Posisble Flashcards
How would each of the following events affect the outcome of mitosis or meiosis.
Mitotic cohesion fails to form early in mitosis?
Improper segregation of chromosomes to daughter cells. (Cohesin necessary to hold the sister chromatids together until anaphase of mitosis, if cohesion fails to perform early in mitosis, the sister chromatids could separate prior to anaphase.)
How would each of the following events affect the outcome of mitosis or meiosis.
Shugoshin is absent during MEIOSIS?
The cohesin at the centromere may be broken, allowing for the separation of sister chromatids along the homologs during anaphase I, leading to improper segregation of chromosomes to daughter cells. ( Shugosin protects cohesin from degradation at the centromeres during MEIOSIS I. Cohesin at the arms of the homologs chromosomes is not protected by shugoshin and is broken in anaphase I, allowing the two homologs to separate.)
What is the probability of rolling one side of a six-sided die and obtaining the following numbers.
1 or 2 ?
The probability of rolling a 1 on a six-sided die is 1/6. Similarly, the probability of rolling a 2 on a six-sided die is 1/6. Because the question asks what is the probability of rolling a 1 or 2, and these are mutually exclusive events, we should use the additive rule of probability to determine the probability of rolling a 1 or 2;
(1/6+1/6= 2/6=1/3)
How would each of the following events affect the outcome of mitosis or meiosis.
Shugoshin does not break down after anaphase I of MEOISIS?
Then the cohesin at the centromere will remain protected from degradation. The intact cohesin will prevent the sister chromatids from separating during anaphase II of MEIOSIS, resulting in an improper separation of sister chromatids and daughter cells with too many or too few chromosomes.
How would each of the following events affect the outcome of mitosis or meiosis.
Separase id defective?
Homologous chromosomes and sister chromatids would not separate in MEIOSIS and MITOSIS, resulting in some cells that have too few chromosomes and some cells that have too many chromosomes.
The fruit fly DROSPHILIA MELANGASTER has four pairs of chromosomes, whereas the house fly MUSCA DOMESTICA has six pairs of chromosomes. All things being equal, in which species would you expect to see more genetic variation among the progeny of a cross? Explain your answer.
The progeny of an organism whose cells contain more homologous pairs of chromosomes should be expected to exhibit more variation. The number of different combinations of chromosomes that are possible in the gametes is 2^n, where n is equal to the number of homologous pairs of chromosomes. For the fruit fly with four pairs of chromosomes, the number of possible combination is 2^4 = 1. For MUSCA DOMESTICA with six pairs of chromosomes, the number of possible combinations is 2^6=64
A horse has 64 chromosomes and a donkey has 62 chromosomes. A cross between a female horse and a male donkey produces a mule, which is usually sterile. How many chromosomes does a mule have? Can you think of any reasons for the fact that most mules are sterile?
The haploid egg produced by the female horse contains 32 chromosomes. The haploid sperm produced by the male donkey contains 31 chromosomes. The union of the horse and donkey gametes will produce a zygote containing 63 chromosomes. From the zygote, the adult mule will develop and will contain cells with a chromosome number of 63.
A color-blind woman and a man with normal vision have three sons and six daughters. All the sons are color blind. Five of the daughter have normal vision, but one of them is color blind. The color blind daughter is 16 years old, is short for her age, and has never undergone puberty. Propose an explanation for how this girl inherited her color blindness.
The trivial explanation for these observations is that this form of color blindness is an AUTOSOMAL RECESSIVE TRAIT. In this case, the father would be a heterozygote, and we would expected equal proportions of color-blind and normal children, of either sex.
In sheep Lustrous fleece(L) results from an allele that is dominant over an allele for normal fleece. The ewe (adult female) with lustrous fleece is mated with a ram (adult male) with a normal fleece. The ewe then gives birth to a single lamb with normal fleece. From this single offspring, is it possible to determine the genotypes of the two parents? If so , what are their genotypes? If not, why not?
Yes it is possible to determine the genotype of each parent, assuming that the dominant lustrous allele(L) exhibits complete penetrance. The ram and the single lamb must be homozygous for the normal allele(l) because both have normal fleece phenotype. Because the lamb receives only a single allele (l) for the ram, the ewe must have contributed the other recessive l allele. Therefore, the ewe must be heterozygous for lustrous fleece.
