Nucleus, membrane-bound organelles, and rod-shaped chromosomes.
Answer:
sorry if it's a bit messy, hope it helps
Answer:
12
Explanation:
Parent 1 : normal petal, long stamen
Parent 2 : short petal, normal stamen
F1 generation has all short petal and normal stamen plants so these two are the dominant traits.
F2 generation will have a mixed progeny. Out of them normal petal, long stamen and short petal, normal stamen will be parental combinations since they were the phenotype of the original parents.
Distance between the two genes is 24 cM and distance between genes = recombination frequency. Hence, recombination frequency here is 0.24.
Out of 100 total plants in F2, 100*0.24 = 24 will be recombinants. Out of 24, 12 will be short petal, long stamen recombinants and other 12 will be normal petal, normal stamen recombinants.
Hence, out of 100 plants 12 are expected to have short petals and long stamens.
2) The recessive genotype is ff; and the phenotype is short fur.
3) If you have a short fur hamster, the possible genotype is ff.
Phenotype Ratio of Punnett Square 1
short fur- 0; long fur- 4 so your ratio is 0:4
Phenotype Ratio of Punnett Square 2
short fur- 1; long fur- 3 so your ratio is 1:3
Phenotype Ratio of Punnett Square 3
short fur- 4; long fur- 0 so your ratio is 4:0
hopefully this helps and if you need any more help just ask me :)
