The object is balanced against standard units of mass
DNA is the basic unit of inheritance
A gene is the basic physical and functional unit of heredity. Genes are made up of DNA. Some genes act as instructions to make molecules called proteins.
Answer:
I = 62.5%
Explanation:
<u>Available data</u>:
- the wild-type DND1 gene confers resistance to attain viruses. Strains are susceptible to viruses
- The EDS1 gene confers resistance to powdery mildew, a type of fungal infection
- The genes are linked, eight map units apart on chromosome
- A third gene, IR 1. imparts resistance to certain lepidopteran leafminers and is 10 map units from EDS1 and 18 map units from DNDI
- The observed double crossover rate is 0.3%
According to this information, we can picture the position of each gene in the chromosome
---DND1----------------------EDS1-----------------------------------IR1---
║---------- 8MU---------║------------------10MU-------------║
║------------------------------18MU------------------------------║
The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.
1% of recombinations = 1 map unit = 1cM. The maximum recombination frequency is always 50%.
So, if we know the distances between genes, we know their recombination percentages.
- DND1-EDS1 = 8 MU = 8% recombination
- EDSI-IR1 = 10 MU = 10% recombination
To calculate interference, first, we need to know the coefficient of coincidence, CC.
CC= observed double recombinant frequency/expected double recombinant frequency
<em>Note: </em>
- observed double recombinant frequency=total number of observed double recombinant individuals/total number of individuals
- expected double recombinant frequency: recombination frequency in region I x recombination frequency in region II.
CC= (0.003)/(0.08 x 0.1)
CC=0.003/0.008
CC=0.375
CC = 37.5%
The coefficient of interference, I, is complementary with CC.
I = 1 - CC
I = 1 - 0.375
I = 0.625
I = 62.5%
Answer:
d. Meiosis II in the father
Explanation:
Since the son has two extra Y chromosomes, the nondisjunction would have occurred in father because females do not have Y chromosome. Before Meiosis DNA replication occurs during which the homologous chromosomes are replicated. During Meiosis I crossing over occurs and homologous chromosomes separate from each other. During Meiosis II sister chromatids of a chromosome separate and move into individual daughter cells.
Here, during Meiosis II the sister chromatids of Y chromosome did not separate from each other (nondisjunction). They collectively moved to one daughter cell. As a result one daughter cell had two Y chromosomes and other daughter cell had none. When the daughter cell containing two Y chromosomes combined with the daughter cell containing one X chromosome from mother, XYY individual was born.
About.1.9 million. Thats just an estimate.