Ayesha M. Khan Spring 2013

Genetic terminology The word “Gene” The term gene was not coined until 1909, when the Danish geneticist Wilhelm Johannsen first used it. In the context of genetic crosses, we will define a gene as an inherited factor that determines a characteristic. Lec-4

Important Genetic terms Lec-4 TermDefinition GeneA genetic factor (region of DNA) that helps determine a characteristic AlleleOne of two or more alternate forms of a gene LocusSpecific place on a chromosome occupied by an allele GenotypeSet of alleles that an individual possesses HeterozygoteAn individual possessing two different alleles at a locus HomozygoteAn individual possessing two of the same alleles at a locus Phenotype or traitThe appearance or manifestation of a character Character or characteristic An attribute or feature

 A phenotype can refer to any type of characteristic: physical, physiological, biochemical, or behavioral.  Thus, the condition of having round seeds is a phenotype, a body weight of 50 kg is a phenotype, and having sickle-cell anemia is a phenotype.  A given phenotype arises from a genotype that develops within a particular environment.  The genotype determines the potential for development; it sets certain limits, or boundaries, on that development. How the phenotype develops within those limits is determined by the effects of other genes and environmental factors, and the balance between these influences varies from character to character. Lec-4

Only the genotype is inherited. Although the phenotype is determined, at least to some extent, by genotype, organisms do not transmit their phenotypes to the next generation. Lec-4

Monohybrid Crosses Mendel began by studying monohybrid crosses—those between parents that differed in a single characteristic. Crosses that look at one trait at a time. Mendel crossed a pea plant homozygous for round seeds with one that was homozygous for wrinkled seeds. This first generation of a cross is the P (parental) generation. The offspring from the parents in the P generation are the F1 (first filial) generation. Lec-4

Question: When peas with two different traits—round and wrinkled seeds—are crossed, will their progeny exhibit one of those traits, both of those traits, or a “blended” intermediate trait?

Results of the monohybrid cross Lec-4  When Mendel examined the F1 of this cross, he found that they expressed only one of the phenotypes present in the parental generation: all the F1 seeds were round. Mendel carried out 60 such crosses and always obtained this result. He also conducted reciprocal crosses: in one cross, pollen (the male gamete) was taken from a plant with round seeds and, in its reciprocal cross, pollen was taken from a plant with wrinkled seeds. Reciprocal crosses gave the same result: all the F1 were round.  The following spring, he planted the F1 seeds, cultivated the plants that germinated from them, and allowed the plants to self-fertilize, producing a second generation (the F2 generation). Both of the traits from the P generation emerged in the F2.

Results of the monohybrid cross Lec-4  First, he reasoned that, although the F1 plants display the phenotype of only one parent, they must inherit genetic factors from both parents because they transmit both phenotypes to the F2 generation.  The presence of both round and wrinkled seeds in the F2 could be explained only if the F1 plants possessed both round and wrinkled genetic factors that they had inherited from the P generation. He concluded that each plant must therefore possess two genetic factors coding for a character.

Lec-4 Mendel’s monohybrid crosses revealed the principle of segregation and the concept of dominance.

Results of the monohybrid cross (contd.)  First conclusion: Each plant must possess two genetic factors coding for a character.  Second conclusion: Two alleles in each plant separate when gametes are formed, and one allele goes into each gamete. Those traits that appeared unchanged in the F1 heterozygous offspring. Mendel called dominant, and those traits that disappeared in the F1 heterozygous offspring he called recessive. When dominant and recessive alleles are present together, the recessive allele is masked, or suppressed; only the trait of the dominant allele is observed in the phenotype.  Third conclusion: The concept of dominance  Fourth conclusion: Two alleles of an individual plant separate with equal probability into the gametes. Lec-4

Principle of Segregation The principle of segregation states that each individual organism possesses two alleles coding for a characteristic. These alleles segregate when gametes are formed, and one allele goes into each gamete. The concept of Dominance The concept of dominance states that, when dominant and recessive alleles are present together, only the trait of the dominant allele is observed.

