UNIT ____: More Complex Genetics Name: _____________________
Essential Idea(s): Genes may be linked or unlinked and are inherited accordingly.
IB Assessment Statements and Class Objectives
10.2.A1: Completion and analysis of Punnett squares for dihybrid traits.
PROBLEM: In flowering plants, flower color is coded for by one gene and plant height is coded for by another gene on a different chromosome. White flowers (F) are dominant over red (f) and short plants (E) are dominant over tall (e). Two double heterozygote plants were crossed.
10.2.S1: Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
PROBLEM: In flowering plants, flower color is coded for by one gene and plant height is coded for by another gene on a different chromosome. White flowers (F) are dominant over red (f) and short plants (E) are dominant over tall (e). A true breeding short white plant is crossed with a true breeding tall red plant. These plants represent the parental generation.
10.2.S3: Use of chi-squared test on data from dihybrid crosses.
PROBLEM: In flowering plants, flower color is coded for by one gene and plant height is coded for by another gene on a different chromosome. White flowers (F) are dominant over red (f) and short plants (E) are dominant over tall (e). Two double heterozygote plants were crossed (as in 10.2.A1) and the following resulting phenotypes were observed:
White, short: 206
Red, short: 83
White, tall: 65
Red, tall: 30
3.4.S2: Comparison of predicted and actual outcomes of genetic crosses using real data.
PROBLEM: In cats, fur color is determined by the codominant, sex linked alleles black (XB) and orange (XO). A calico female (mix of black and orange) is bred (many times) with a black male. They produce the following offspring:
Black female: 78
Calico female: 65
Black male: 81
Orange male: 45
10.2.U5: Chi-squared tests are used to determine whether the difference between an observed and expected frequency distribution is statistically significant.
PROBLEM: Color blindness is a sex-linked trait in Bombats. A female who is a carrier of the color blind allele mates with a male who is color blind. The phenotypes of the offspring are:
Normal female: 132
Color blind female: 124
Normal male: 126
Color blind male: 136
10.1.NOS: Making careful observations- careful observations and record keeping turned up anomalous data that Mendel’s law of independent assortment could not account for. Thomas Hunt Morgan developed the notion of linked genes to account for the anomalies.
10.2.NOS: Looking for patterns, trends and discrepancies- Mendel used observations of the natural world to find and explain patterns and trends. Since then, scientists have looked for discrepancies and asked questions based on further observations to show exceptions to the rules. For example, Morgan discovered non-Mendelian ratios in his experiments with Drosophila.
AND
10.2.A2: Morgan's discovery of non-Mendelian ratios in Drosophila.
10.2.U2: Gene loci are said to be linked if on the same chromosome (Oxford Biology Course Companion page 448).
10.2.S2: Identification of recombinants in crosses involving two linked genes.
PROBLEM: The genes for pollen shape and flower color are located on the same chromosome as each other, thus are inherited together. Purple (P) and long pollen (L) are dominant to red (p) and round pollen (l). What are the possible genotypes and phenotypes of the F1 offspring if the parents are PPLL × ppll? If two F1 plants are crossed, calculate the genotypes and phenotypes in the F2 generation (without crossing over). Calculate the relative genotypes and phenotypes in the F2 generation with crossing over in the F1.
10.2.U4: The phenotypes of polygenic characteristics tend to show continuous variation.
10.2.A3: Polygenic traits such as human height may be influenced by environmental factors.
Linked vs Unlinked Genes
In this unit, we will examine the simultaneous inheritance of two (or more) genes. The genes we examine can be “unlinked” or “linked.”
Unlinked Genes | Linked Genes |
Genes found on: Inherited __________________________________ because the chromosomes assort independently during meiosis. Notation: Gametes made:
25% 25% 25% 25% Phenotypic ratio of a cross AaBb X AaBb: | Genes found on: Inherited __________________________________ unless __________________ occurs during meiosis. . Notation: Gametes made:
More More Fewer Fewer
Phenotypic ratio of a cross AaBb X AaBb: |
Unlinked Genes follow Mendel’s Law of Independent Assortment:
The amazing did these experiments too (yes, he used peas). We’ll examine one of his experiments closely.
