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In snakes, being rude ( R ) is dominant to being respectful (r), and being sneaky (S) is dominant to being sincere (s). A female snake that is homozygous recessive for both the r and s trait is mated with a male snake that is homozygous dominant for both traits (R and S). Both the R and S traits assort independently from one another. What is the probability that they will have an offspring that is respectful and sincere?

asked
User Meteore
by
8.5k points

2 Answers

3 votes

Answer:

0

Step-by-step explanation:

trust me its correct

answered
User Umutyerebakmaz
by
8.6k points
3 votes

Answer:

Zero (0).

Step-by-step explanation:

As per the question, the male snake will have genotype RRSS while the female snake will have genotype rrss. The male snake will produce 4 identical gametes which will have "RS" genetic combination while female snake will produce 4 identical gametes with "rs" genetic combination.

The cross depicting the genotypes of their off-springs is attached.

It is clear from the "Punnett square diagram" that all their progeny will have "RrSs genotype" which simply means that all the progeny will be heterozygous dominant and therefore phenotypically they all will be rude and sneaky. It also implies that no progeny will have rrss genotype which could make it respectful and sincere so the probability is zero.

In snakes, being rude ( R ) is dominant to being respectful (r), and being sneaky-example-1
answered
User TheTXI
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7.7k points
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