Ageneticist crossed pure breeding black mice with pure breeding brown mice. all the 992 mice in f1 generation had black coats. when these mice were crossed, the yielded 961 black coated mice and 317 brown coated mice. how could you account for the ratio of black coated to brown coated mice in the f2 generation?

Answer/Explanation:

The pure breeding (homozygous) black mice are crossed with pure breeding (homozygous) brown mice. The F1 offspring are all black. A punnett square is shown, with B representing the black allele, and b representing the brown allele. The cross is therefore BB x bb

The offspring are 100% heterozygous black (Bb)

Crossing two F1 heterozygous black mice (Bb x Bb) to give the F2 generation. The genotypes are 1 BB: 2 Bb: 1 bb. Therefore, the genotypes are 3 black : 1 brown.

This accounts for the real ratios shown in the F2 generation (992:317 is roughly 3:1

The branch of biology which deals with genes and inheritance is called genetics. There are two types of genes present, these genes are as follows:-

These genes are responsible for the formation of hormones and protein in the human body.

In the F1 generation, all the mice were black coats which signifies the law of dominance.

In the F2 generation, The ratio is different due to crossing over the genes and the law of segregation. This law, it state that the genes are independently crossed with any genes and lead to the formation of the new characters.

For more information, refer to the link:-

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a scenario where a cell may be needed to perform a form of endocytosis is when transporting large molecules. a scenario where a cell may be needed to perform a form of exocytosis is when releasing the large molecule from the cell.