Transcript Bio 2970 Lab 1

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
3
vs
Genotype
Purple
PP
(homozygous)
Purple
Pp
(heterozygous)
1
2
1
Purple
Pp
(heterozygous)
White
pp
(homozygous)
Ratio 3:1
Ratio 1:2:1
1
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
Locus for flower-color gene
Allele for white flowers
Homologous
pair of
chromosomes
Genotypic Ratios in Simple Crosses
Dihybrid crosses
Law of Independent Assortment
Result of dihybrid cross (F1
cross between two
heterozygotes with two
different traits)
Dihybrid Crosses
Principle of Independent
Assortment
Each 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
/
3/4 Round
/ \
/
1/4 constricted = 9/64
/
3/4 Yellow
/ \
/
\
3/4 Full
= 9/64
/
\ /
/
1/4 Wrinkled
/
\
/
1/4 constricted = 3/64
/
YyRrFf x YyRrFf
\
\
3/4 Full
= 9/64
\
/
\
3/4 Round
\
/ \
\
/
1/4 constricted = 3/64
\ /
1/4 green
\
\
3/4 Full
= 3/64
\ /
1/4 wrinkled
\
1/4 constricted = 1/64
Forked line diagram
vs
Mathematical calculations
YRF phenotype
YRf phenotype
YrF phenotype
Yrf phenotype
yRF phenotype
yRf phenotype
yrF phenotype
yrf phenotype
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….
9:3:4 =
Recessive
Epistasis
BbCc
Sperm
1/
4 BC
1/

4 bC
BbCc
1/
4 Bc
1/
4 bc
Eggs
• e.g., coat color in mice
• The gene B encodes for
the pigment, but gene C
encodes for transporting
that pigment into hair
follicles.
1/
4 BC
BBCC
BbCC
BBCc
BbCc
BbCC
bbCC
BbCc
bbCc
BBCc
BbCc
BBcc
Bbcc
BbCc
9 B_C_ = black (pigment + hair)
3 B_cc = white (pigment, not in hair)
9
3 bbC_ = brown (No pigment)
1 bbcc = white (No pigment, not in hair)
bbCc
Bbcc
bbcc
1/
1/
1/
4 bC
4 Bc
4 bc
: 3
: 4
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 Test
What is the chance that the difference between “expected” and
“observed” is due to chance?
Chi-Squared Example