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Lecture

Probability, chi square calculations and Inheritance Types

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Department
Biology
Course
BIO207H5
Professor
Steven M Short
Semester
Winter

Description
Lecture 4▯ From the last lecture, we followed gene segregation in a cross of a true▯ breedingshibire fly with a wild type fly. Shibire x wild type ↓▯ F1: all not paralyzed ↓▯ F2: 3 not paralyzed : 1 paralyzed This is the segregation pattern expected for a single gene. But in an actual experiment how do we know that the phenotypic ratio is really 3 : 1 ? There is no logical way to prove that we have a 3 :1 ratio. Nevertheless, we can think of an alternative hypothesis then show that the alternative hypothesis does▯ not fit the data. Usually, we then adopt the simplest hypothesis that still fits the data.▯ A possible alternative hypothesis is that recessive mutations in two dif▯ferent genes are needed to get a paralyzed fly. In this case a true breeding paralyzed fly would have geno/ ,e:/ a b AA B Whereas wild type would have genotype:/A , /B F1: A a B b not paralyzed F2: p( a and / b = (/4 ) = 1/16 p(/aand / )–= / 4 / = 4 3 /16 p(/band A/–) = /16 p( /–and / – = the rest = / 16 This is the classic ratio for two gene segreg9 : 3 : 3 : 1 paralyzed For our hypothesis we should see a phenotypic ratio of 15 not paralyzed ▯: 1 paralyzed. Therefore, to distinguish one-gene segregation from two-gene segregation▯ we need a statistical test to distinguish 3 : 1 from 15 : 1. Intuitively, we know that in order to get statistical significance, we need to look at a sufficient number of indi▯viduals. For a chi-square ttestyou start with a specific hypothesis that gives a precise expectation. The test is then applied to the actual experimental results and will giv▯e the probability of obtaining the results under the hypothesis. The test is useful for ruling out hypotheses that would be very unlikely to give the actual results. Say we look at 16 flies in the2F and observe 14 not paralyzed and 2 paralyzed flies. Under the hypothesis of two genes we expect 15 not paralyzed flies and 1▯ paralyzed fly. We calculate the value χ▯ using the formula below. Where O is the number of individuals observed in each class and E is the number of individuals expected for e▯ach class. 2 2 2 χ2 = (O–E) = 1 + 1 = 0.067 + 1 = 1.067 Σ▯ E 15 1 (all classes) degrees of freedom (df) = number of classes – 1 From the table using 1 df, 0.05 < p < 0.5 The convention we use is that p ≤0.05 constitutes a deviation from expectation that is significant enough to reject the hypothesis. Therefore, on the basis of this sample of 16 flies we can’t rule out the hypothesis that two genes are required. Say we look at 64 F2 flies and find that 12 are paralyzed. For the hypothesis of two genes the expectation is that 4 would be paralyzed. The χ 2for this data: 2 2 χ2 = 8 + 8 = 1.07 + 16 = 17.1 60 4 From the table p < 0.005 so we reject the two-gene hypothesis. Let’s use this data to test the hypothesis of one gene segregation wh▯ich would be expected to give 16 paralyzed flies from 64 F2 flies, 42 42 χ2 = + = 0.33 + 1 = 1.33 48 16 From the table using 1 df, 0.5 < p < 0.5. Thus the data still fits the hypothesis of one- gene segregation. So far, the hypothesis that one gene is responsible for the paralyzed tr▯ait is the simplest explanation that fits the data. The way to distinguish most easily between a heterozygote and a homozygo▯te expressing a dominant trait is to cross to a homozygous recessive test strain. Test cross: cross to homozygote recessive: A/A x / a gives allA/a. i.e. all offspring will express the dominant trait. A a 1 A 1 a /a x / aives /2 aa and / 2 /a. i.e. one half of the offspring will express the dominant trait. Meendelan inheriannce n humaans For humans we can’t do test crosses, of course, but by following inhe▯ritance of a trait for several generations the modes of inheritance can usually be identified b▯y applying basic principles of Mendel. The following are guidelines for identifying different modes of inheritance in pedigrees. Autosomaaldoominant )i)Affected individuals must have at least one affected parent Exceptions to this rule will occur if a new mutation arises in one of th▯e parents (in real life a more likely explanation is extramarital paternity). Another possibility is incomplete penetrance, where other genetic or environmental factors prev▯ent the trait from being expressed in one of the parents. Autosomaalrecesssve )i)hen both parents are carriers, on average 4 of the children will be affected. i)) When both parents are affected, then all of the children will be affect▯ed. ii)iii) the trait is very rare then consanguinity is likely. That is, it is likely that parents of affected children are themselves related (e.g. cousins). X-llnked nhhertancee O X X+ x O X Y▯ (carrier)▯ ↓ O X X , O X X , O X Y, O X Y▯ (carrier) (color blind)▯ )i)When parents are a carrier O and an unaffected O , then on average, 1/2 of the daughters will be carriers and 1/2 of the sons will be affected. If the trait is rare then the vast majority of affected individuals will▯ be male which is the hallmark of X-linked traits. i)) Affected sons inherit the allele from mother • Maternal uncles often affected • Since inherited only from mother, inbreeding doesn’t increase the probability of an affected O . Connditonnalprrobabiiites Consider the following pedigree of a recessive trait. = female = male ▯▯ p(affected child) = p(mother carrier and father carrier and affected child) = / 3 / x /3 1 4 = / 9 However, if they have a child that is affected we must reassess the prob▯ability that their next child will be affected
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