Let’s look at two genes: one for seed and one for seed .
Each of the genes has (versions of the gene):
Round (R) | Wrinkled (r) |
Yellow (Y) | Green (y) |
The genotype for pure breeding round yellow is written as RRYY while pure breeding wrinkled green is written as rryy.
Round Yellow (RRYY) | Wrinkled Green (rryy) |
Only ________________________________________________________________________ (because of the ____________________________________________), so the gametes for the pure breeding round yellow must be and for the pure breeding wrinkled green .
Round Yellow (RRYY)
| Wrinkled Green (rryy) |
When the pure breeding smooth yellow RRYY are crossed with the pure breeding wrinkled green rryy, all the _______________________________________________ and have the genotype ________________.
Round Yellow (RRYY) X Wrinkled Green (rryy) X |
Then the _______________ plants were crossed (_______________ x _______________ )
There are four possible gametes created from a RrYy plant
The potential combinations of gametes of a RrYy x RrYy cross can be shown in a 16 square Punnett square:
RY | Ry | rY | ry | |
RY | ||||
Ry | ||||
rY | ||||
ry |
The results in the F2 generation show a ______________________ ratio of phenotypes
New combinations of traits different from those shown in the parent line are called ___________________.
In the RrYy x RrYy the parents were round and yellow. So, the recombinant offspring would be the ones that are:
Chi Square Goodness of Fit (X2)
Pronounced “kai square”
The null hypothesis predicts that there will be no real significant difference between our observed results and the expected results. In other words, any observed variation between what we expect and what we actually observe is due to chance alone.
Chi Square Calculation Table
Categories | Observed Values | Expected Values | (O-E)2 / E |
Total |
(O – E)2
E
Degrees of Freedom =
Critical X2 value =
“The X2 obtained was __________. The critical value X2 (found in the distribution table) was ________. The calculated value is ____________ (lower or higher) than the 5% level of significance, so we can _____________ (accept or reject) the null hypothesis. This means that …………………”
Variation
Discrete Variation Define: Cause: Examples: | Continuous Variation Define: Cause: Examples: |
Polygenic Inheritance
Definition:
Example:
Linked Genes DO NOT follow Mendel’s Law of Independent Assortment
William Bateson and Reginald Punnett conducted a cross in sweet pea involving two different traits; flower color and pollen shape.
F2 Phenotype | Observed | Expected Ratio | Expected Number | (O-E)2 / E |
Purple long | 284 | |||
Purple round | 21 | |||
Red long | 21 | |||
Red round | 55 | |||
TOTAL |
Conclusion: Crosses produced deviations from the predicted Mendelian Independent Assortment ratios. They suggested that the transmission of the two traits from the parents was somehow coupled.
Thomas Hunt Morgan’s Discovery of Linked Genes
Thomas Hunt Morgan designed a series of experiments using fruit flies (Drosophila melanogaster) that:
What Scientists Knew at the Time: | What Thomas Hunt Morgan Figured Out: |
BIG IDEA…
Thomas Hunt Morgan’s Crosses
CROSS | CONCLUSION | |
1 | ||
2 | ||
3 | ||
4 |
Linkage is the tendency for a group of genes, on the same chromosome, to be inherited together
Example:
Gene for _______________ is on one chromosome Dominant allele: ______= ___________ Recessive allele: ______= ___________ | Gene for _______________ is on the same chromosome Dominant allele: ______= ___________ Recessive allele: ______= ___________ |
When the genes are linked, the possible games from a heterozygous individual are _______ and _______. Therefore, if you cross two parents, both heterozygous for two genes AND THE GENES ARE LINKED, the Punnett Square looks like:
What if crossing over occurs between LINKED GENES during meiosis?
Example Problem:
Parent Generation: | ||
Possible gametes: | ||
Impact of crossing over: |
F1 Generation: | ||
Possible gametes without crossing over: | MANY | MANY |
Possible gametes with crossing over: | A FEW | A FEW |
Cross F1 X F1
Identify recombinants in the F2 generation