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$I^Ai$ | | B antibodies. | +| B | $I^BI^B$
$I^Bi$ | | A antibodies. | +| AB | $I^AI^B$ | | Neither A or B antibodies. | +| O | $ii$ | | Both A and B antibodies. | +Blood carries antibodies against the protein markers that are not present on the red blood cells. Antibodies are responsible for clumping or agglutination of blood cells when incompatible blood groups are mixed. Agglutination can clog vessels, causing sever damage. Agglutination is good for an immune response. + +Rhesus factor (Rh) determines if blood is plus or minus. It's another glycoprotein found on the red blood cell. + +Two alleles determine your Rh blood type. +Dominant D represents the allele for $RH^+$. +Recessive d represents the allele for $RH^-$. + +| Blood phenotypes | Genotype(s) | Protein markers | Antibodies produced by immune system | +| ---------------- | ----------- | --------------- | ------------------------------------ | +| $RH^+$ | DD,Dd | | No Rh antibodies. | +| $RH^-$ | dd | | Rh antibodies. | +Rh antibodies are only produced when you are exposed to Rh antigens. +Rh negative mothers can produce Rh+ antibodies if baby is Rh+. + +Blood is tested using antiserums (antibodies in a bottle). The blood is checked for agglutination. \ No newline at end of file diff --git a/Chi Squared.md b/Chi Squared.md new file mode 100644 index 0000000..1da5d8e --- /dev/null +++ b/Chi Squared.md @@ -0,0 +1,79 @@ +[[G11 Pre-AP Biology]] +Punnet squares determines if there is a significant relationship between observed data and expected data. + +Example begin: +let yellow be rep recessive allel y +let green rep allele for Y + +p1: Yy x Yy + +gametes: Y, y x Y, y + +Example end. + +Chi square tutorial. +Start off with two hypothesis. + +Null hypothesis ($H_0$): +"There is not significant difference between observed and expected values of 75 yellow, 25 green peas. The differences are due to random chance alone." + +Alternative hypothesis ($H_A$): +"There is a significant difference between the observed and expected values of 85, 25 green peas" + +Find the degrees of freedom and critical value. +Degree of freedom: Number of categories/classes subtract 1. +In the example, the 2 categories would have a degree of freedom of 1. + +Find the critical value. Find p=0.05 and degree of freedom. +If the Chi Square is less than the critical value, accept the null hypothesis. +Else accept alternative hypothesis. + +Chi square formula:$$\Sigma\frac{(o-e)^2}{e}$$ +![[Pasted image 20261026103929.png]]For the example, the formula results in 1.92, which is less than 3.84, thus it's failed to reject null hypothesis. + +An example conclusion for the example is "Fail to reject null hypothesis. The $x^2$ is smaller than the critical value of 3.841. There is not significant difference between the observed and expected values of yellow to green peas." + +Example begin: +Let C represent purple. +Let c represent corn. +Let W represent wrinkled. +Let w represent smooth. + +P1: CcWw x ccww +Gametes: +C, c, W, w x c,c,w,w + +| | CW | Cw | cW | cw | +| --- | ---- | ---- | ---- | ---- | +| cw | CcWw | Ccww | ccWw | ccww | +Genotypes: +25% Ccww +25% ccWw +25% ccww +25% CcWw +Phenotypes: +25% purple smooth corn. +25% yellow wrinkled corn. +25% yellow smooth corn. +25% purple wrinkled corn. + +Hypothesis: +null +not null + +Degree of freedom is 3. +Critical value is 7.82. + +Expected 200 purple smooth, 200 yellow wrinkled, 200 yellow smooth 200 purple wrinkled. +Results are 200 purple wrinkled 