Addition principle sum multiplication principle
1. addition principle
There are n ways to do one thing and finish it. The first one has m 1, the second one has m2, …, and the nth one has mn, so there is a method of N=m 1+m2+…+mn.
2. Multiplication principle
It takes n consecutive steps to do one thing. There are m 1 different ways to do the first step, m2 different ways to do the second step, ... and mn different ways to complete the n step, so there are * * * ways to complete it.
N=m 1×m2×…mn methods.
Steps to solve common genetic problems
Mathematical explanation of genetic law in (1) Mendel pea experiment
1. Research on a pair of related characters
p:
Phenotype
High stem
Dwarf stem
genotype
Direct injury
×
Deadline Date (abbreviation for deadline date)
↓
F 1:
Phenotype
High stem
genotype
Deadline Date (abbreviation for deadline date)
F 1 selfing
Phenotype
High stem
High stem
genotype
Deadline Date (abbreviation for deadline date)
×
Deadline Date (abbreviation for deadline date)
↓
F2:
Phenotype
3/4 tall stem+1/4 short stem
genotype
1/4DD + 2/4Dd + 1/4dd
2. Research on two pairs of related characters.
Phenotype
P
F 1 selfing
Second generation
Consider separately
colour
Yellow × Green →
Yellow →
Yellow+1/4 green
Two phenotypes 3: 1
Consider separately
form
Round grain × wrinkled grain →
Round particles →
3/4 round grain+1/4 wrinkled grain
Two phenotypes 3: 1
Consider at the same time
Two pairs of features
Yellow circle × green wrinkle →
Yellow circle →
(3/4 yellow+1/4 green) ×(3/4 round grain+1/4 wrinkle grain)
=9/ 16 yellow circle +3/ 16 yellow wrinkle +3/ 16 green circle+116 green wrinkle
2×2=4 phenotypes
(3: 1)2=9:3:3: 1
genotype
P
F 1 selfing
Second generation
Consider separately
colour
YY×yy→
Yy→
1/4+2/4+ 1/4
Three genotypes 1:2: 1.
Consider separately
form
RR×rr→
Rr→
1/4RR+2/4Rr+ 1/4rr
Three genotypes 1:2: 1.
Consider at the same time
Two pairs of features
YYRR×yyrr→
YyRr→
( 1/4YY+2/4YY+ 1/4YY)×( 1/4RR+2/4RR+ 1/4RR)= 1
1/ 16 yyrr+ 1/ 16 yyrr+ 1/ 16 yyrr+ 1/ 16 yyrr
+2/ 16 yyrr+2/ 16 yyrr+2/ 16 yyrr+2/ 16 yyrr
+4/ 16YyRr
3×3=9 genotypes
( 1:2: 1)2
= 1: 1:2:2:4:2:2: 1: 1
3. Cross test of two pairs of related traits-F1yellow circle and recessive homozygous green wrinkle;
yyrr×yyrr →( 1/2Yy+ 1/2Yy)×( 1/2Rr+ 1/4YyRr+ 1/4YyRr+ 1/4YyRr+ 1/4YyRr+ 1/4YyRr+ 1/4YyRr
(2) Examples of solving problems
Example: 1 YyRr individuals hybridize with YyRr individuals and inherit according to the law of free combination. How many genotypes are there in the offspring?
Analysis: For the study of more than two pairs of relative traits, we suggest that students first regard them as two independent events, and then comprehensively summarize them through binomial theorem.
Steps to solve the problem:
YY× YY→ (1/2yy+1/2yy) ① Two genotypes RR× RR→ (1/4rr+2/4rr+1/4rr) ② Three genotypes.
So * * * has 2×3=6 genotypes.
The six genotypes were ①× ② =1/8yyrr+1/8yyrr+2/8yyrr+1/8yyrr respectively.
Example 2 A couple with earlobe and double eyelids gave birth to a son without earlobe and double eyelids for the first time, and the probability of having another son with earlobe and double eyelids is _ _ _ _ _; The probability of regenerating a son with earlobe and double eyelids is _ _ _ _ _; The probability of having another child with an earlobe and a single eyelid is _ _ _ _ _ _.
Analysis: It is known that the double eyelid with earlobe is an autosomal dominant gene, which is represented by A and B respectively. The genes controlling the single eyelid without earlobe are A and B, and the couple genotypes are AaBb and AaBb. Because we are familiar with this combination, the ratio of each trait in the offspring accords with 9:3:3: 1, so the probability of each combination can be directly deduced when considering two traits.
Steps to solve the problem: AABB× AABB→ (9/16a _ b+3/16a _ bb+3/16aab _+16aab).
1) Double eyelids with earlobes 9/16a _ b _; The probability of sons is1/29/16a _ b _×1/2 sons =9/32.
2) The son is known, regardless of his probability. Double eyelid with earlobe 9/ 16.
3) Children with earlobe and single eyelid 3/ 16 A_bb