Punnett Square Lec-4 Short-hand method of predicting the genotypic and phenotypic ratios of progeny from a genetic cross.  Rules of probability for determining the outcome of a genetic cross:  Probability expresses the likelihood of a particular event occurring.  Multiplication Rule The probability of two or more independent events occurring together is calculated by multiplying their independent probabilities.  Addition Rule The probability that any one of two or more mutually exclusive events occurring is calculated by adding their probabilities.

Lec-4

The application of probability to genetic crosses Example: a cross between two pea plants heterozygous for the locus that determines height, Tt xTt. Half of the gametes produced by each plant have a T allele, and the other half have a t allele; so the probability for each type of gamete is 1/2. We can use the addition rule to determine the overall phenotypic ratios. Because of dominance, a tall plant can have genotype TT, Tt, or tT; so, using the addition rule, we find the probability of tall progeny to be ¼+ ¼ + ¼ = ¾

Binomial expansion Lec-4 When probability is used, it is important to recognize that there may be several different ways in which a set of events can occur. Example: Albinism Consider two parents who are both heterozygous for albinism, a recessive condition in humans that causes reduced pigmentation in the skin, hair, and eyes. When two parents heterozygous for albinism mate (AaxAa), the probability of their having a child with albinism (aa) is ¼ and the probability of having a child with normal pigmentation (AA or Aa) is ¾.

Binomial expansion (contd.) Lec-4 1 st scenario: What is the probability of this couple having three children with albinism? Apply multiplication rule: ¼ x ¼ x ¼ = 1/64 2 nd scenario: What is the probability of this couple having three children, one with albinism and two with normal pigmentation? 3 ways to do this: ¼ x ¾ x ¾ = 9/64 ¾ x ¼ x ¾ = 9/64 ¾ x ¾ x ¼ = 9/64 Because either the first sequence or the second sequence or the third sequence produces one child with albinism and two with normal pigmentation, we apply the addition rule and add the probabilities: 9/64 + 9/64 + 9/64 = 27/64

Binomial expansion (contd.) Lec-4  In order to find out the probability of this couple having five children, two with albinism and three with normal pigmentation, figuring out the different combinations of children and their probabilities becomes more difficult.  The binomial takes the form (a+b) n, where a equals the probability of one event, b equals the probability of the alternative event, and n equals the number of times the event occurs. The expansion:

Lec-4 How did we expand the binomial in this example? 1. In general, the expansion of any binomial (a+b) n consists of a series of n + 1 terms. In the preceding example, n= 5; so there are 5+ 1= 6 terms: a 5, 5a 4 b, 10a 3 b 2, 10a 2 b 3, 5ab 4, and b To write out the terms, first figure out their exponents. The exponent of a in the first term always begins with the power to which the binomial is raised, or n. In our example, n equals 5, so our first term is a The exponent of a decreases by one in each successive term; so the exponent of a is 4 in the second term (a 4 ), 3 in the third term (a 3 ), and so forth. The exponent of b is 0 (no b) in the first term and increases by 1 in each successive term, increasing from 0 to 5 in our example. 4. Next, determine the coefficient of each term. The coefficient of the first term is always 1; so in our example the first term is 1a 5, or just a 5. The coefficient of the second term is always the same as the power to which the binomial is raised; in our example this coefficient is 5 and the term is 5a 4 b. For the coefficient of the third term, look back at the preceding term; multiply the coefficient of the preceding term (5 in our example) by the exponent of a in that term (4) and then divide by the number of that term (second term, or 2). So the coefficient of the third term in our example is (5 x 4)/2=10 and the term is 10a 3 b 2. Follow this same procedure for each successive term.

Lec-4 where P equals the overall probability of event X with probability a occurring s times and event Y with probability b occurring t times. For our albinism example, event X would be the occurrence of a child with albinism and event Y the occurrence of a child with normal pigmentation; s would equal the number of children with albinism (2) and t, the number of children with normal pigmentation (3). The ! symbol is termed factorial, and it means the product of all the integers from n to 1. Another method :

Lec-4 1 st method (binomial expansion) 2 nd method