Answer: 1:2:1
Explanation:
The genotypic ratio is 1:2:1 whereas, the phenotypic ratio is 3:1 in the monohybrid cross. 2. Dihybrid cross: It is simple defined as the cross between two pairs of contrasting characters or parental alleles. Pea shape and color are used to conduct the dihybrid cross test.
Use these 2 links to help you with more problems like this
you can just click on it if you can't copy and paste it
https://www.chegg.com/learn/biology/introduction-to-biology/cross-between-two-pea-plants
https://www.albert.io/blog/dihybrid-cross-ap-biology-crash-course/#:~:text=A%20cross%20between%20two%20plants%20that%20are%20heterozygous,alleles%29%20parents%20this%20will%20always%20be%20the%20ratio.
Answer:
please mark as brainliest answer as it will also give you 3 points
Explanation:
Cyclin-dependent kinases (CDKs) are the families of protein kinases first discovered for their role in regulating the cell cycle. They are also involved in regulating transcription, mRNA processing, and the differentiation of nerve cells.[1] They are present in all known eukaryotes, and their regulatory function in the cell cycle has been evolutionarily conserved. In fact, yeast cells can proliferate normally when their CDK gene has been replaced with the homologous human gene.[1][2] CDKs are relatively small proteins, with molecular weights ranging from 34 to 40 kDa, and contain little more than the kinase domain.[1] By definition, a CDK binds a regulatory protein called a cyclin. Without cyclin, CDK has little kinase activity; only the cyclin-CDK complex is an active kinase but its activity can be typically further modulated by phosphorylation and other binding proteins, like p27. CDKs phosphorylate their substrates on serines and threonines, so they are serine-threonine kinases.[1] The consensus sequence for the phosphorylation site in the amino acid sequence of a CDK substrate is [S/T*]PX[K/R], where S/T* is the phosphorylated serine or threonine, P is proline, X is any amino acid, K is lysine, and R is arginine.[1]
A divergent boundary plates moves away from each other.
HOPE THIS HELPS! ^_^
<span>The cytoplasm of the cell.</span>
Correct question:
if a nondisjunction occurs at anaphase I of the first meiotic division, what will the proportion of abnormal gametes (for the chromosomes involved in the nondisjunction)?
Answer:
100%
Explanation:
Nondisjunction at meiosis-I means that two homologous chromosomes of at least one homologous pair fail to separate from each other during anaphase-I. This would result in the formation of one cell with one extra chromosome and the other with one less chromosome by the end of meiosis-I. Meiosis-II in these two cells would maintain this chromosome number in the daughter cells. Therefore, out of the total four gametes formed by the end of the meiosis, two would have one extra chromosome and would be denoted as "n+1". The rest of the two gametes would have one less chromosome and would be denoted as "n-1".
