In chickens, a condition referred to as 'creeper' exists whereby ...

Question

In chickens, a condition referred to as 'creeper' exists whereby the bird has very short legs and wings and appears to be creeping when it walks. If creepers are bred to normal chickens, one-half of the offspring are normal and one-half are creepers. Creepers never breed true. If bred together, they yield two-thirds creepers and one-third normal. Propose an explanation for the inheritance of this condition.

Accepted Answer

everyone. Let's take a look at this question together in guinea pigs for color is an autism. A recessive Mandali in trait a dominant gene Bigby and guinea pigs is responsible for making the for black, whereas a recessive allele little B makes the for white two guinea pigs were made to produce the following number of offspring. Guinea pigs with black for nine and guinea pigs with white for three. What is the possible genotype of the parents of this offspring? So looking at what we have produced, we have nine guinea pigs that have black fur and we have three guinea pigs with white. For we know that the black for is the one that is dominant and the white fur is the one that is recessive and looking at these phenotype. So we can see that we have a 3 to 1 ratio of black to white because we have black fur is nine and we have white Is three, which when we reduce these we have three and one because three goes into 93 times and one goes into 33 times. And so looking at this 3-1 ratio, we can think back to the different types of pundit squares that we have drawn. And what would give us a 3 to 1 ratio of that dominant black for two. That recessive white fur. And so we know that we're dealing with two parents who are heterosexual or for color because when we have both of those hetero ziggy's for color gina types of the parents we end up with this, our punnett square and we know that this yields us that 3 to ratio because here we have 123 which have that dominant allele for the black. For so we have three out of four and then we have this one right here, which is that one out of four. So we have three out of four are dominant and then we have that one out of four which is recessive, thus giving us that 3 to 1 ratio. So that means both parent Jenna types are that hetero sickness. And so answer choice B is the correct answer. Because answer choice B is the only possible genotype for the parent where when we do that cross we end up with that 3 to 1 ratio of black fur to white fur. So answer choice B is the correct answer. I hope you found this video to be helpful. Thank you and goodbye.

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