About the question:
You will find the diagram used to answer the question in the attached files.
Answer:
D) Some of the bacterial population was resistant to this antibiotic.
Explanation:
The term resistance refers to an inheritable change in a population sensitivity, reflected through the consecutive failure of the chemical effects, correctly used to reach a desired effect on the target population.
The excessive use of antibiotics on the bacteria population leads to the fixation of new mutated genes -by natural selection-. The mutated genes make these cells even more resistant to the drug.
In the exposed example we have,
- Day 1 ⇒ a big population of white bacteria and only one black cell, probably <em>carrying mutated genetic material</em>.
The population was treated with antibiotics.
- Day 2 ⇒ there are fewer white bacteria and three black cells. This means that white bacteria are dying, while the black ones are reproducing.
The mutated phenotype gets to survive under the effects of the chemical.
- Day 3 ⇒ There are no more white bacteria. Only the black population survived and keeps reproducing.
The mutated bacteria reproduce with the capability of tolerating the antibiotic dose that is usually used to destroy a population of white bacteria.
- Day 4 ⇒ Black bacteria got to survive a produce a big population. Remember that <em>bacteria follow the exponential growth model, </em>meaning that their reproductive rate is too high. This black population survived the antibiotic action and reproduced at high rates.
Plant cells contain a cell wall. Plants provide us will the food and nutrients we need to carry on daily life
Answer:
The probability that their first child will both be ab/ab = 41%.
Explanation:
<u>Available data:</u>
- Two linked genes (A) and (B) ⇒ 18 centiMorgans apart
- Man ⇒ Aa Bb
- Man´s father ⇒ AA BB
- Woman ⇒ aabb
The recombination frequency is given by the distance between genes. We have to know that 1% of recombinations = 1 map unit = 1cm.
Recombination frequency = 0.18
Man: AaBb
Gametes) AB Parental
ab Parental
Ab Recombinant
aB recombinant
18 centi morgan = 18 % of <u>recombination</u> in total
= % aB + % Ab
= 9% aB + 9% Ab
100% - 18% = 82% of <u>parental</u> in total
= % of AB + % ab
= 41% AB + 41% ab
The frequency for each gamete is:
AB 41%
ab 41%
Ab 9%
aB 9%
Cross: man x woman
Parental) AaBb x aabb
Gametes) AB, Ab, aB, ab ab, ab, ab, ab
Punnett square) AB ( 41%) Ab (9%) aB (9%) ab (41%)
ab AaBb Aabb aaBb aabb
ab AaBb Aabb aaBb aabb
ab AaBb Aabb aaBb aabb
ab AaBb Aabb aaBb aabb
F1) 41% of the progeny is expected to be AaBb
9% of the progeny is expected to be Aabb
9% of the progeny is expected to be aaBb
41% of the progeny is expected to be aabb
The six major early civilizations were the Sumerians, Egyptians, Norte Chico, Chinese, Indus Valley, and Olmec. A civilization refers to a composite culture in which huge numbers of human beings share various common elements. The six of the most essential characteristics are government, cities, social structure, religion, writing, and art.
The early civilizations would have most likely cooperated about enhancing trade, and blending cultures.
If the parental genotypes are homzygous, the F1 birds would be heterozygous while all the F2 offspring would have the wild-type phenotype.
<h3>Monohybrid crossing</h3>
Based on the assumption that the wild-type genotype is AA and the slow feathering genotype is aa.
A cross between AA and aa:
AA x aa
F1 Aa Aa Aa Aa
One F1 male (Aa) is then crossed with a wild-type female (AA):
Aa x AA
AA AA Aa Aa
Genotypic proportion: 2 AA:2 Aa
Phenotypic proportion = 100% wild-type
More on monohybrid crossing can be found here: brainly.com/question/1185199
