Answer:
<u>4 types </u>of phenotypes with <u>phenotypic ratio 9:3:3:1 </u>will be obtained.
Explanation:
The Punnett square diagram for the cross between mice with genotype BbCc & BbCc is attached.
It is clear from the diagram attached that the <u>phenotypic ratio</u> will be 9:3:3:1
- The following<u> 9 genotypes</u> will show dominant trait for both the characters i.e. characters represented by B & C genes.
BBCC, BBCc, BbCC, BbCc, BBCc, BbCc, BbCC, BbCc, BbCc
- The following <u>3 genotypes</u> will show dominant trait for one character represented by gene B and recessive trait for other character represented by gene C.
BBcc, Bbcc, Bbcc
- The following <u>3 genotypes</u> will show recessive trait for one character represented by gene B and dominant trait for other character represented by gene C.
bbCC, bbCc, bbCc
- The following genotype is the <u>only genotype </u>which will show recessive trait for both the characters represented by gene B & C.
bbcc
Answer:
The phylum Hemichordata is the one that could form an evolutionary connection between the chordates and non-chordates.
Explanation:
To the phylum Hemichordata belongs a genus that scientists believe could explain how the chordates could evolve from the non-chordates. This genus is Balanoglossus, of the class Enteropneusta.
Balanoglossus is similar to a worm, whose habitat is the seabed, and like other hemicordates it has a stomach, a structure that forms part of its digestive system and fulfills the functions of a spine.
Learn more:
Chordates and non-chordates brainly.com/question/1387264
If they were to be no separation <span>of a pair of chromosomes during meiosis resulting in formation of a sperm or ovum with 24 chromosomes instead of 23.</span>
ATP, exocytosis is active
<span>Active and passive transport, process where waste is released from the cell is both
and the rest are passive
</span>