10 -(UEFS/2018)Genes zigóticos são expressos durante o desenvolvimento e mbrionário . Em moscas Drosophila melanogaster existe um gene zigótico que letal em homozigose recessiva . Um cruzamento entre moscas heterozigotas para o gene zigótico letal gerou 120 moscas adultas . Dentre essas 120 moscas adultas, 0 número esperado de moscas heterozigotas e a) 30 b) 40 c) 60 d) 80 e) 120

<br />Let's break down the problem step by step step:<br /><br />## Step 1: Understanding the problem<br />We are given that in Drosophila melanogaster, there is a gene that is lethal in homozygous recessive form. This means that if an individual has two copies of this gene, it will not survive. We are also given that a cross between heterozygous flies for this lethal gene resulted in 120 adult flies.<br /><br />## Step 2: Determining the genotypes of the flies<br />Let's denote the lethal gene as "L" and the normal gene as "l". Since the lethal gene is recessive, it will only be expressed if an individual has two copies of it (LL). The heterozygous flies will have one copy of the lethal gene and one copy of the normal gene (Ll). <br /><br />## Step 3: Setting up the Punnett square<br />To determine the expected number of heterozygous flies, we can set up a Punnett square. Let's use "P" to represent the parent flies and "F1" to represent the offspring flies.<br /><br />```<br />P: Ll x Ll<br />F1: LL Ll Ll ll<br />```<br /><br />## Step 4: Analyzing the Punnett square<br />From the Punnett square, we can see that there are 2 LL flies, 4 Ll flies, and 2 ll flies in the F1 generation. However, we are only interested in the heterozygous flies (Ll). Therefore, the expected number of heterozygous flies is 4.<br /><br />## Step 5: Answering the question<br />The question asks for the expected number of heterozygous flies among the 120 adult flies. From our analysis in Step 4, we determined that the expected number of heterozygous flies is 4.<br /><br /> <br />The expected number of heterozygous flies among the 120 adult flies is 4.