Apologia Biology Module 8 Study Guide
Define autosomes.
Chromosomes that do not determine the sex of an individual
If a gamete has two alleles for the same genetic trait, what type of genetic disorder will result in a zygote formed with this gamete?
If a gamete has two alleles for the same trait, it must have two of the same chromosome. In the fertilization process, then, there will be three chromosomes. Thus, a genetic disorder from a change in the chromosome number will result.
Define sex-linked inheritence.
Inheritance of a genetic trait located on the sex chromosomes
Define genotype.
Two-letter set that represents the alleles an organism possesses for a certain trait.
Define heterozygous genotype.
A genotype with two different alleles
List Mendel's principle's of genetics in updated terminology.
1)The traits of an organism are determined by its genes. 2)Each organism has two alleles that make up the genotype for a given trait. 3)In sexual reproduction, each parent contributes ONLY ONE of its alleles to its offspring. 4)In each genotype, there is a dominant allele. If it exists in an organism, the phenotype is determined by that allele.
Define dominant allele.
An allele that will determine phenotype if just one is present in the phenotype.
Define allele.
One of a pair of genes that occupies the same position on homologous chromosomes.
A pea plant which is homozygous in the dominant, axial flower allele ("A")is crosse with a pea plant that is heterozygous in that allele. What are the possible genotypes and phenotypes, along with their percentage chances, for the offspring?
One parent is homozygous dominant, so its genotype is "AA." The other is heterozygous, so its genotype is "Aa." The Punnet square looks like: A A A AA AA a Aa Aa Thus, 50% of the offspring have the "AA" genotype and 50% have an "Aa" genotype. Since each offspring has at least one of the dominant allele,however, 100% have the axial flower phenotype.
Give the possible phenotypes and the percentage chance for each in the dihybrid cross between a pea plant that is homozygous in producing smooth, yellow peas and a pea plant that produces wrinkled, green peas. The smooth and yellow alleles are dominant.
Since the parent with smooth, yellow peas is homozygous, its genotype is "SSYY." SInce the other expresses both recessive alleles, it must be homozygous in the recessive alleles. Thus, its genotype is "ssyy." Both of these parents can only produce one type of gamete each. The one parent can only produce a SY allele and the other can only produce a sy. This gives us a 1x1 Punnett square. SY sy SsYy Since there is only one possible genotype, 100% of the offspring have the "SsYy" genotype and the smooth, yellow phenotype.
Give the possible phenotypes and the percentage chance for each in the dihybrid cross between a pea plant that is heterozygous in producing smooth, yellow peas and another with the same genotype.
Since the parents are both heterozygous in each allele, their genotypes are "SsYy." There are 4 possible gametes: SY, Sy, sY, sy. The resulting Punnett square, then, is: SY Sy sY sy SY SSYY SSYy SsYY SsYy Sy SSyy SSyy SsYy Ssyy sY SsYY SsYy ssYY ssYy sy SsYy Ssyy ssYy ssyy Smooth, yellow peas (genotypes SSYY, SsYy, SSYy, SsYY) 9 out of 16, which equals 9/16 or 56.25% Smooth, green peas (genotypes SSyy, Ssyy) 3 out of 16, which equals 3/16 or 18.75% Wrinkled, yellow peas (genotypes ssYY, ssYy) 3 out 16, which equals 3/16 or 18.75% Wrinkled, green peas (genotype ssyy) 1 out of 16, which equals 1/16 or 6.25%
If a person has type B- blood, what are the possible genotypes for that person? Include the possible genotypes related both to the type of blood as well as the Rh-factor.
Since the person is type B, the genotype must be either BB or BO. For the Rh-factor, the person person expresses the recessive allele. Thus, her genotype must be homozygous in the recessive allele. Thus, her genotype must be homozygous in the recessive allele, which we called "pp" in "On Your Own" problem 8.10.
A woman is heterozygous in the ability to roll her tongue when extended. If she marries a man who cannot roll his tongue, what percentage of their children will be able to roll their tongues? Remember, the allele for being able to roll your tongue is dominant.
Since the woman is heterozygous, her genotype is "Rr." That man cannot roll his tongue. Since the inability to roll your tongue is recessive, his genotype must be "RR." The resulting Punnet square is: R r r Rr rr r Rr rr Since having even one dominant allele allows you to be able to roll your tongue, 50% of the children will be able to roll their tongues.
A woman with type O blood marries a man with type AB blood. What blood types are possible for their children? What is the percentage chance for each blood type?
Since the woman is type O, her genotype must be OO, as that would be the only way the recessive allele could be expressed. The man is Type AB, so his genotype is "AB." The Punnet square,then, is given below: A B O AO BO O AO BO Since the O allele is recessive, the possible blood types for the children are: A (50%) and B (50%).
Define phenotype.
The observable expression of an organism's genes