number of monohybrid formed in F2 generation of dihybrid cross
Answers
Explanation:
In the Mendel dihybrid cross ,in the F2 generation 16 progeny are formed . In this 9 are yellow round 3 yellow wrinkled ,3 green round and 1 green wrinkled. So, in the 16 progeny, 15 progeny show both dominant phenotypes or one phenotype.
Answer:
Total no. of monohybrids formed in F2 dihybrid cross are 4.
Explanation:
Lets take an example of 2 types of true- breeding pea plants that were different in regarding to their 2 characters.
1st plant had round & yellow seeds- RRYY
2nd plant had wrinkled & green seeds - rryy
- F1 generation- cross the
true breeding plants to each other RRYY X rryy
↓ ↓
- Gametes RY ry
↓
- Collection of many seeds and RrYy
Phenotype is recorded ↓
- F1 seeds are planted and grown and then RrYy X RrYy
further are allowed to self- fertilized or Self- crossed producing the seeds of F2 generation
Punnett square of F2 generation is attached below-
- F1 generation- All round & yellow seeds
- F2 generation- 315-round & yellow seeds
108- round & green seeds
101- wrinkled & yellow seeds
32- wrinkled & green seeds
- If we divide each of these numbers by 32 (the no. of pants with wrinkled & green seeds)
- The phenotypic ratio of F2 generation will be - 9.8: 3.2: 3.4: 1.0
- If we count the experimental error then approximated ratio will be- 9:3:3:1 (In total 16 combinations are possible by randomly combining male and female gametes).
F2 of all offspring of all offspring combined probabilities
↓ ↓ ↓
- 3/4 are yellow →3/4 are round → (3/4) (3/4)= 9/16 yellow, round →1/4 are wrinkled → (3/4)(1/4)=3/16 yellow, wrinkled
- 1/4 are green → 3/4 are round → (1/4)(3/4)=3/16 green, round
→1/4 are wrinkled → (1/4)(1/4)=1/16 green, wrinkled
In F2 generation of monohybrid cross the phenotypic ratio is always 3:1 for different types of individuals. Therefore, out of 16 progeny, total no. of monohybrids formed in F2 dihybrid cross are 4.
