Chapter 11: Simple Inheritance and Meiosis

Flashcards covering simple Mendelian inheritance, monohybrid and dihybrid crosses, meiosis, recombination, independent assortment, and single-gene genetic disorders such as cystic fibrosis and Huntington disease.

What specific mutation in the CFTR gene causes the form of cystic fibrosis (CF) discussed in the study guide, and what is its effect on the resulting protein?

The deletion of an entire codon in the CFTR gene, which causes the omission of an amino acid in the CFTR protein, making it nonfunctional.

How are chromosomes, genes, and alleles defined and ranked from largest to smallest in size?

From largest to smallest: chromosome (a single, long DNA molecule wrapped around proteins), gene (a section of DNA containing instructions for assembling a polypeptide chain), and allele (an alternate version of a gene).

How many total chromosomes and homologous pairs are found in a nonreproductive body cell of a human male karyotype?

A human nonreproductive body cell contains $46$ total chromosomes, organized into $23$ homologous pairs.

In the provided meiosis diagram, which genetic materials are separated in the stages represented by the blue box (5) and the green box (6)?

The blue box represents Meiosis I, where homologous chromosomes are separated. The green box represents Meiosis II, where sister chromatids are separated.

How many times does a cell divide during meiosis, and how does the chromosome count of the daughter cells compare to the initial diploid cell?

The cell divides $2$ times, yielding $4$ haploid daughter cells that each contain half the chromosome number of the original diploid starting cell.

What is independent assortment, and how does it generate genetic variation during meiosis?

It is the random orientation and distribution of maternal and paternal homologous chromosomes during meiosis I, resulting in varied combinations of chromosomes in the resulting gametes.

How does the process of recombination in Meiosis I, as illustrated in the provided diagram, alter the genetic makeup of the chromosomes?

Homologous non-sister chromatids exchange segments of DNA during Meiosis I, creating new combinations of maternal and paternal alleles along the chromosomes.

Why is an individual with one mutated allele for a recessive disorder like cystic fibrosis referred to as a "carrier"?

Because the individual has one recessive disease allele ($a$) and one dominant normal allele ($A$); they carry and can transmit the mutation without displaying symptoms of the disease.

Based on the provided comparison table, what is the probability of two heterozygous parents ($Aa \times Aa$ or $Tt \times Tt$) having a child affected by Cystic Fibrosis versus Huntington disease?

For Cystic Fibrosis (recessive), the chance of having an affected child ($aa$) is $25\%$. For Huntington disease (dominant), the chance of having an affected child ($TT$ or $Tt$) is $75\%$.

If one parent is a cystic fibrosis carrier ($Aa$) and the other has cystic fibrosis ($aa$), what are the probabilities of having an unaffected child and a child with cystic fibrosis?

There is a $50\%$ probability of having an unaffected child (carrier, genotype $Aa$) and a $50\%$ probability of having a child with cystic fibrosis (genotype $aa$).

If one parent is homozygous dominant for Huntington disease ($TT$) and the other is homozygous recessive ($tt$), what is the probability that their child will have Huntington disease?

The probability is $100\%$, because all offspring inherit the dominant disease allele, resulting in genotype $Tt$.

Tongue rolling is a dominant trait governed by the $R$ allele, while non-rolling is governed by $r$. If a woman can roll her tongue but her father cannot, what is her genotype?

Her genotype is heterozygous ($Rr$), because she inherits the dominant $R$ allele from her mother and must inherit a recessive $r$ allele from her father ($rr$).

If a woman has Huntington disease and her husband does not ($tt$), what is the probability that their child will have Huntington disease?

Either $50\%$ or $100\%$, depending on the mother's genotype: $50\%$ if she is heterozygous ($Tt$) and $100\%$ if she is homozygous dominant ($TT$).

In the provided Punnett square for a father with genotype $ffAa$ and a mother with genotype $FfAa$, what are the types of gametes produced by each parent?

The father produces sperm cells with genotypes $fA$ and $fa$. The mother produces egg cells with genotypes $FA$, $fA$, $Fa$, and $fa$.

For parents with genotypes $ffAa$ (no freckles, CF carrier) and $FfAa$ (freckles, CF carrier), what is the probability that a child will have freckles and the probability that a child will have cystic fibrosis?

The probability of having freckles ($F\_$) is $50\%$ ($1/2$), and the probability of having cystic fibrosis ($aa$) is $25\%$ ($1/4$).

For parents with genotypes $ffAa$ and $FfAa$, what is the probability that a child will have freckles AND be a carrier of cystic fibrosis?

The probability is $25\%$ (or $1/4$), calculated by multiplying the probability of having freckles ($1/2$) by the probability of being a CF carrier ($1/2$).

How does the modifier gene TGFβ1 influence the phenotype of cystic fibrosis in individuals with genotype $aa$?

The dominant allele ($D$) of TGFβ1 results in a more severe form of cystic fibrosis, whereas the homozygous recessive genotype ($dd$) leads to a milder form of the disease.

If two parents are heterozygous for both cystic fibrosis ($Aa$) and TGFβ1 ($Dd$), what is the probability that their child will have severe cystic fibrosis?

The probability is $3/16$ (or $18.75\%$), derived from multiplying the probability of having cystic fibrosis ($aa = 1/4$) by the probability of inheriting the dominant severe modifier ($D\_ = 3/4$).

A man and a woman are both heterozygous for dimpled chin ($Dd$) and tongue-rolling ($Rr$). What is the probability that their child will NOT have a dimpled chin AND will NOT be able to roll their tongue?

The probability is $1/16$ (or $6.25\%$), calculated by multiplying the probability of no dimpled chin ($dd = 1/4$) by the probability of non-rolling ($rr = 1/4$).

How can sequencing the BRCA II gene aid an individual in making personalized healthcare decisions?

Detecting a mutation in the BRCA II gene, which is linked to cancer onset, allows patients to consider proactive medical choices, such as preventive surgery to remove tissue likely to develop cancer.