Answer:
3. An even number of chromosomes are required for synapsis during prophase I and proper pairing during metaphase
Explanation:
Mules are hybrids of a cross between a female horse and a male donkey. Horses contain 64 chromosomes while donkeys contain 63 chromosomes in their somatic cells respectively. This means that they each produce 32 and 31 chromosomes respectively during meiosis. A mule, hence, contains 32+31= 63 chromosomes in their somatic cells.
This chromosome number in mules are uneven for meiosis to occur because meiotic division requires that an even number of homologous chromosomes pair together in a process called SYNAPSIS during prophase I of meiosis I. This is impossible in a mule because of the uneven number of chromosomes in its cells.
Also, during metaphase of meiosis, the homologous chromosomes need to be properly aligned at the equator for separation to occur. This is also impossible in a mule considering the number of chromosomes that don't add up.
Due to this reason of unevenness in the number of chromosomes present in a mule, meiosis will not occur and if meiosis (gamete formation) does not occur, reproduction cannot take place. Therefore, the mule is a sterile species i.e. cannot produce offsprings via sexual reproduction.
The outer periphery of the inter-vertebral disk is composed of strong fibrous tissue called the Annulus fibrosus. This is a tough circular exterior of the inter-vertebral disc that surrounds the soft inner core, the nucleus pulposus. The outer portion is composed of a ring of ligament fibers that encases the inner core of the disc and securely connects the spinal vertebrae above and below the disc.
Answer:
1) The alleles for homozygous brown hair will be BB.
2) The alleles for heterozygous brown hair will be Bb
3) Let's make a punnet square to check for the outcomes:
B b
B BB Bb
B BB Bb
4) The results from the punnet square depict that the phenotype of all the children produced will be brown hair.
5) The punnet square depicts that there will be a 50% probability of the child to carry the heterozygous genome (Bb) and there is also 50% probability for the child to carry homozygous dominant genome (BB).
The answer is b.
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People with central body obesity are more likely to develop iron-deficiency anemia than people who do not have excess fat in the mid section of their bodies.
Hope it helped you.
-Charlie
