Choose vacuoles are largest in ___________ cells. the incorrect statement concerning centrioles.
1 answer:
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As this family has an Autosomal Recessive Disorder, we must know that this disorder will only be expressed, in the family members who are homozygous recessive (aa), in this way, to know the probability of IV-4 and IV-5 to present the disorder, we must evaluate the probabilities of combinations of their parents' alleles in a diagram called a punnett square.
In the pedigree above, we can see that the dark colored balls and squares represent the members of the family who have the disorder and therefore have the "aa" genotype.
Thus, when observing the crossings throughout the family, we know that the father of IV-4, has the genotype "Aa", however we do not know which genotype of the mother of IV-4, but as she does not have the disorder, she may have the genotype "Aa", or "AA", for that reason, we must make the punnett square for both situations and see what the probabilities of IV-4 will present the disorder.
As you can see in the figure attached below, if IV-4's mother has the "Aa" genotype, IV-4 will have a 25% chance of presenting the disorder. If IV-4's mother has the "AA" genotype, IV-4 has a 0% chance of inheriting the disorder.
In the case of IV-5, we can see that his mother has the disorder and therefore has the genotype "aa", we know that his father does not have the disorder, but we do not know if the father's genotype is "AA" or "Aa "and therefore we must evaluate both situations.
As you can see in the figure below called "IV-5", if IV-5's father has the "AA" genotype, IV-5 will have a 0% chance of inheriting the disorder. However, if IV-5's father has the "aa" genotype, IV-5 will have a 50% chance of inheriting the disorder.
Answer:
0.035
Explanation:
<u>cv+ is the wild-type dominant allele over cv, therefore:</u>
- cv+cv+ and cv+cv cause wild-type phenotype for crossveinless
- cv cv causes the crossveinless phenotype
<u>Sb is a dominant mutant allele over wild-type Sb+, therefore:</u>
- Sb Sb and Sb Sb+ cause Stubble phenotype
- Sb+ Sb+ causes wild type phenotype for Stubble
It's the cross between the heterozygous female with a homozygous recessive male. Remember that cv and Sb+ are the recessive alleles.
X
-The male produces only 1 type of gamete: cv Sb+
-The female produces 4 types of gametes:
- cv Sb+ ] Parental
- cv+ Sb ] Parental
- cv Sb ] Recombinant
- cv+ Sb+ ] Recombinant
The genes are linked and separated by 7 map units. A distance of 7 mu means that 7% of the resulting gametes will be recombinant. Because there are 2 possible recombinant gametes, each of them will appear in 3.5% of the cases.
The genotypes and proportions of the offspring resulting from the test cross can be seen in the Punnett Square. The phenotypically wild-type individuals will have the genotype cv+ Sb+ / cv Sb+ (heterozygous for crossveinless and homozygous recessive for Stubble) and a 0.035 proportion.
