Bio 2970 Lab 1: Mendelian Genetics in Zea mays Sarah Chavez [email protected] Rebstock 131 314-935-4782 Office Hours: Thursdays 11-12.

Bio 2970 Lab 1:Mendelian Genetics in Zea mays

Sarah Chavez

[email protected]

Rebstock 131

314-935-4782

Office Hours: Thursdays 11-12

Results:

Traits do not blend.

F1 plants display the phenotype of one parent.

Both traits are passed to F2 in a 3:1 ratio

Phenotype vs Genotype

Purple

Purple3

Purple

1 White

Ratio 3:1

(homozygous)

(homozygous)

(heterozygous)

(heterozygous)

PP

Pp

Pp

pp

Ratio 1:2:1

1

1

2

Monohybrid Crosses: Conclusions• Although the F1 plants display the phenotype of one parent, they must inherit

genes from both parents because they transmit both phenotypes to the next generation. So, each plant must have two genetic factors encoding one character.

• The two alleles separate into the gametes, with one allele per gamete.

• Those traits that appeared unchanged in the F1 heterozygous offspring as dominant, and those that disappear in F1 heterozygotes as recessive.

• The two alleles of an individual plant separate with equal probability into the gametes. Allele for purple flowers

Homologouspair ofchromosomes

Locus for flower-color gene

Allele for white flowers

Genotypic Ratios in Simple Crosses

Dihybrid crosses

Law of Independent AssortmentResult of dihybrid cross (F1 cross between two heterozygotes with two different traits)

Dihybrid Crosses

Principle of Independent AssortmentEach unlinked gene pair assorts independently into the offspring. Thus, among offspring that have received a particular allele at one locus, there will be a random assortment of alleles at a second unlinked locus. (This is only valid for genes that are not on the same chromosome and for genes on the same chromosome but so far apart that crossing over makes them appear unlinked).

Trihybrid Crosses

How to keep them straight…

3/4 Full = 27/64 YRF phenotype / 3/4 Round / \ / 1/4 constricted = 9/64 YRf phenotype / 3/4 Yellow / \ / \ 3/4 Full = 9/64 YrF phenotype / \ / / 1/4 Wrinkled / \ / 1/4 constricted = 3/64 Yrf phenotype /YyRrFf x YyRrFf \ \ 3/4 Full = 9/64 yRF phenotype \ / \ 3/4 Round \ / \ \ / 1/4 constricted = 3/64 yRf phenotype \ / 1/4 green \ \ 3/4 Full = 3/64 yrF phenotype \ / 1/4 wrinkled \ 1/4 constricted = 1/64 yrf phenotype

Forked line diagram vs

Mathematical calculations

So,

PP and yy = multiply

PPyy or Ppyy = add

Epistasis

• In epistasis, a gene at one locus alters the phenotypic expression of a gene at a second locus

• Look for a ratio that is a variation on the 9:3:3:1 ratio….

BbCc BbCc

Sperm

EggsBC bC Bc bc

BC

bC

Bc

bc

BBCC

1/41/4

1/41/4

1/4

1/4

1/4

1/4

BbCC BBCc BbCc

BbCC bbCC BbCc bbCc

BBCc BbCc

BbCc bbCc

BBcc Bbcc

Bbcc bbcc

9 : 3 : 4

9:3:4 =

Recessive Epistasis

• e.g., coat color in mice

• The gene B encodes for the pigment, but gene C encodes for transporting that pigment into hair follicles.

9 B_C_ = black (pigment + hair)3 B_cc = white (pigment, not in hair)3 bbC_ = brown (No pigment)1 bbcc = white (No pigment, not in hair)

15:1 = Duplicate Gene

• e.g., wheat kernel color

• For this type of pathway a functional enzyme A or B can produce a product from a common precursor. The product gives color to the wheat kernel. Therefore, only one dominant allele at either of the two loci is required to generate the product.

• Thus, if a pure line wheat plant with a colored kernel (genotype = AABB) is crossed to plant with white kernels (genotype = aabb) and the resulting F1 plants are selfed, a modification of the dihybrid 9:3:3:1 ratio will be produced. The following table provides a biochemical explanation for the 15:1 ratio.

9 A_B_ = color (A and B work)3 A_bb = color (A works)3 aaB_ = color (B works)1 aabb = colorless (neither work)

9:7 = Complementary Gene

• e.g., flower color in sweet peas

• If two genes are involved in a specific pathway and functional products from both are required for expression, then one recessive allelic pair at either allelic pair would result in the mutant phenotype.

• If a pure line pea plant with colored flowers (genotype = CCPP) is crossed to pure line, homozygous recessive plant with white flowers, the F1 plant will have colored flowers and a CcPp genotype.

9 C_P_ = color (both enzymes)3 C_pp = white (no enzyme C)3 ccP_ = white (no enzyme P)1 ccpp = white (no enzymes P or C)

12:3:1 = Dominant Epistasis• e.g., fruit color in squash

• With this interaction, color is recessive to no color at one allelic pair. This recessive allele must be expressed before the specific color allele at a second locus is expressed. At the first gene white colored squash is dominant to colored squash, and the gene symbols are W=white and w=colored. At the second gene yellow is dominant to green, and the symbols used are G=yellow, g=green. If the dihybrid is selfed, three phenotypes are produced in a 12:3:1 ratio.

• Because the presence of the dominant W allele masks the effects of either the G or g allele, this type of interaction is called dominant epistasis. 9 W_G_ = white (white allele negates green)

3 W_gg = white (white allele negates green)3 wwG_ = yellow (recessive allows yellow color)1 wwgg = green (recessive allows green color)

13:3 = Dominant Suppression Epistasis• e.g., malvidin production in Primula

• Certain genes have the ability to suppress the expression of a gene at a second locus. The production of the chemical malvidin in the plant Primula is an example. Both the synthesis of the chemical (controlled by the K gene) and the suppression of synthesis at the K gene (controlled by the D gene) are dominant traits. The F1 plant with the genotype KkDd will not produce malvidin because of the presence of the dominant D allele.

• The ratio from the above table is 13 no malvidin production to 3 malvidin production. Because the action of the dominant D allele masks the genes at the K locus, this interaction is termed dominant suppression epistasis.

9 K_D_ = none (dominant D)3 K_dd = malvidin produced (dominant K, no D)3 kkD_ = none (dominant D)1 kkdd = none (recessive k only)

Chi-Square TestWhat is the chance that the difference between “expected” and “observed” is due to chance?

Chi-Squared Example