200 purple smooth 190 yellow wrinkled 189 yellow smooth. + +| Phenotype | Observed | Expected | $\frac{(o-e)^2}{e}$ | +| --------------- | -------- | -------- | ------------------- | +| purple wrinkled | 200 | 195 | 0.1282 | +| purple smooth | 200 | 195 | 0.1282 | +| yellow wrinkled | 190 | 195 | 0.1282 | +| yellow smooth | 189 | 195 | 0.1846 | +| | 779 | | 0.5692 | + +0.5692 is less than the critical value of 7.82, thus null hypothesis.12 + diff --git a/Pasted image 20261026103929.png b/Pasted image 20261026103929.png new file mode 100644 index 0000000..7cf6d18 Binary files /dev/null and b/Pasted image 20261026103929.png differ diff --git a/Pasted image 20261102104200.png b/Pasted image 20261102104200.png new file mode 100644 index 0000000..37e7255 Binary files /dev/null and b/Pasted image 20261102104200.png differ diff --git a/Pedigree Analysis.md b/Pedigree Analysis.md new file mode 100644 index 0000000..2aa31f2 --- /dev/null +++ b/Pedigree Analysis.md @@ -0,0 +1,5 @@ +[[G11 Pre-AP Biology]] + +Pedigree is a diagram of family relationships. + +![[Pasted image 20261102104200.png]] \ No newline at end of file diff --git a/biology/G11 Pre-AP Biology/8 Levels of Biodiversity.md b/biology/G11 Pre-AP Biology/8 Levels of Biodiversity.md index 9248c2f..82f02af 100644 --- a/biology/G11 Pre-AP Biology/8 Levels of Biodiversity.md +++ b/biology/G11 Pre-AP Biology/8 Levels of Biodiversity.md @@ -1,4 +1,4 @@ -[[G11 Pre-AP Biology]] +6[[G11 Pre-AP Biology]] Types of diversity: - Genetic diversity. diff --git a/biology/G11 Pre-AP Biology/Gregor Mendel and Monohybrid Cross.md b/biology/G11 Pre-AP Biology/Gregor Mendel and Monohybrid Cross.md index 9f321bb..737e13c 100644 --- a/biology/G11 Pre-AP Biology/Gregor Mendel and Monohybrid Cross.md +++ b/biology/G11 Pre-AP Biology/Gregor Mendel and Monohybrid Cross.md @@ -58,7 +58,7 @@ $P_2$: Tt x Tt Example concluded. Autosomal traits are any traits found on chromosome 1-22. -Carriers are people having a recessive allele. +Carriers are people having a recessive allele and a dominant allele. Sex linked traits are any traits found on sex chromosomes. X+Y are not homologous. X-linked means it's found on X chromosome. diff --git a/biology/G11 Pre-AP Biology/Other patterns of inheritance.md b/biology/G11 Pre-AP Biology/Other patterns of inheritance.md index ddd6d7e..b6d1653 100644 --- a/biology/G11 Pre-AP Biology/Other patterns of inheritance.md +++ b/biology/G11 Pre-AP Biology/Other patterns of inheritance.md @@ -5,4 +5,27 @@ Incomplete dominance is when neither allele is dominant, therefore a heterozygou Example being two flours homozygous red flowers and homozygous white flowers producing pink flowers. If you then breed these pink flowers together, there's 25% chance of red flowers, 25% chance for white flowers and 50% chance of pink flowers. -Co dominance is when both alleles are dominant. Therefore a heterozygous will have both alleles fully expressed. \ No newline at end of file +Co dominance is when both alleles are dominant. Therefore a heterozygous will have both alleles fully expressed. + +For blood types, A and B are codominant, O is recessive. + +Example dihybrid cross begin: + +Let $I^a$ rep A blood allele. } codominant +Let $I^a$ rep B blood allele. } ^ +Let i rep O blood allele. } recessive +$P_1$: $I^AI^A * ii$ +Gamete: $I^A I^A * i i$ + +$F_1$: + +| | $I^A$ | $I^A$ | +| --- | ------ | ------ | +| i | $I^A$i | $I^A$i | +| i | $I^A$i | $I^A$i | +Phenotypes: +100% $I^Ai$ +Genotype: +100% A blood. + +Example end. \ No newline at end of file