats who are carriers of TMS recessive allele is 1:2. Through this diagram we can predict that the among the given population of 400 rats the ratio of number of rats who are homozygous dominant of TMS gene to the number of rats who are carriers of one TMS recessive allele is 1:2. If we calculate according to this ratio than we find that out of population of 400 approximately 235 rats are carriers of a TMS recessive allele and are heterozygotes while 117 rats are pure bred for dominant allele. b) To eradicate TMS from the rats’ colony, only rats homozygous for dominant allele should be allowed to breed with each other to produce offspring. To test which of the rats are homozygous dominant and which are heterozygotes all non-sufferer rats of the population should be first test crossed with rats that are homozygous recessive of TMS allele. If the cross breeding produces no sufferers in the F1 generation than the rat that is being tested is homozygous dominant and could be taken as potential parent to breed further population of rats without the TMS disease. However, if the test -cross produces even one offspring who is homozygous recessive for the TMS allele than the rat who is being tested is a heterozygous for TMS allele and should not be taken as a potential parent to breed future population of rats. This way by selecting only those rats that are pure bred for dominant allele as parents to breed future population of rats every time, the disease of TMS can be effectively eradicated from the colony of rats as the recessive allele of TMS which causes the disease will disappear from the gene pool of the colony. References Freese, Ernst (1959). The Difference between Spontaneous and Base-Analogue Induced Mutations of Phage T4. Proc of NAS 45 (4): 622–633.  Maki H. 2002. Origins of spontaneous mutations: specificity and directionality of base-substitution, frameshift, and sequence-substitution mutageneses. Annual Review of Genetics 36:279-303. Emigh, T.H. (1980). A comparison of tests for Hardy-Weinberg equilibrium. Biometrics 36: 627 – 642. Stern, C. (1943). "The Hardy–Weinberg law". Science 97: 137–138 Wigginton, J.E., Cutler, D.J., Abecasis, G.R. (2005). A note on exact tests of Hardy-Weinberg equilibrium. American Journal of Human Genetics 76: 887 – 893. Read More