Answer:
The correct answer is -3/16
Explanation:
The cross between BBss and bbSS In cocker spaniels will form gametes that are - Bs and bS. As it is given that black coat color ( B) is dominant over red ( b), and solid color ( S) is dominant over spotted ( s) then f1 gen would be :
Bs Bs
bS BbSs BbSs
bS BbSs BbSs
Now F2 generation;
BS Bs bS bs
BS BBSS BBSs BbSS BbSs
Bs BBSs BBss BbSs Bbss
bS BbSS BbSs bbSS bbSs
bs BbSs Bbss bbSs bbss
Here black-spotted that is possible only in two cases, BBss or Bbss, are 3/16
6 In the condition hyperthyroidism, patients have elevated levels of both T3 and 14 due to a malfunction of the
immune system that causes the thyroid gland to overproduce thyroid hormones. Do you think the negative
feedback loops would lead to high levels of TSH or low levels of TSH? Explain.
which products of photosynthesis are necessary for human life ?
a) carbon dioxide for us to breathe in, as well as sunlight and water.
b) oxygen for us to breathe in , as well as sunlight and water.
c) oxygen for us to breathe in , as well as glucose forcus to consume for energy.
d) carbon dioxide for us to breathe in, as well as glucose for us to consume for energy.
The movement of glucose across a plasma membrane is achieved by
Answer:
facilitated diffusion
Explanation:
If you tossed a coin a total of ten times and got a total of three
heads, what is the probability of tossing heads in your next
toss?
Answer:
The probability of exactly 3 heads in 10 tosses is 1201024.
The probability that result will be head in the next toss is 1/2.
What is Probability?This is a mathematical term which is used to describe how likely an event is to occur or how true a proposition is. In this scenario, we have two variables which are head and tail.
The probability probability of tossing heads in your next toss is therefore
1/2 = 50% chance.
Read more about Probability here brainly.com/question/24756209
Which process is represented by the picture below?
A. Vegetative Propagation
B. Meiosis
C. Fertilization
D. Binary Fission
HELPP MEEE
Answer:
Meiosis is the correct answer
The process of water moving from the Earth to the atmosphere and back again is called the
A. Earth cycle
B. Atmosphere cycle
C. life cycle
D. water cycle
The...........is also known as the tympanum.
1)eardrum
2)stapes
3)cochlea
4)malleus
Answer and I will give you brainiliest
Answer:
1
djdhdbbdbdbsbdbdbsb
‼️‼️‼️‼️HELP HELP HELP HELP HELP WHOEVER CAN PLEASE‼️‼️‼️‼️
Answer: i would say b!
Explanation:
hope i helped! :D
The diagram below shows a cell placed in a solution.
A cell is shown placed inside a beaker. It is labeled Cell. The solution inside the beaker is labeled 2% salt solution and the solution inside the cell is labeled 4% salt solution.
What will most likely happen to the cell?
It will expand as water moves out of it.
It will shrink as water moves out of it.
It will expand as water moves into it.
It will shrink as water moves into it.
The most likely to happen to the cell is ; ( C ) It will expand as water moves into it
The movement of water in and out of a cell is based on osmosis .The cell membrane of the cell will regulate the movement of water molecules in and out of the cell. based on the solute concentration inside the cell and outside the cell.
When a cell is placed in a hypotonic solution ( i.e. a solution with a lower solute concentration as seen in the 2% salt solution ) the water molecules from the hypotonic solution will move into the cell which will cause the cell to be Turgid/expand over time.
Hence we can conclude that the most likely thing to happen to the cell is ; It will expand as water moves into it
Learn more : https://brainly.com/question/82248
which list shows the phase of mitosis in the correct order
Answer: The answer is B. prophase, metaphase, anaphase, telophase
Explanation:
HELPPPPP
How many grams of LiCl are in 235.0 mL of 1.80 M lithium chloride solution?
1
Show Your Work
Answer:
Mass of LiCi = 17.9352 gm
Explanation:
Given:
Volume of LiCi = 235 ml
Molarity = 1.80 M
Find:
Mass of LiCi
Computation:
Atomic mass of LiCi = 6.9 + 365.5 = 42.4 gm
Mole(n) = Weight / Atomic mass
1.80 = [W / (42.4 x 235)]1000
Mass of LiCi = 17.9352 gm
A geneticist is mapping the chromosomes of the newly captured gremlin. Stripe is heterozygous for three linked genes with alleles Ee, Hh, and Bb, that determine if gremlins are evil (E), have hair (H), and biting teeth (B). In order to determine if the three genes are linked, a standard testcross was done, and the 1000 offspring had the following genotypes: 48 ee Hh bb 36 ee hh Bb 400 ee Hh Bb 4 Ee Hh Bb 426 Ee hh bb 46 Ee hh Bb 38 Ee Hh bb 2 ee hh bb What is the recombination frequency between genes E and H
Recombination frequency determines if two genes are linked or if they segregate independently from each other. The recombination frequency between E and H is 8%
What are linked genes?Linked genes are those that are too close to each other in the same chromosome and they do not segregate independently.
These are the linked genes that do not exhibit an independent distribution, and they are inherited together more frequently.
What is recombination frequency?Recombination frequency can be defined as the crossing over events frequency occurring between two genes.
The value of recombination frequency indicates weather genes are linked or not. The maximum recombination frequency is always 50%.
When RF is 50% or more, we can assume that genes are not linked.
In the exposed example, we can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent.
Parental)
400 ee Hh Bb
426 Ee hh bb
Double recombinant)
4 Ee Hh Bb
2 ee hh bb
Simple recombinant)
48 ee Hh bb
36 ee hh Bb
46 Ee hh Bb
38 Ee Hh bb
We will call Region I to the area between genes E and H and Region II to the area between genes H and B.
------------E------------H--------------B-----------
/---RI------//------RII-----/
To calculate the recombination frequency between E and H we just need to consider region I. We will call P1 to the recombination frequency in this region.
P1 = (R + DR) / N
Where:
R is the number of simple recombinants in this region, DR is the number of double recombinants in each region, and N is the total number of individuals.Now, let us analyze the information.
We know that individuals carrying double recombinant gametes are
Ee Hh Bbee hh bbWe know that the recessive parent from the testcross could only provide the following gametes ⇒ ehb
This leads us to assume that the followings are the double recombinant gametes from the trihybrid parent,
EHBehbSince these gametes are the product of a double recombination event in region I and II, we can assume that the parental gametes of the trihybrid individual are
eHbEhBHence the simple recombinant gametes in region I are
EHbehBNow, according to this analysis, we know that
Simple recombinants in region I
36 ee hh Bb
38 Ee Hh bb
Double recombinants
4 Ee Hh Bb
2 ee hh bb
Total number of individuals
N = 1000
Recombination frequency in Region I
P1 = (R + DR) / N
P1 = (38 + 36 + 4 + 2) / 1000
P1 = 80 / 1000
P1 = 0.08 = 8%
The recombination frequency between genes E and H is P1 = 8%
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