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Exercise 11: Human development

Fertilization

When the oocyte is released from the ovary during ovulation, it is scooped up by the infundibulum of the fallopian tube. Here, peristaltic contractions of the smooth muscle layer of the tube and cilia of the cells lining the tube move the oocyte along. If sperm enters the woman’s body, it will make its way to the oocyte thanks to the chemotactic signals the corona radiata cells release. The sperm usually meet the oocyte in the ampulla of the fallopian tube. Sperm must push through the corona radiata cells and then penetrate the zona pellucida. Chemicals in the acrosome of the sperm help the sperm penetrate this layer. As soon as a single sperm makes it through the zona pellucida, the zona pellucida hardens and no more sperm can enter. In the event that multiple sperm penetrate the zona pellucida, the sperm and oocyte die as this would result in multiple sets of chromosome and therefore is incompatible with life.

Penetration of zona pellucida by a single sperm signals for the oocyte to complete meiosis II. The oocyte divides, produces second polar body which will ultimately die, and the nucleus of the ovum and the nucleus of the sperm fuse, producing the first diploid cell of the new organism, zygote. Zygote will divide after about 24-30 hours by process known as cleavage – a type of cell division which results in increase in number of cells, but the structure doesn’t grow in size. When the structure reaches 16-cell stage, it is called morula, a solid ball of cells. It is now getting closer to the uterus and continues dividing. Zona pellucida begins to get weaker, and when the morula reaches uterus, it disintegrates, fluid fills the morula and pushes the cells apart, forming a hollow ball of cells blastocyst. The outside of the blastocyst is a single layer of cells known as trophoblast. Inside of the blastocyst is a cluster of cells known as embryoblast. This cluster of cells will differentiate into embryo, yolk sac, and amnion.

Blastocyst begins to burrow into the endometrium around day 7 after fertilization. Structure known as chorion starts developing within the endometrium, this will eventually give rise to the placenta. Yolk sac is forming, embryoblast develops a bilaminar embryonic disc, and amnion is also developing. Third layer differentiates between the two original layers of the bilaminar embryonic disc. There is now trilaminar embryonic disc made of three primary germ layers – ectoderm, endoderm, and mesoderm.

Fill the table: consult fig. 29.11

Layer	Gives rise to:

Practice

See Fig. 29.4, 29.5, and be able to identify the following structures:

· Sperm

· Secondary oocyte

· Ovum

· Zygote

· Morula

· Blastocyst

· Trophoblast

· Embryoblast

· Blastocyst cavity

· Cytotrophoblat

· Syncytotrophoblast

· Yolk sac

· Hypoblast

· Epiblast

· Bilaminar embryonic disc

· Uterine gland

· Amnion

· Amniotic cavity

Practice

See Fig. 29.7 and be able to identify the following structures:

Name_____________________________________ Class___________

· Yolk sac

· Connecting stalk

· Amnion

· Amniotic cavity

· Chorion

· Functional layer of endometrium

· Placenta

· Embryo

· Maternal artery

· Maternal vein

· Maternal blood

· Trophoblast cells

· Branch of umbilical vein

· Branch of umbilical artery

· Chorionic villus

· Umbilical chord

Exercise 12: Heredity

Basics of genetics

Human body consists of two main types of cells somatic cells (regular body cells, diploid), and gametes (sperm and egg cells, haploid). Each diploid cell has two sets of chromosomes, one set from mom and one set from dad. The corresponding chromosomes are called homologous chromosomes; these chromosomes have same genes, but not necessarily the same versions of these genes. The different forms of genes are called alleles, and each diploid cell has two alleles for each gene. If an individual has two identical alleles they are said to be homozygous for that particular trait. If they have two different alleles, they are said to be heterozygous.

Strict-dominance inheritance and Punnett squares

We can use information from the parents genotype (genetic make up) to predict the genotype and phenotype (the presentation of the trait) of the offspring. In strict dominance inheritance, some alleles are dominant to others, called recessive. We use letter abbreviations to represent genotype. Since every individual has two alleles, a genotype is represented by a two letter combination. Dominant allele is represented by the capital letter (usually the first letter of the trait), and recessive allele is represented by the same letter but lowercase.

Fill in the table:

Trait: Tongue rolling

Tongue rolling (R) is dominant to no tongue rolling (r).

