In a cross of Lymnaea, the snail contributing the eggs w...

Question

In a cross of Lymnaea, the snail contributing the eggs was dextral but of unknown genotype. Both the genotype and the phenotype of the other snail are unknown. All F₁ offspring exhibited dextral coiling. Ten of the F₁ snails were allowed to undergo self-fertilization. One-half produced only dextrally coiled offspring, whereas the other half produced only sinistrally coiled offspring. What were the genotypes of the original parents?

Accepted Answer

Welcome back everyone. Let's look at our next question. What does it imply if a test cross yields 50% hetero sickness and 50% homesickness recessive offspring. So let's recall what you're doing in a test cross. In a test cross, you're trying to determine the genotype of an parent with an unknown genotype but showing the dominant phenotype. So you want to know are they Hamas? I guess dominant or hetero sickness. And this is of course done by doing a test cross, you cross the unknown parent with a homogeneous recessive parent. The recessive parent can only pass on recessive genes. And so you can look at the test cross to determine reason backwards to the genotype of the unknown parent. So let's think about what kind of test cross would yield 50% hetero sickness and 50% homesickness recessive offspring. To put up our little punnett square. Well imagine a gene that has alleles Bigby and little B. So our test parent, the homos, I guess recessive is going to be little B. Little B. We know we have at least one big B. Because our parents showing the dominant phenotype. But we don't know what this second alil is. Well we know that our offspring are 50% hetero sickness. So that's going to be this here Big B. Little B. Big B. Little B. And 50% home is recessive. But in order for these two to be little B. Little B. They must inherit a second recessive allele from the other parent. So our unknown parent Must be a heterocyclic, big b. Little B. Because they have to be able to pass on a recessive, valuable for any of the offspring to have a recessive phenotype. If this parent were homesick is dominant, all of the offspring would be 100% hetero. So let's look at our answer choices. We know the unknown parent must be had. So that's going to be choice A. The unknown parent is a hetero. So we'll select that. Let's look through other answer choices. Choice B. Is the unknown parent is a dominant homicide. Well, that would result in 100% hetero zegas offspring. So that's not what we have here. So that's not an answer choice. Joyce C. Says the unknown parent has a dominant recessive genotype. Well dominant recessive genotype doesn't even make sense. So we're going to cross that out and then choice D The unknown parent is a wild type. Well, this is incorrect. Wild type refers to the un mutated form. So if you're comparing maybe a novel mutation, you want to study how this mutation is inherited versus the wild type genotype, but we don't know which is the wild type here. Is it dominant and recessive? We're just looking at a dominant versus recessive gene. So again, what does it imply if a test cross yields 50% hetero sickos and 50% homesickness. Recessive offspring choice a. The unknown parent is a heterocyclic. See in the next video

Key Concepts

Genotype and Phenotype

Inheritance Patterns

Self-Fertilization and Punnett Squares

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