# Motlow College - Bob Reeder - BIO1110 - Human Genetics

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```					BIOL 1110                                                                                REEDER
HUMAN GENETICS

I.    Expressing Outcome in Genetic Crosses
A. Probability in Genetics: ratio of the number of ways a particular event might occur to
the total number of events possible. The probability (P) of an event occurring (X)
equals the number of events (f) divided by the total number of events that can occur
(n). P (x) = f/n Example: In humans, albinism (condition where hair and skin lack
pigment) is produced by a recessive gene.

Key: A = Normal              Parents: Aa x Aa
a = albino

What is the probability of having an albino child?

Gametes A a A a              P (albino child) = f/n=1/4=25%

Offspring possibilities: AA Aa Aa aa

Example: What is probability of having two albino children in a row?

P (two albino children) = f/n x f/n = 1/4 x 1/4 = 1/6

B.    Ratio: genotypic phenotypic
1. Genotypic Ratio = 1:2:1
2. Phenotypic Ratio = 3:1

II.    Incomplete Dominance: the pairs of alleles studied by Mendel all exhibited a dominant
recessive relationship. Mendel chose his pairs because they behaved as distinct
alternatives. Mendel never became aware of the fact that the inheritance of many traits
in a wide variety of organisms involve pairs of alleles where one member of the pair
may not be completely dominant to the other. Instead, many allelic pairs exhibit
incomplete dominance or codominance in which the heterozygote has a different
phenotype (as well as genotype) from that of either homozygote. In incomplete
dominance each of the two different alleles if present in the heterozygote have an equal
effect on the trait resulting in an intermediate variation between those of homozygotes.

A. Examples:
1) Human blood type AB (IAIB = AB): individual produces both types of protein antigens
(A and B = codominant)

2) Shorthorn cattle and horses:
a) if red bull crossed with a white cow (or vice versa)
b) offspring is neither red nor white but roan = white hairs intermingled with red.
c) HR = Red HW = white
P HRHR x HWHW
G HR HW
F1 HRHW = Roan phenotype

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3) In snapdragons: red flowered crossed with white flowered results in pink

4) In humans: straight hair crossed with curly results in wavy

5) Sickle cell anemia in humans:
Hbs = abnormal hemoglobin, HbA = normal
a. Genotype of Hbs Hbs= sickle cell anemia (bone and kidney damage, ulceration of
lower legs, breaks down or clogs blood vessels, poor resistance to infection;
infants die, few live beyond age 40.)

b. Genotype of HbA Hbs = sickle cell trait (few outward symptoms unless oxygen
levels decrease those individuals with the abnormal allele are protected against
malaria, especially during ages 2 and 4).

B.   Overdominant: a heterozygote that is more vigorous than either of the parent
homozygotes; hybrid vigor. Example: typical corn plant may have a higher grain
yield than either of the parent strains from which it was derived; true of mixed-breed
dogs where mongrels are healthier than pure bred dogs.

III.   Genetic Determination of Sex: An exception to the general rule that all homologous pairs
of chromosomes are identical in size and shape is the sex chromosomes. The cells of
the females of most species contain two identical sex chromosomes (X) while in males,
the set of two chromosomes contains a single X and a smaller y. The remaining 22 pair
of homologous chromosomes in cells of males and females are autosomes.

In humans and other mammals, maleness is determined in large part by the presence of the Y
chromosome. The normal male produces two kinds of sperm: half contain an X, and half
contain a Y (therefore it is the male parent's gamete that determines the sex of the offspring).
All eggs contain only a single X.

IV. Sex-limited Trait:
A. Defined: affects a part or function of the body that is present in one sex but not the
other
1. This type of inheritance is important in animal breeding as in milk yield and horn
development (only males have horns) in cattle and sheep where the specific trait
affects only one sex, but the genes controlling them can be transmitted by either
parent.
2. Human examples:
a. Beard or facial hair where the son has a heavy beard and the father does not:
mother does not have a beard because her hormones do not encourage the
growth of facial hair; however, she may still pass on to her son genes controlling
beard growth as well as autosomal genes passed from the father; in this situation
the father's alleles do not foster heavy beard growth
b. Breast size: a young woman with large breasts whose mother is flat-chested

V.     Sex-Influenced Inheritance:
A. Defined: an allele is dominant in one sex but recessive in the other
1. Hormonal differences between the sexes can be the cause of the differences in the
trait expressions

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a. A gene for hair growth pattern has two alleles, one that produces hair all over the
head and another that causes pattern baldness
b. The baldness allele is dominant in males but recessive in females
c. A heterozygous male is bald, but a heterozygous female is not