Genotype	Genotype abbreviation	Phenotype
Homozygous _______________
Homozygous _______________
Heterozygous

Practice

Strict dominance pattern: Construct Punnett squares and predict possible genotypes and phenotypes for the following scenarios:

A. Widow’s peak (W) is dominant over straight hairline (w).

Widow’s peak (homozygous): _______

Widow’s peak (heterozygous): ______

Straight hairline_______

Dad with widow’s peak is homozygous and mom has straight hairline

Dad_______ Mom_______

Genotype % chance:

% homozygous dominant _____

% homozygous recessive _____

% heterozygous _____

Phenotype % chance

% with widow’s peak _____

% with straight hairline _____

B. Dimples (D) are dominant over no dimples (d).

Dimples (homozygous): _______

Dimples (heterozygous): ______

No dimples_______

Both parents have dimples and are heterozygous for this trait.

Dad_______ Mom_______

Genotype % chance:

% homozygous dominant _____

% homozygous recessive _____

% heterozygous _____

Phenotype % chance

% with dimples _____

% with no dimples_____

Incomplete dominance

Some traits exhibit an incomplete dominance pattern of inheritance. This means that neither of the alleles is dominant, and heterozygous phenotype will be a blend of the two homozygous phenotypes.

Practice

Incomplete dominance: Construct Punnett squares and predict possible genotypes and phenotypes for the following scenarios:

A. Straight hair allele (S) and curly hair allele (s)will produce wavy heterozygous phenotype.

Straight hair : _______

Curly hair: ______

Wavy hair_______

Dad with curly hair and mom with straight hair

Dad_______ Mom_______

Genotype % chance:

% homozygous dominant _____

% homozygous recessive _____

% heterozygous _____

Phenotype % chance

% straight hair _____

% curly hair _____

% wavy hair _____

B. Red flowering (RR) and white flowering plant (rr) produces pink flowering plants.

Red flower : _______

White flower: ______

Pink flower: _______

Red flower crossed with white flower:

Genotype % chance:

% homozygous dominant _____

% homozygous recessive _____

% heterozygous _____

Phenotype % chance

% red flower _____

% white flower _____

% pink flower _____

2nd generation: Cross two pink flowers and see possibilities for 2nd generation

Red flower : _______

White flower: ______

Pink flower: _______

Pink flower crossed with pink flower:

Genotype % chance:

% homozygous dominant _____

% homozygous recessive _____

% heterozygous _____

Phenotype % chance

% red flower _____

% white flower _____

% pink flower _____

Codominance

Other traits exhibit codominance pattern of inheritance. This means that both of the alleles are dominant, and heterozygous phenotype will consist of both alleles being expressed.

Practice

Codominance: Construct Punnett squares and predict possible genotypes and phenotypes for the following scenarios:

A. Blood types.

As we already discussed in the chapter on blood typing, there are 4 main types of blood based on the presence or absence of A and B antigens. If one has an allele for antigen A, they will express this antigen on their erythrocyte no matter what the other allele is.

Type A genotype: ________ (homozygous) or ________ (heterozygous)

Type B genotype: ________ (homozygous) or ________ (heterozygous)

Type AB genotype: ________ (this is ________zygous)

Type O genotype: ________ (this is ________zygous)

Homozygous A crossed with homozygous B

Genotype % chance:

% A _____ % AB _____

% B _____ % O _____

Heterozygus A crossed with heterozygous B

Genotype % chance:

% A _____ % AB _____

% B _____ % O _____

X-linked inheritance

Traits that are inherited through the X chromosomes are part of the sex linked inheritance, and the pattern is different for males than females. Results of Punnett square are interpreted for male and for female. Generally, one normal X chromosome is enough to overcome the abnormal X chromosome. In order for the abnormal X trait to actually present in the phenotype, two abnormal X chromosomes are required. It is very similar to strict dominance inheritance pattern, but it is specific to the X chromosome.

Practice:

Assume that XN is normal X chromosome and Xn is affected X chromosome.

Complete the genotypes:

Affected female: __________ Affected male: __________

Unaffected female: __________ Unaffected male: __________

Carrier female: __________

Complete the Punnett squares for each scenario:

2. Carrier female, affected male:

3. Affected female unaffected male:

4. Unaffected female, affected male:

5. Carrier female, unaffected male:

6. Unaffected female ,unaffected male:

7. Affected female, affected male:

Analysis:

For each of the 6 Punnett squares answer the following questions:

1. a. What is the probability of a female offspring being affected? __________