The factors of 2^5 are 1 and 2
Answer:
20 chromosomes
Explanation:
Mitosis is a kind of cell division that results in daughter cells with same number of chromosomes as the parent cell. It involves stages including Prophase, Metaphase, Anaphase and Telophase. In prophase, the Chromatin condenses into Chromosomes.
In Anaphase stage, the 10 chromosomes as mentioned in the question divides into opposite poles of the cell. One individual chromosome contains two sister chromatids, which actually separates in the Anaphase stage. Hence, at the end of the Anaphase stage, there will be 10 chromatids each at opposite poles of the cell. Each chromatid at this stage is considered a chromosome.
Hence, a cell with 10 chromosomes will contain 20 chromosomes (10+10 chromatids on each pole) in the Anaphase stage just before the cell divides into two in a process called CYTOKINESIS.
Answer:
A titin mutation that occurs in muscular dystrophy with myositis (mdm) mice results in a predicted 83 amino acid deletion in the N2A and PEVK regions of the titin protein. Muscles from mdm mice are actively more compliant possibly owing to the deletion in titin's I-band region. This suggests that modulation of titin stiffness in active sarcomeres by the proposed titin–thin filament interaction may be affected by the mdm mutation. The answer is YES I believe.
Explanation:
I believe the answer is yes from my deep reaserch. You may want to research in your texts book/lesson or courses and review what your teacher/professer has given you.
False, because nucleotides are what make up a persons DNA so each person has its own unique order.
Getting the probability for each event can easily be done using Punnett Square (say, Hh x Hh for the couple since they are both carriers of the disease). Doing so will result to: 0.25 probability of an offspring not having the disease and is not a carrier of the gene, 0.50 probability of an offspring not having the disease and is a carrier of the gene, and a 0.25 probability of an offspring having the disease. The probability of these events apply to each offspring, and will not depend on how many children they want to have. This means each offspring has a 0.25-0.50-0.25 chance of not being a carrier, being a carrier, and having the disease, respectively.
