A pea plant that has the genotype RrGgwwdd is crossed to a plant that has the rrGgWwDd genotype. The R gene controls round versus wrinkled seed, the G gene controls yellow versus green seed, the W gene controls purple versus white flower, and the D gene controls tall versus short plants. Determine the following;

What proportion of the progeny are expected to have the genotype rrggwwdd?

Question

A pea plant that has the genotype RrGgwwdd is crossed to a plant that has the rrGgWwDd genotype. The R gene controls round versus wrinkled seed, the G gene controls yellow versus green seed, the W gene controls purple versus white flower, and the D gene controls tall versus short plants. Determine the following;

What proportion of the progeny are expected to have the genotype rrggwwdd?

What proportion of the progeny are expected to have the genotype rrggwwdd?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem together. If A P plant with genotype heterozygous gene, A homozygous dominant gene B, homozygous dominant gene C and Homozygous recessive gene D is crossed with another P plant with genotype homozygous dominant gene A homozygous recessive gene B, heterozygous gene C and heterozygous gene D. What is the probability of producing offspring with genotype homozygous dominant gene A heterozygous gene B, homozygous dominant gene C and Homozygous recessive gene D or genotype heterozygous for all four genes. The answer options are a one out of four, B, one out of eight, C, three out of 4 ND, 5 out of six. So let's discuss how to tackle this problem first for this problem. Let's say that the offspring that is heterozygous for all four genes we will refer to as offspring number two and the offspring that is homozygous dominant gene A ho heterozygous gene B. Homozygous dominant gene C and Homozygous recessive gene D. We will refer to as offspring number one to solve this. We need to use both the probability product rule and some rule. We'll first apply the product rule to each offspring and then we will use the sum rule to determine the final answer. Now, how do we know to use both of those rules? The, or in the question is important. Think of the sum rule as the or rule. It's a small word but or tells us that we will use that sum rule. And by the end of this, you'll have a good grasp on the product rule, probability. So to begin, let's determine the possible offspring genotypes for each of the genes, I've already filled out those punt squares for each of the genes. And we will review those results together. The results are on the screen. Gene. A is a cross between a heterozygous parent and a homozygous dominant parent gene. A results in offspring, two homozygous dominant and two heterozygous gene B is a cross between a homozygous dominant parent and a homozygous recessive parent. All offspring of this cross are heterozygous gene C is a cross between a homozygous dominant parent and a heterozygous parent. Offspring are two homozygous dominant and two heterozygous gene D is a cross between homozygous recessive parent and a heterozygous parent offspring results in two heterozygous offspring and two homozygous recessive offspring. So now that we know those results, let's focus on the probability of having offspring. No 1. So we need to go through and determine the probability of each genotype for each of the genes and multiply them together. So, offspring, number one is homozygous dominant for gene A, the probability of having homozygous dominant, gene, a offspring is two offspring out of four total. So two divided by four. And we multiply that by the probability of gene B being hetero zit and that probability is that all offspring of that cross were heterozygous. So four divided by four, multiplied by the probability of gene C being homozygous dominant. And there are two homozygous dominant out of four offspring for gene C two divided by four, multiplied by the probability of gene D being homozygous recessive. And that probability is also two offspring out of four were homozygous recessive. Now, if we reduce this calculation, we will have one divided by two multiplied by one divided by one multiplied by one, divided by two, multiplied by one, divided by two. And this equals one divided by eight or one out of eight offspring is the probability of having offspring. # 1's genotype. Let's repeat that for offspring. # 2. So the probability of offspring # is a heterozygous gene. A, the probability of that occurring is to divided by four, multiplied by one, divided by one for gene B. Remember all the offspring of gene B are heterozygous. Then for gene C, two out of four offspring are heterozygous. And lastly, gene D two out of four offspring are also heterozygous. So let's simplify this as well. One divided by two, multiplied by one, divided by one multiplied by one, divided by two, multiplied by one divided by two. This also is one out of a probability that offspring, number two's genotype will occur. Now, we know the probability of having offspring with genotype, homozygous dominant hetero homozygous dominant gene, a heterozygous gene B, homozygous dominant gene C and homozygous recessive gene D. And we know the probability of having an offspring that is heterozygous for all four genes. Now we use the sum rule. So the overall probability of having offspring, number one or number two equals one, divided by eight plus one, divided by eight, that equals two, divided by eight equals one, divided by four or one out of four. Offspring will have genotype homozygous dominant gene A heterozygous gene B, homozygous dominant gene C heter homozygous recessive gene D or genotype heterozygous for all four genes. So the correct answer is option. A, all right, everyone. Thanks for hanging in with me. This one's a bit of a longer one. I hope you found this helpful and I'll see you soon for the next practice problem.

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Key Concepts

Genotype and Phenotype

Punnett Square

Independent Assortment