A. Defined: refers certain traits (genes) carried on the X chromosome, whereas most
traits are autosomal. Remember that the sexes have different sex chromosome
pairs (XX,XY), and genes on the X chromosome are inherited differently in males
than they are in females. Any gene on the X chromosome of a male is exposed in
his phenotype, because there is no second allele for that gene (XRY, XrY =
hemizygous). An allele on an X chromosomes in a female may or may not be
expressed (may be dominant or recessive = XRXR, XRXr, XrXr) depending on the
nature of the allele for that gene contributed by the second X chromosomes. Sex
linked variations from the normal are recessive alleles.
1. Trends noted in sex-linked transmission:
a. Expect sex-linked traits to occur more often in males than in females; males will
show the trait with a hemizygous recessive genotype; females must be
homozygous recessive which is very rare.
b. Only females are carriers of the traits with a heterozygous genotype (XRXr);
carriers only carry the recessive gene and normally do not express the recessive
trait; rarely, a female who is heterozygous for a sex-linked gene expresses the
associated condition: for example, in a female carrier of hemophilia, she
expresses the trait in only a mild from and is called a manifesting heterozygote.
c. Sex-linked examples: hemophilia A (blood clots slowly or not at all; red-green
color blindness; cleft palate (does not close normally during fetal development);
form of diabetes insipidus (inability of kidneys to respond to ADH); Duchenne
muscular dystrophy (childhood muscle degeneration-characterized by a waddling
gait, toe-walking; frequent falls where difficulty in rising may appear as soon as
child starts to walk; muscle weakness intensifies until wheelchair confined with
death during the teen years); Lesch-Nyhan syndrome (deficiency of the enzyme
HGPRT causes mental retardation, spastic cerebral palsy, urinary stones, and
self-mutilative behavior); testicular feminization (male embryo does not respond
to male hormones, so female phenotype develops, although the genotype is
male).

3. Example: Homozygous, normal XRXR and color blind husband XrY will produce
sons with normal vision and carrier daughters.

Key = R = Normal
r = colorblind

P XR XR x Xr Y              XrY = would be colorblind

XR XR XR Y      XR Xr XR Y                      XR Xr = Normal vision, but
Normal Normal Normal Normal                     carrier
carrier

B.   Holandric Genes

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1. There are some linked traits carried by genes on the Y chromosome.
a. Examples include testis-determining factor, spermatogenesis control, some for
heighth and slower maturation, and "hairy ear" trait seen in men of some families
in Caylon, India, Israel and aboriginal Australia.

VII. X Inactivation - Equaling Out the Sexes
A. DISCUSSION
1. Because females are XX, they carry two alleles for every gene, whereas males carry
only one with a XY genotype resulting in a potential inequality in the number of sex-
2. Balancing of this inequality occurs by a mechanism called X inactivation, which
operates in all mammals.
a. Early in the development of females, one of the X chromosomes in each somatic
cell is randomly inactivated resulting in a female mammal expressing the X
chromosome genes she inherited from her father in some cells and those from
her mother's X in others.
b. Once an X chromosome is inactivated in a cell, all of the cells that form
mitotically from it will have the same X chromosome turned off.
c. Inactivation occurs early in development, and the adult female will have patches
of tissue that are phenotypically different with respect to the expression of sex-
d. Result of inactivation is that the female is now genetically equivalent to the male
since each cell in her body has only one active X chromosome.
3. The female's heterozygosity (XRXr) can still offer her protection from sex-linked
disorders in spite of X inactivation.
a. For example: if she inherits one allele that specifies a vital enzyme, and another
allele that specifies an inactive version, she will still probably be healthy because
some of her cells manufacture the enzyme; a male who has the defective allele
would not survive.
4. X inactivation may be observed in female cells because the turned-off X absorbs a
specific stain much faster than does the active X.
a. This turned-off X, called a Barr body, can be seen in the nucleus of a female cell
in interphase as a dark stained body.
b. A normal male cell has no Barr body because his one X remains active.
5. A good example of X inactivation is the coat color pattern of the calico cat.
a. Shapes and positions of this trait's characteristic black and orange patches are
b. Result is that cells in which the orange allele is inactivated develop into orange
patches.
6. X inactivation has a significant medical application in detecting carriers for Lesch-
Nyhan syndrome: sex-linked disorder characterized by a child that bites his or her
fingers and lip to the point of mutilation, cerebral palsy, mental retardation, and
passes painful urinary stones.
a. The mutant allele causes deficiency in the enzyme HGPRT which can be
detected in a carrier mother by testing hairs taken from widely separated part of
her head and testing them for the pressure of HGPRT
1) If some hairs contain the enzyme, but others do not, them the woman is a
carrier.

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