Joe has classic Hemophilia, an X-linked recessive disease. Could JOe have inherited the gene for this disease from the following pesons?
His mother’s mother?
Yes, X-linked traits are passed on from mother to son. Therefore, Joe must have inherited the hemophilia trait from his mother. His mother could have inherited the trait from either her mother or her father.
Joe has classic Hemophilia, an X-linked recessive disease. Could Joe have inherited the gene for this disease from the following persons?
His mother’s father?
Yes, X-linked traits are passed on from mother to son. Therefore, Joe must have inherited the hemophilia trait from his mother. His mother could have inherited the trait from either her mother or her father.
Joe has classic Hemophilia, an X-linked recessive disease. Could Joe have inherited the gene for this disease from the following persons?
His father’s mother?
No,Because Joe could not have inherited the trait from his father (Joe inherited the Y chromosome from his father), he could not have inherited hemophilia from either.
Joe has classic Hemophilia, an X-linked recessive disease. Could Joe have inherited the gene for this disease from the following persons?
His father’s father?
No, Because Joe could not have inherited the trait from his father (Joe inherited the Y chromosome from his father), he could not have inherited hemophilia from either.
Red-Green color blindness in humans is due to an X-Linked recessive gene. A woman whose father is color blind possesses one eye with normal color vision and one eye with color blindness.
Explain for this woman’s vision pattern.
The woman is HETEROZYGOUS, with one X-chromosome bearing the allele for normal vision and one X-chromosome with the allele for color blindness. One of the two X-chromosomes is INACTIVATED at random during early embryogenesis. If one eye, derived exclusively from progenitor cells that inactivated the normal X, then that eye would be color blind,whereas the other eye may be derived from progenitor cells that inactivate the color-blind X, or is a mosaic with sufficient normal retinal cells to permit color vision.
Red-Green color blindness in humans is due to an X-Linked recessive gene. A woman whose father is color blind possesses one eye with normal color vision and one eye with color blindness.
Would it be possible for a man to have one eye with normal color vision and one eye with color blindness?
One way would be for the man to be XXY, and the answer to part (a) would apply. Another rare possibility would be a somatic mutation in the progenitor cells for one retina, but not the other.
Bob has XXY chromosome(Klinefelter syndrome) and is color blind. His mother and father have normal color vision, but his maternal grandfather is colorblind. Assume that Bob’s chromosome abnormality arose from nondisjunction in meiosis. In which parent and in which meiotic division did non-junction occur?
Explain your answer.
Because Bob must have inherited the Y chromosome from his father, and his father has normal color-vision, there is no way non-disjunction event from the paternal lineage could account for Bob’s genotype. Bob’s mother must be heterozygous X^+X^c because she has normal color vision, and she must have inherited a color-blind X-chromosome from her color-blind father. For Bob to inherited two color-blind X-chromosomes from his mother, the egg must have arisen from a NON-DISJUNCTION in MEIOSIS II. In meiosis I, the homologous X-chromosomes separate, so one cell has the X^+ and the other has X^c. Failure of sister chromatids to separate in meiosis II would then result in an egg with two copies of X^c
Pedigrees with AUTOSOMAL RECESSIVE TRAITS
Will show affected males and females arising with equal frequency from unaffected parents. The trait often appears to skip generations. Unaffected people with an affected parent will be carriers. Mating between two affected individuals will result in 100% affected offspring. If the genetic condition is rare, affected individuals will often result from consanguineous matings.
Pedigrees with AUTOSOMAL DOMINANT TRAIT
Will show affected males and females arising with equal frequency from a single affected parent. The trait does not usually skip generations. If the trait is fully penetrant,there will never be affected individuals resulting from two unaffected parents;however, mating between two affected parents can result in unaffected offspring.
X-LINKED RECESSIVE TRAITS
Affect males predominantly and will be passed from an affected male through his unaffected daughter to his grandson. X-linked recessive traits are not passed from father to son. Any affected female must have an affected father.
X-LINKED DOMINANT TRAIT
Will affect males and females and will be passed from an affected male to all his daughters, but not to his sons. An affected woman(usually heterozygous for a rare dominant trait) will pass on the trait equally to half her daughters and half her sons.
Y-LINKED TRAITS
Will show up exclusively in males, passed from father to all of his sons.
