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LuckyWell [14K]
2 years ago
10

Explain how these tail lengths were inherited by describing the mode of inheritance, indicating how many gene pairs were at work

, and designating the genotypes of Harvey, Irma, and their 18- cm -tail offspring. Drag the terms on the left to the appropriate blanks on the right to complete the sentences.
Biology
1 answer:
emmainna [20.7K]2 years ago
8 0

Here we can see that 7 phenotypes appeared in F2 generation, so the number of genes involved is 3 and alleles is 6. Contribution per allele = 4 cm Base contribution = 6 cm

Genotype of Harvey = aa bb cc Genotype of Erma = AA BB CC

Number of dominant alleles if length is

18 cm = 18-6 / 4 = 12 / 4 = 3.

F1 generation genotype = Aa Bb Cc This clarifies why the length is 18 cm, as there are three dominant alleles.

seven 1. It is likely that there are gene pairs in this cross. There are

categories of phenotypes. The genotypes of the parents would be combinations of alleles that would produce a 6 cm ( aabbcc ) tail and a 30 cm) tail, whereas the 18 cm offspring would have a genotype of AaBbcc.

learn more about phenotypes here: brainly.com/question/902712

#SPJ4

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According to the question, the snapping turtle possesses 52 chromosomes in its cell. This means that the chromosomal number of its body cell is 52. However, the turtle, which reproduces sexually will produce gametes or sex cells via meiosis. Meiosis is the kind of cell division that results in daughter cells (gametes) that have a reduced number of chromosomes (by half).

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N.B: Diploid (2n) means two complete sets of chromosomes while haploid (1n) means one complete set of chromosome.

The body cell of the turtle contains two set of chromosomes received from each parent, hence, making it Diploid (2n). However, meiosis will reduce the chromosome number to one set in each gamete i.e. haploid (n).

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